RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	20
Exercise	20.4
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals

Evaluate of each of the following integrals (1-16):

Question 1. $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

Solution:

We know that $\int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx$

so,

$\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{sin(2π-x)}}{e^{sin(2π-x)}+e^{-sin(2π-x)}}dx$

we know,

$\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$

I = $\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$

then

I = $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

2I = $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx +\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx$

$2I=\int\limits_0^{2π}(\frac{e^{sinx}}{e^{sinx}+e^{-sinx}} +\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}})dx$

$2I=\int\limits_0^{2π}(\frac{e^{sinx}+e^{-sinx}}{e^{sinx}+e^{-sinx}} )dx$

$2I=\int\limits_0^{2π}dx$

l=π

Question 2. $\int\limits_0^{2π} log (secx+tanx)dx$

Solution:

We know that $\int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx$

So,

$\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (sec(2π-x)+tan(2π-x)dx$

$\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (secx-tanx)dx$

I = $\int\limits_0^{2π} log (secx-tanx)dx$

then

I = $\int\limits_0^{2π} log (secx+tanx)dx$

2I = $\int\limits_0^{2π} log (secx+tanx)dx+ \int\limits_0^{2π} log (secx-tanx)dx$

2I= $\int\limits_0^{2π} log (sec^2x-tan^2x)dx$

2I = $\int\limits_0^{2π} log (1)dx$

2I=0

I=0

Question 3. $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

So,

$\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{tan(π/2-x)}}{\sqrt{tan(π/2-x)}+\sqrt{cot(π/2-x)}}dx$

$\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$

I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

then

I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$

2I= $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$

2I = $\int\limits_{π/6}^{π/3}(\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}} +\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}})dx$

2I= $\int\limits_{π/6}^{π/3}dx$

2I=π/6

I=π/12

Question 4. $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

So,

$\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{sin(π/2-x)}}{\sqrt{sin(π/2-x)}+\sqrt{cos(π/2-x)}} dx$

$\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}} dx$

I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

then,

I = $\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

2I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

2I = $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

2I = $\int\limits_{π/6}^{π/3}dx$

2I=π/6

I=π/12

Question 5. $\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

so,

$\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx =\int\limits_{-π/4}^{π/4} \frac{tan^2{-x}}{1+e^{-x}}dx$

$\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$

$I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

then,

$I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{e^xtan^2x}{1+e^x}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x+e^xtan^2x}{1+e^x}dx$

$2I=\int\limits_{-π/4}^{π/4} \frac{(e^x+1)tan^2x}{1+e^x}dx$

$2I=\int\limits_{-π/4}^{π/4} tan^2xdx$

$I=\frac{\int\limits_{-π/4}^{π/4} tan^2x}{2}dx$

we know if

f(x) is even $\int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx$

f(x) is odd $\int\limits_{-a}^af(x)dx=0$

Here, f(x) = tan²x which is even

hence,

$I=\int\limits_{0}^{π/4}tan^2xdx$

$I=\int\limits_{0}^{π/4}(sec^2x-1)dx$

$I=\left.({tanx-x})\right|_0^{π/4}$

I =

Question 6. $\int\limits_{-a}^a \frac{1}{1+a^x}dx,a>0″ height=”60″ width=”190″><span class=$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

So,

$\int\limits_{-a}^a \frac{1}{1+a^x}dx = \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$

if, $I = \int\limits_{-a}^a \frac{1}{1+a^x}dx$ then $I=\int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$

So,

$2I=\int\limits_{-a}^a \frac{1}{1+a^x}dx+ \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx$

$2I=\int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{1}{1+a^{-x}}dx$

$2I= \int\limits_{-a}^a \frac{1+a^x}{1+a^x}dx$

$2I= \int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{a^x}{1+a^x}dx$

$2I= \int\limits_{-a}^a dx$

Question 7. $\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

Hence,

$\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx =\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$

if, $I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

then $I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$

so,

$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx+\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx$

$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{1}{1+e^{-tanx}} dx$

$2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{e^{tanx}}{1+e^{tanx}} dx$

$2I=\int\limits_{-π/3}^{π/3} dx$

Question 8. $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

Solution:

We know $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

hence, $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx =\int\limits_{-π/2}^{π/2} \frac{cos^2(-x)}{1+e^{-x}}dx$

$\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx = \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$

if $I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

Then, $I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$

So, $2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx +\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx$

$2I= \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}+ \frac{e^xcos^2x}{1+e^x}dx$

$2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x+e^xcos^2x}{1+e^x}dx$

$2I=\int\limits_{-π/2}^{π/2} \frac{(1+e^x)cos^2x}{1+e^x}dx$

$2I=\int\limits_{-π/2}^{π/2} cos^2x dx$

$2I=\int\limits_{-π/2}^{π/2} \frac{1+cos2x}{2}dx$

$I= \left.\frac{1}{4}(x+\frac{sin2x}2)\right|_{-π/2}^{π/2}$

$I=\frac{1}{4}[\frac{π}2-(-\frac{π}2)]$

$I=\frac{π}4$

Question 9. $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

Solution:

$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\frac{1}{cos^2x} dx$

$\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\int\limits_{-π/4}^{π/4}sec^2x dx$

if f(x) is even

$\int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx$

if f(x) is odd

$\int\limits_{-a}^af(x)dx=0$

here, $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}dx$ is odd and $\int\limits_{-π/4}^{π/4}sec^2x dx$

is even

Hence,

$0+2\int\limits_0^{π/4}sec^2x dx$

$\left.2tanx\right|_0^{\frac{π}{4}}$

Question 10. $\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx,n\in N,n\geq 2$

Solution:

$I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

then,

$I=\int\limits_{a}^{b} \frac{{(a+b-x)}^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

$2I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx +\int\limits_{a}^{b} \frac{(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

$2I=\int\limits_{a}^{b} \frac{x^{1/n}+(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx$

$2I=\int\limits_{a}^{b} dx$

$I=\frac{b-a}{2}$

Question 11. $\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

Solution:

let, $I=\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

$I=\int\limits_0^{π/2} (log\frac{cos^2x}{sinx2x})dx$

$I=\int\limits_0^{π/2} (log\frac{cos^2x}{2sinxcosx})dx$

$I=\int\limits_0^{π/2} (log\frac{cosx}{2sinx})dx$

$I=\int\limits_0^{π/2} logcosxdx-\int\limits_0^{π/2} logsinxdx-\int\limits_0^{π/2}log2dx$

we know that, $\int\limits_0^{π/2}logcosxdx=\int\limits_0^{π/2}logsinxdx$

hence,

$I=-\int\limits_0^{π/2}log2dx =\frac{-π}{2}log2$

Question 12. $\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

Solution:

Let, $I=\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

we know that , $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$

so, $I=\int\limits_0^a \frac{\sqrt {a-x}}{\sqrt{a-x}+\sqrt {x}} dx$

then,

$2I= \int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx + \int\limits_0^a \frac{\sqrt{ a-x}}{\sqrt {a-x}+\sqrt x} dx$

$2I= \int\limits_0^a \frac{\sqrt x+\sqrt{a-x}}{\sqrt x+\sqrt {a-x}} dx$

$2I= \int\limits_0^a dx$

$I=\frac{a}{2}$

Question 13. $\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

Solution:

We know that, $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$

So,

$I=\int\limits_0^5 \frac {\sqrt[4]{(5-x)+4} }{\sqrt[4]{(5-x)+4} +\sqrt[4]{9-(5-x)} }dx$

$I=\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx$

then,

$2I=\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx +\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx$

$2I= \int\limits_0^5 \frac{\sqrt[4]{x+4}\sqrt[4]{9-x} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

$2I= \int\limits_0^5 dx$

I $I=\frac{5}{2}$

Question 14. $\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx$

Solution:

We know that, $\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx$

so,

$I=\int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx$

then,

$2I=\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx+ \int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx$

$2I=\int\limits_0^7 dx$

$I=\frac{7}{2}$

Question 15. $\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

Solution:

We know that, $\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

Let, $I=\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

hence,

$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cos(\frac{π}{2}- x)}}{\sqrt{cos(\frac{π}{2}-x)} +\sqrt {sin(\frac{π}{2}-x)}}dx$

$I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

$2I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx + \int\limits_{π/6}^{π/3} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

$2I= \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}+\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx$

$2I= \int\limits_{π/6}^{π/3} dx$

$I=\frac{π}{12}$

Question 16. If f(a+b-x)=f(x), then prove that $\int\limits_a^b xf(x)dx=\frac{a+b}{2} \int\limits_a^bf(x)dx$

Solution:

$I=\int\limits_a^bf(x)dx$

$I=\int\limits_a^b(a+b-x)f(a+b-x)dx$

$I=\int\limits_a^b(a+b-x)f(x)dx........... [\because f(a+b-x)=f(x)]$

$I=\int\limits_a^b(a+b)f(x)dx-\int\limits_a^bf(x)dx$

$I=(a+b)\int\limits_a^bf(x)dx-I$

$2I=(a+b)\int\limits_a^bf(x)dx$

$I=\frac{(a+b)}{2}\int\limits_a^bf(x)dx$

$\therefore \int\limits_a^bxf(x)dx=\frac{(a+b)}{2}\int\limits_a^bf(x)dx$

Evaluate of each of the following integrals (1-46):

Question 1. $\int\limits_0^{\frac{π}{2}} \frac{1}{1+tanx}dx$

Solution:

We have,

$\int\limits_0^{\frac{π}{2}} \frac{1}{1+tanx}dx = \int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx$

$\int\limits_0^{\frac{π}{2}} \frac{1}{1+tanx}dx =\int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx$

Let

$I=\int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx$ —— 1

So, $I= \int\limits_0^{\frac{π}{2}} \frac{cos(\frac{π}{2}-x)} {cos(\frac{π}{2}-x)+sin(\frac{π}{2}-x)}dx$ , $[\because \int\limits_0^af(x)= \int\limits_0^af(a-x)dx]$

$\int\limits_0^{π/2}\frac{sinx}{sinx+cosx}dx$ ———— 2

Hence, by adding 1 and 2 ..

$2I=\int\limits_0^{π/2} \frac{cosx}{cosx+sinx}dx+ \int\limits_0^{π/2} \frac{sinx}{cosx+sinx}dx$

$2I=\int\limits_0^{π/2} \frac{cosx+sinx}{cosx+sinx}dx$

$2I=\int\limits_0^{π/2}dx$

$2I=\frac{π}{2}$

$I=\frac{π}{4}$

Question 2. $\int\limits_0^{\frac{π}{2}} \frac{1}{1+cotx}dx$ ,

Solution:

We have,

$\int\limits_0^{\frac{π}{2}} \frac{1}{1+cotx}dx =\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx$

Let, I= $\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx$ —– 1

So, $I= \int\limits_0^{\frac{π}{2}} \frac{sin(\frac{π}{2}-x)}{sin(\frac{π}{2}-x)+ cos(\frac{π}{2}-x)}dx$ —— 2

Adding 1 & 2 ——-

$2I=\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx +\int\limits_0^{\frac{π}{2}} \frac{cosx}{sinx+cosx}dx$

$2I=\int\limits_0^{\frac{π}{2}} \frac{sinx+cosx}{sinx+cosx}dx$

$2I=\int\limits_0^{\frac{π}{2}} dx$

$2I=\frac{π}{2}$

$I=\frac{π}{4}$

Question 3. $\int\limits_0^{\frac{π}{2}} \frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$ ,

Solution:

We have , $\int\limits_0^{\frac{π}{2}} \frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx = \int\limits_0^{\frac{π}{2}} \frac{\sqrt{\frac{cosx}{sinx}}} {\sqrt{\frac{cosx}{sinx}}+\sqrt{\frac{sinx}{cosx}}}dx = \frac{cosx}{cosx+sinx}$

$\therefore \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx = \int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx$

Let, $\int\limits_0^{\frac{π}{2}} \frac{cosx}{cosx+sinx}dx$

So,

$I=\int\limits_0^{\frac{π}{2}} \frac{cos(\frac{π}{2}-x)}{sin(\frac{π}{2}-x)+cos(\frac{π}{2}-x)}dx$

$I=\int\limits_0^{\frac{π}{2}} \frac{sinx}{sinx+cosx}dx$ ————— 2

Adding 1 and 2 ——–

$2I=\int\limits_0^{\frac{π}{2}} \frac{sinx+cosx}{sinx+cosx}dx$

$2I=\int\limits_0^{\frac{π}{2}} dx$

$I=\frac{π}{4}$

Question 4. $\int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx$

Solution:

Let, $I= \int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx$ —— 1

$I=\int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}(\frac{π}{2}-x)} {sin^{3/2}(\frac{π}{2}-x)+cos^{3/2}(\frac{π}{2}-x)}dx$

$I=\int\limits_0^{\frac{π}{2}} \frac{cos^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx$ ———- 2

Adding 1 & 2 —–

$2I=\int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}x+cos^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx$

$2I=\int\limits_0^{\frac{π}{2}} dx$

$I=\frac{π}{4}$

Question 5. $\int\limits_0^{\frac{π}{2}} \frac{sin^nx}{sin^nx+cos^nx}dx$

Solution:

Let, $\int\limits_0^{\frac{π}{2}} \frac{sin^nx}{sin^nx+cos^nx}dx$ ———– (1)

So,

$I=\int\limits_0^{\frac{π}{2}} \frac{sin^n(\frac{π}{2}-x)}{sin^n(\frac{π}{2}-x)+cos^n(\frac{π}{2}-x)}dx$

$I= \int\limits_0^{\frac{π}{2}} \frac{cos^nx}{sin^nx+cos^nx}dx$

$I=\frac{π}{4}$

Question 6. $\int\limits_0^{\frac{π}{2}} \frac{1}{1+\sqrt{tanx}}dx$

Solution:

$\int\limits_0^{\frac{π}{2}} \frac{1}{1+\sqrt{tanx}}dx = \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx$

$I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx$ ———- 1

$\int\limits_0^{\frac{π}{2}} \frac{\sqrt{cos(\frac{π}{2}-x)}} {\sqrt{cos(\frac{π}{2}-x)}+\sqrt{sin(\frac{π}{2}-x)}}dx$

$I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx$ —– 2

Adding 1 & 2 ——-

$2I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx + \int\limits_0^{\frac{π}{2}} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx$

$2I= \int\limits_0^{\frac{π}{2}} \frac{\sqrt{cosx}+\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx$

$2I= \int\limits_0^{\frac{π}{2}} dx$

$2I=\frac{π}{2}$

$I=\frac{π}{4}$

Question 7. $\int\limits_0^{a} \frac{1}{x+\sqrt{a^2-x^2}}dx$

Solution:

Let, $I=\int\limits_0^{a} \frac{1}{x+\sqrt{a^2-x^2}}dx$

Let $x = sin\theta$ , $dx=acos\theta d\theta$

Now, x=0 , $\theta = 0$ , then $x=a , \theta = π/2$

$\int\limits_0^{π/2} \frac{acos\theta }{asin\theta+acos\theta}d\theta$

$I=\int\limits_0^{π/2} \frac{cos\theta }{sin\theta+cos\theta}d\theta$ ——————– 1

So,

$I=\int\limits_0^{π/2} \frac{cos(\frac{π}{2}-\theta )} {sin(\frac{π}{2}-\theta )+cos(\frac{π}{2}-\theta )}d\theta$

$I=\int\limits_0^{π/2} \frac{sin\theta }{sin\theta+cos\theta}d\theta$ ————- 2

Adding (1) and (2) —————-

$2I=\int\limits_0^{π/2} \frac{cos\theta +sin\theta}{sin\theta+cos\theta}d\theta$

$2I=\int\limits_0^{π/2} d\theta$

$I=\frac{π}{4}$

Question 8. $\int\limits_0^{\infty} \frac{logx}{1+x^2}dx$

Solution:

Let, $x= tan \theta, dx=sec^2\theta d\theta$

then $x=0, \theta=0 ; x=\infty , \theta=\frac{π}{2}$

$\therefore I=\int\limits_0^{\infty} \frac{logx}{1+x^2}dx$

$I=\int\limits_0^{\frac{π}{2}} \frac{logtan\theta sec^2\theta d\theta}{1+tan^2\theta}dx$

$I=\int\limits_0^{\frac{π}{2}} {logtan\theta d\theta}$ ———— (1)

$I=\int\limits_0^{\frac{π}{2}} {logtan(\frac{π}{2}-\theta) d\theta}$

$I=\int\limits_0^{\frac{π}{2}} {logcot\theta d\theta}$ ——————- (2)

Adding (1) & (2) ————

$2I=\int\limits_0^{\frac{π}{2}} {(logtan\theta +logcot\theta )d\theta}$

$2I=\int\limits_0^{\frac{π}{2}} log1x dx=\int\limits_0^{\frac{π}{2}} 0x dx=0$

Question 9. $\int\limits_0^{1} \frac{log(1+x)}{1+x^2}dx$

Solution:

Let, $x= tan \theta, dx=sec^2\theta d\theta$

$x=0, \theta=0 ; x=1 , \theta=\frac{π}{4}$

$\therefore \int\limits_0^{1} \frac{log(1+x)}{1+x^2}dx$

$I=\int\limits_0^{\frac{π}{4}} log(1+tan\theta) d \theta$

$I=\int\limits_0^{\frac{π}{4}} log(1+tan(\frac{π}{4}-\theta)) d \theta$

$I=\int\limits_0^{\frac{π}{4}} log(1+\frac{1-tan\theta}{1+tan\theta}) d \theta$

$I=\int\limits_0^{\frac{π}{4}} log(\frac{2}{1+tan\theta}) d \theta$

$I=\int\limits_0^{\frac{π}{4}} \{ log2-log(1+tan\theta)\} d \theta$

$I=\frac{π}{8}log2$

Question 10. $\int\limits_0^{\infty} \frac{x}{(1+x)(1+x^2)}dx$

Solution:

$I=\int\limits_0^{\infty} \frac{x}{(1+x)(1+x^2)}dx$

Let,

$\frac{x}{(1+x)(1+x^2)} = \frac{A}{1+x}+\frac{Bx+C}{1+x}$

Equating coefficients, we get

$\therefore A=-\frac{1}{2},B=\frac{1}{2},C=\frac{1}{2}$

So,

$I=\int\limits_0^{\infty} \frac{-\frac{1}{2}}{(1+x)}+\frac{1}{2}\frac{1+x}{(1+x^2)}dx$

$I=\int\limits_0^{\infty} -\frac{1}{2(1+x)}dx +\frac{1}{2} \int\limits_0^{\infty} \frac{x}{(1+x^2)}dx +\frac{1}{2} \int\limits_0^{\infty} \frac{1}{(1+x^2)}dx$

$=0+0+\frac{π}{4}+0-0-0$

$I=\frac{π}{4}$

Question 11. $\int\limits_0^{π} \frac{xtanx}{secx cosecx}dx$

Solution:

$I=\int\limits_0^{π} \frac{xtanx}{secx cosecx}dx$

$I=\int\limits_0^{π} \frac{x\frac{sinx}{cosx}}{\frac{1}{cosx} \frac{1}{sinx}}dx$

$I=\int\limits_0^{π} xsin^2xdx$ ————- (1)

$I=\int\limits_0^{π} (π-x)sin^2(π-x)dx$

$I=\int\limits_0^{π} (π-x)sin^2xdx$ —————— (2)

Adding (1) & (2) —————-

$2I=\int\limits_0^{π} (π-x)sin^2xdx =π\int\limits_0^{π} \frac{1-cos2x}{2}dx =\frac{π}{2}[π-0-0+0]=\frac{π^2}{2}$

$\therefore \int\limits_0^{π} \frac{xtanx}{secx cosecx}dx= \frac{π^2}{4}$

Question 12. $\int\limits_0^{π}{xsinxcos^4x}dx$

Solution:

Let , $I=\int\limits_0^{π}{xsinxcos^4x}dx$ ————– 1

So,

$I=\int\limits_0^{π}{(π-x)sin(π-x)cos^4(π-x)}dx$

$I=\int\limits_0^{π}{(π-x) sinx.cos^4x}dx -1$

$2I=π\int\limits_0^{π}{sin(x)cos^4(x)}dx$

Let,

As, x=0, t=1 ; x=π , t=-1

Hence,

$2I=π\int\limits_{-1}^{+1}t^4dt = π[\frac{1}{5}+\frac{1}{5}]$

$I=\frac{π}{5}$

Question 13. $\int\limits_0^{π}{xsin^3xdx}$

Solution:

Let, $I=\int\limits_0^{π} {(π-x)sin^3(π-x)dx}$

$=\int\limits_0^{π}πsin^3dx -\int\limits_0^{π}xsin^3dx$

$\therefore I=\int\limits_0^ππsin^3dx - I$

$2I=π\int\limits_0^π\frac{3sinx-sin3x}{4}dx$

$2I=\frac{π}{4}\int\limits_0^π(3sinx-sin3x)dx$

Question 14. $\int\limits_0^{π} {xlogsinx}dx$

Solution:

we have, $\int\limits_0^{π} {xlogsinx}dx$

= $\int\limits_0^{π} {(π-x)logsin(π-x)}dx$

$I=π\int\limits_0^π logsin(x)dx-\int\limits_0^π xlogsinxdx$

$2I=π\int\limits_0^π log sinx dx$

Since, f(x) = f(-x) , f(x) is an even function.

$\therefore 2I=2π\int\limits_0^\frac{π}{2} log sinx dx$

$I=π\int\limits_0^\frac{π}{2} log sinx dx$ ————– 1

$I=π\int\limits_0^\frac{π}{2} log sin(\frac{π}{2}-x) dx =π\int\limits_0^\frac{π}{2} log cosx dx$

$I=π\int\limits_0^\frac{π}{2} log cosx dx$ —————- 2

Adding 1 and 2 —————-

$2I=π\int\limits_0^\frac{π}{2} log sinx dx +π\int\limits_0^\frac{π}{2} log cosxdx$

$2I=π\int\limits_0^\frac{π}{2} log (sinx +cosx) dx = π\int\limits_0^\frac{π}{2} log sinx cosx dx$

$2I=π\int\limits_0^\frac{π}{2} log \frac{2sinxcosx}{2} dx =π\int\limits_0^\frac{π}{2} log \frac{sin2x}{2} dx =π\int\limits_0^\frac{π}{2} log sin2x dx +π\int\limits_0^\frac{π}{2} log 2dx$

Now, let $I=\int\limits_0^\frac{π}{2} log sin2x dx$

Putting 2x=t, we get

$2I=I-π\frac{π}{2}log2$

$I=- \frac{π^2}{2}log2$

Question 15. $\int\limits_0^{π} \frac{xsinx}{1+sinx}dx$

Solution:

Let, $I=\int\limits_0^{π} \frac{xsinx}{1+sinx}dx$

$=\int\limits_0^{π} \frac{(π-x)sin(π-x)}{1+sin(π-x)}dx$

$I=\int\limits_0^π \frac{πsinx}{1+sinx}dx -\int\limits_0^π \frac{xsinx}{1+sinx}dx$

$I=\int\limits_0^π \frac{πsinx}{1+sinx}dx -I$

$2I=\int\limits_0^π \frac{πsinx}{1+sinx}dx$

$2I=π\int\limits_0^π \frac{sinx}{1+sinx}\frac{(1-sinx)}{(1-sinx)}dx$

$2I=π\int\limits_0^π \frac{sinx-sin^2x}{1+sin^2x}dx$

$2I=π\int\limits_0^π \frac{sinx-sin^2x}{cos^2x}dx$

$2I=π\int\limits_0^π {tanxsecx}{-tan^2x}dx$

$2I=π\int\limits_0^π [{tanxsecx}{-(sec^2x-1)}]dx$

$2I=π\int\limits_0^π [{tanxsecx}{-sec^2x+1)}]dx$

$I=\frac{π}{2}(π-2)$

Question 16. $\int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx, 0<\alpha <π$

Solution:

We have , I=\int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx ———- 1

$I=\int\limits_0^{π} \frac{(π-x)}{1+cos\alpha sin(π-x)}dx$ [Tex]=\int\limits_0^{π} \frac{(π-x)}{1+cos\alpha sinx}dx [/Tex]——- 2

Adding 1 and 2 —-

$2I=\int\limits_0^{π} \frac{(π)}{1+cos\alpha sin(π-x)}dx$

substituting s $sinx= \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$

$2I = π\int\limits_0^π \frac{sec^2xdx}{1+tan^2\frac{x}{2} 2cos\alpha tan\frac{x}{2}}$

$2I = π\int\limits_0^π \frac{sec^2xdx} {1-cos^2\alpha+ (cos\alpha tan\frac{x}{2})^2}$

$tan\frac{x}{2} = t; \frac{1}{2} sec^2 \frac{x}{2}dx=dt$

when x=0 , t=0 ; x=π , $t=\alpha$

$2I = \int\limits_0^\alpha \frac{dt.dx} {(1+cos^2\alpha)+ (cos\alpha+t )^2}$

$I =\frac{\alphaπ}{sin\alpha}$

Question 17. $\int\limits_0^{π} xcos^2xdx$

Solution:

Let, $I=\int\limits_0^{π} xcos^2xdx$

$I=\int\limits_0^{π}(π-x)cos^2(π-x)dx$

$I=\int\limits_0^{π} cos^2(x)dx - \int\limits_0^{π}x cos^2(x)dx$

$2I=π \int\limits_0^{π} cos^2(x)dx$

$=π\int\limits_0^π(\frac{1+cos2x}{2})dx$

$=\frac{π}2\int\limits_0^π(1+cos2x)dx$

$I=\frac{π^2}{4}$

Question 18. $\int\limits_{π/6}^{π/3} \frac{1}{1+cot^{3/2}x}dx$

Solution:

$I=\int\limits_{π/6}^{π/3} \frac{1}{1+cot^{3/2}x}dx$

$I=\int\limits_{π/6}^{π/3} \frac{sin^{3/2}(\frac{π}{2}-x)} {sin^{3/2}(\frac{π}{2}-x)+cos^{3/2}(\frac{π}{2}-x)}dx$

$I=\int\limits_{π/6}^{π/3} \frac{cos^{3/2}x} {cos^{3/2}x+sin^{3/2}x}dx$

$2I=\int\limits_{π/6}^{π/3} \frac{sin^{3/2}x} {sin^{3/2}x+cos^{3/2}x}dx+\int\limits_{π/6}^{π/3} \frac{cos^{3/2}x} {cos^{3/2}x+sin^{3/2}x}dx$

$2I=\int\limits_{π/6}^{π/3} \frac{cos^{3/2}x+sin^{3/2}x} {cos^{3/2}x+sin^{3/2}x}dx$

$I=\frac{π}{12}$

Question 19. $\int\limits_{0}^{π/2} \frac{tan^7x}{tan^7x+cot^7x}dx$

Solution:

$I=\int\limits_{0}^{π/2} \frac{tan^7x}{tan^7x+cot^7x}dx$

$I=\int\limits_{0}^{π/2} \frac{tan^7(\frac{π}{2}-x)}{tan^7(\frac{π}{2}-x)+cot^7(\frac{π}{2}-x)}dx$

$I=\int\limits_{0}^{π/2} \frac{cot^7x}{tan^7x+cot^7x}dx$

$2I=\int\limits_{0}^{π/2} \frac{cot^7x}{tan^7x+cot^7x}dx+\int\limits_{0}^{π/2} \frac{tan^7x}{tan^7x+cot^7x}dx$

$2I=\int\limits_{0}^{π/2} dx$

$2I=\frac{π}{2}$

$I=\frac{π}{4}$

Question 20. $\int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x} +\sqrt{10-x}}dx$

Solution:

$I=\int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x} +\sqrt{10-x}}dx$

$I=\int\limits_{2}^{8} \frac{\sqrt{10-(8+2-x)}} {\sqrt{(8+2-x)} +\sqrt{10-(8+2-x)}}dx$

$I=\int\limits_{2}^{8} \frac{\sqrt{x}}{\sqrt{x} +\sqrt{10-x}}dx$

$2I=\int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x} +\sqrt{10-x}}dx+ \int\limits_{2}^{8} \frac{\sqrt{x}}{\sqrt{x} +\sqrt{10-x}}dx$

$2I=\int\limits_{2}^{8} dx$

Question 21. $\int\limits_{0}^{π} {xsinxcos^2x}dx$

Solution:

$\int\limits_{0}^{π} {xsinxcos^2x}dx =\int\limits_{0}^{π} {(π-x)sin(π-x)cos^2(π-x)}dx$

$=\int\limits_{0}^{π} {(π-x)sinxcos^2x}dx$

$2\int\limits_{0}^{π} {xsinxcos^2x}dx= \int\limits_{0}^{π} {πsinxcos^2x}dx$

$\int\limits_{0}^{π} {xsinxcos^2x}dx =\frac{π}{2} \int\limits_{0}^{π}sinxcos^2xdx$

Now,

$\int\limits_{0}^{π} sinxcos^2xdx$

Let cosx=t

sinx dx=-dt

$-\int\limits_{1}^{-1} t^2dt$

$\int\limits_{1}^{-1} t^2dt$

$\int\limits_{0}^{π} {xsinxcos^2x}dx =\frac{π}{3}$ [Tex]I= \frac{π }{8}[\fracπ 4+\fracπ 4] dt[/Tex]

Question 22. $\int\limits_{0}^{π/2} \frac{xsinxcosx}{sin^4x+cos^4x}dx$

Solution:

$I=\int\limits_{0}^{π/2} \frac{xsinxcosx}{sin^4x+cos^4x}dx$ —————— 1

$\int\limits_{0}^{π/2} \frac{(\frac{π}{2}-x)sinxcosx}{sin^4x+cos^4x}dx$ ———— 2

Adding 1 & 2 ————-

$2I= \frac{π }{2} \int\limits_0^{π/2} \frac{sinxcosx}{cos^4x+sin^4x}dx$

$2I= \frac{π }{4} \int\limits_0^{π/2} \frac{2sinxcosx}{cos^4x+sin^4x}dx$

Let ,

$2I= \frac{π }{4} \int\limits_0^1 \frac{1}{(1-t^2)+t^2}dt$

$2I= \frac{π }{8} \int\limits_0^1 \frac{1}{(t-\frac{1}{2})^2+\frac12^2}dt$

$I=\frac{π^2}{16}$

Question 23. $\int\limits_{-π/2}^{π/2} sin^3xdx$

Solution:

Let, $I=\int\limits_{-π/2}^{π/2} sin^3xdx$

$f(-x)=\int\limits_{-π/2}^{π/2} sin^3(-x)dx$

$=\int\limits_{-π/2}^{π/2} sin^3xdx$

Here, f(x)=-f(x)

Hence, f(x) is odd function

Question 24. $\int\limits_{-π/2}^{π/2} sin^4xdx$

Solution:

We have, $I=\int\limits_{-π/2}^{π/2} sin^4xdx = 2\int\limits_{-π/2}^{π/2} sin^4xdx \because sin^4x$ is an even function.

$=2 \int\limits_{0}^{π/2} (sin^2x)^2dx$

$=2\int\limits_{0}^{π/2}( \frac{1-cos2x}{2})^2 dx$

$=\frac12 \int\limits_{0}^{π/2} ({1-cos2x})^2 dx$

$=\frac12[ \int\limits_{0}^{π/2}( {1+cos^22x}-2cos2x) ]dx$

$=\frac12[ \int\limits_{0}^{π/2} ( {1-2cos2x}+\frac{1+cos4x}{2}) ]dx$

$=\frac14[ \int\limits_{0}^{π/2}( {3-4cos2x}+cos4x) ]dx$

$\int\limits_{-π/2}^{π/2} sin^4xdx=\frac{3π }{8}$

Question 25. $\int\limits_{-1}^{1} log(\frac{2-x}{2+x})dx$

Solution:

we have, $I=\int\limits_{-1}^{1} log(\frac{2-x}{2+x})dx$

Since, $\int\limits_{-1}^{1} log(\frac{2-(-x)}{2+(-x)})dx= \int\limits_{-1}^{1} log(\frac{2-x}{2+x})dx$

$\therefore$ this is an odd function

Question 26. $\int\limits_{-π/4}^{π/4} sin^2xdx$

Solution:

we have, $I=\int\limits_{-π/4}^{π/4} sin^2xdx$

sin²x is even function

Hence,

$I=2\int\limits_{0}^{π/4} sin^2xdx =2\int\limits_{0}^{π/4} \frac{1-cos2x}{2}dx$

$\therefore \int\limits_{-π/4}^{π/4} sin^2xdx =\fracπ 4-\frac12$

Question 27. $\int\limits_{0}^{π} log(1-cosx)dx$

Solution:

$I=\int\limits_{0}^{π} log(1-cosx)dx$

$=\int\limits_{0}^{π} log(2sin^2\frac x 2)dx$

$=\int\limits_{0}^{π} log(2)dx +\int\limits_{0}^{π} log(sin^2\frac x 2)dx$

$=\int\limits_{0}^{π} log(2)dx +2\int\limits_{0}^{π} log(sin\frac x 2)dx$

Question 28. $\int\limits_{-π/4}^{π/4} log(\frac {2-sinx}{2+sinx})dx$

Solution:

we have , $I=\int\limits_{-π/4}^{π/4} log(\frac {2-sinx}{2+sinx})dx$

Let, $f(x)=log(\frac{2-sinx}{2+sinx})dx$

Then,

$f(-x)=log (\frac{2-sin(-x)}{2+sin(-x)})= -log(\frac{2-sinx}{2+sinx})=-f(x)$

$\therefore \int\limits_{-π /4}^{π /4} log(\frac{2-sinx}{2+sinx})dx=0$

Question 29. $\int\limits_{-π}^{π} \frac{2x(1+sinx)}{1+cos^2x}dx$

Solution:

$I=\int\limits_{-π}^{π} \frac{2x(1+sinx)}{1+cos^2x}dx$

$I=\int\limits_{-π}^{π} \frac{2x}{1+cos^2x}dx +\int\limits_{-π}^{π} \frac{2xsinx}{1+cos^2x}dx$

$I=0+\int\limits_{-π}^{π} \frac{2xsinx}{1+cos^2x}dx$

$I=2\int\limits_0^π\frac{2xsinx}{1+cos^2x}dx$

$I=4\int\limits_0^π \frac{xsinx}{1+cos^2}dx$

$I=2π\int\limits_0^π \frac{sinx}{1+cos^2x}$

Put cosx = t then -sinx dx = dt

$I=-2π\int\limits_{-1}^{1} \frac{1}{1+t^2}dt$

Question 30. $\int\limits_{-a}^{a} log(\frac{a-sin\theta}{a+sin\theta})d\theta ,a>0″ height=”60″ width=”261″><span class=$

Solution:

$I=\int\limits_{-a}^{a} log(\frac{a-sin\theta}{a+sin\theta})d\theta$

Let $f(\theta)= log(\frac{a-sin\theta}{a+sin\theta})$

$f(-\theta)= log(\frac{a-sin(-\theta)}{a+sin(-\theta)}) =-log(\frac{a-sin\theta}{a+sin\theta})=-f(\theta)$

$\therefore f(\theta)= log(\frac{a-sin\theta}{a+sin\theta})$ is an odd function

$\therefore \int\limits_{-a}^a log (\frac{a-sin\theta}{a+sin\theta}) d\theta =0$

Question 31. $\int\limits_{-2}^{2} \frac{3x^3+2| x |+1}{x^2+| x |+1}dx$

Solution:

$I=\int\limits_{-2}^{2} \frac{3x^3+2| x |+1}{x^2+| x |+1}dx$

$I=\int\limits_{-2}^{2} \frac{3x^3}{x^2+| x |+1}dx +\int\limits_{-2}^{2} \frac{2| x |+1}{x^2+| x |+1}dx$

$I=0 +\int\limits_{-2}^{2} \frac{2| x |+1}{x^2+| x |+1}dx$

$I=2\int\limits_{-2}^{2} \frac{2| x |+1}{x^2+| x |+1}dx$

Question 32. $\int\limits_{-3π/2}^{-π/2} sin^2(3π+x)+(π+x)^3)dx$

Solution:

$I=\int\limits_{-3π/2}^{-π/2} sin^2(3π+x)+(π+x)^3)dx$

Substitute π+x=u then dx=du

$I=\int\limits_{-3π/2}^{-π/2} sin^2(2π+u)+(u)^3)du$

$I=\int\limits_{-3π/2}^{-π/2} sin^2(u)+(u)^3)dx$

$I=\frac{π}{2}$

Question 33. $\int\limits_{0}^{2} x\sqrt{2-x} dx$

Solution:

Let, $I=\int\limits_{0}^{2} x\sqrt{2-x} dx$

$I=\int\limits_{0}^{2} ({2-x})\sqrt x dx$

$=\frac43 (2)^{\frac32} - \frac25x^{\frac52}$

$= \frac{4*2\sqrt 2}{3} - \frac{2}{5}*4\sqrt 2$

$=\frac{8 \sqrt 2}{3} - \frac{8\sqrt 2}{5}$

$=\frac{40\sqrt 2 - 24\sqrt 2} {15}$

$=\frac{16\sqrt 2}{15}$

Question 34. $\int\limits_{0}^{1} log(\frac{1}{x}-1)dx$

Solution:

$I=\int\limits_{0}^{1} log(\frac{1}{x}-1)dx$

$=\int\limits_{0}^{1} log(\frac{1-x}{x})dx$

$=\int\limits_{0}^{1} log(1-x)dx - \int\limits_{0}^{1} log(x)dx$

Applying the property , $\int\limits_0^a f(x)dx=\int\limits_0^a f(a-x)dx$

Thus, $I=\int\limits_0^1 log(1-(1-x))dx-\int\limits_0^1 log(x)dx$

$=\int\limits_0^1 log(1-1+x)dx -\int\limits_0^1log(x)dx$ –

Question 35. $\int\limits_{-1}^{1} | xcosπx |dx$

Solution:

$I= \int\limits_{-1}^{1} | xcosπx |dx$

Let,

$\therefore \int\limits_{-1}^{1} | xcosπx |dx=2\int\limits_{-1}^{1}| xcosπx |dx$

$f(x) = | xcosπx |dx= \{xcosπx, if 0\leq x\leq \frac12 ;-xcosπx, if \frac12< x< 1$

$\therefore I=2\int\limits_0^1 | xcosπx |dx$

$I= \frac{2}{π}$

Question 36. $\int\limits_{0}^{π} (\frac{x}{1+sin^2x}+cos^7x) dx$

Solution:

$I=\int\limits_{0}^{π} (\frac{π-x}{1+sin^2(π-x)}+cos^7(π-x) dx$

$I=\int\limits_{0}^{π} (\frac{π-x}{1+sin^2x}-cos^7x) dx$

$2I=\int\limits_{0}^{π} (\frac{π}{1+sin^2x}) dx$

$2I=π\int\limits_{0}^{π} (\frac{1}{1+sin^2x}) dx$

$2I=π\int\limits_{0}^{π} (\frac{1}{1+tan^2x})sec^2x dx$

$I=π\int\limits_{0}^{π/2} (\frac{1}{1+2tan^2x})sec^2x dx$ [Tex][\because \int\limits_0^{2a} f(x)dx= 2\int\limits_0^{a} f(x)dx, f(2a-x)=f(x)][/Tex]

let tanx = v

dv = sec²xdx

$I=π\int\limits_0^\infty \frac{1}{1+2v^2} dv$

$I=\frac{π^2}{2\sqrt2}$

Question 37. $\int\limits_{0}^{π} (\frac{x}{1+cos\alpha sinx}) dx$

Solution:

$I=\int\limits_{0}^{π} (\frac{x}{1+cos\alpha sinx}) dx$

$I=\int\limits_{0}^{π} (\frac{(π-x)}{1+cos\alpha sin(π-x)}) dx$

$I=\int\limits_{0}^{π} (\frac{π-x}{1+cos\alpha sinx}) dx$

$2I=π\int\limits_{0}^{π} (\frac{1}{1+cos\alpha sinx}) dx$

$2I=π\int\limits_{0}^{π} (\frac{1+tan^2(\frac{x}2)} {( 1+tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}}) dx$

$2I=π\int\limits_{0}^{π} (\frac{sec^2(\frac{x}2)} {( tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}+1}) dx$

$I=\frac{π}2\int\limits_{0}^{π} (\frac{sec^2(\frac{x}2)} {( tan^2(\frac{x}2))+2cos\alpha tan\frac{x}{2}+1}) dx$

Put $tan(\frac{x}2)=t$ then $sec^2(\frac{x}2)dx=2dt$

x=0 ⇒ t=0 and x=π ⇒ $t=\infty$

$I=\frac{π}{2} \int\limits_0^\infty \frac{2}{t^2+2tcos\alpha +1} dt$

$I={π} \int\limits_0^\infty \frac{1}{(t+cos\alpha)^2+(1-cos^2\alpha)} dt$

$I={π} \int\limits_0^\infty \frac{1}{(t+cos\alpha)^2+(sin^2\alpha)} dt$

$I=\frac{π\alpha}{sin\alpha}$

Question 38. $\int\limits_{0}^{2π} (sin^{100}xcos^{101}x) dx$

Solution:

we know, $\int\limits_{0}^{2a} f(x)= \int\limits_{0}^{a}f(x)+\int\limits_{0}^{a}f(2a-x)dx$

Also here,

f(x) = f(2π -x)

So, $I=\int\limits_{0}^{2π} (sin^{100}xcos^{101}x) dx =2\int\limits_{0}^{2π} (sin^{100}xcos^{101}x) dx$

$I=2\int\limits_{0}^{2π} (sin^{100}(π -x)cos^{101}(π -x)) dx$

$I=-2\int\limits_0^π sin^{100}x cos^{101}xdx$

Question 39. $\int\limits_{0}^{π/2} \frac{(asinx+bcosx)}{sinx+cosx} dx$

Solution:

$I=\int\limits_{0}^{π/2} \frac{(asinx+bcosx)}{sinx+cosx} dx$

then,

$\int\limits_{0}^{π/2} \frac{(asin(\frac{π}2-x)+bcos(\frac{π}2-x))} {sin(\frac{π}2-x)+cos(\frac{π}2-x)} dx$

$I=\int\limits_{0}^{π/2} \frac{(acosx+bsinx)}{sinx+cosx} dx$

$2I=\int\limits_{0}^{π/2} \frac{(asinx+bcosx)}{sinx+cosx} dx+\int\limits_{0}^{π/2} \frac{(acosx+bsinx)}{sinx+cosx} dx$

$2I=(a+b)\int\limits_{0}^{π/2} \frac{(sinx+cosx)}{sinx+cosx} dx$

$I=\frac{(a+b)}{2}\int\limits_{0}^{π/2} 1 dx$

$I=\frac{(a+b)π}{4}$

Question 40. If f is an integrable function such that f(2a-x)=f(x), then prove that $\int\limits_{0}^{2a} f(x)dx= 2\int\limits_{0}^{a} f(x)dx$

Solution:

We have , $I=\int\limits_{0}^{2a} f(x)dx$

Then,

$I=\int\limits_{0}^{a} f(x)dx +I(1)$

Let , 2a-t =x then dx=-dt

if t=a ⇒x=a

if t=2a ⇒ x=0

$I(1)=\int\limits_{0}^{2a} f(x)dx= \int\limits_{a}^{0} f(2a-t)(-dt) =-\int\limits_{a}^{0} f(2a-t)dt$

$I(1)=\int\limits_{0}^{a}f(2a-t)dt= \int\limits_{0}^{a}f(2a-x)dx$

$\therefore I=\int\limits_{0}^{a}f(x)dx +\int\limits_{0}^{a}f(2a-x)dx$

$I=\int\limits_{0}^{a}f(x)dx +\int\limits_{0}^{a}f(x)dx$ [Tex]=2\int\limits_{0}^{a}f(x)dx[/Tex]

Hence Proved.

Question 41. If, prove that $\int\limits_{0}^{2a} f(x)dx =0$

Solution:

We have, $I=\int\limits_{0}^{2a} f(x)dx = \int\limits_{0}^{2a} f(x)dx+\int\limits_{a}^{2a} f(x)dx$

$I=\int\limits_{0}^{2a} f(x)dx+I(1)$

Let 2a-t=x then dx=-dt

t=a , x=a ; t=2a , x=0

$I(1) = \int\limits_0^{2a}f(x)dx=\int\limits_a^0f(2a-t)(-dt)$

$= \int\limits_a^0 f(2a-t)dt$

$I(1)=\int\limits_0^{2a}f(2a-t)dt=\int\limits_0^{2a}f(2a-x)dx$

$I=\int\limits_0^a f(x)dx + \int\limits_0^a f(2a-x)dx$

$I=\int\limits_0^a f(x)dx - \int\limits_0^a f(x)$

Question 42. If f is an integrable function, show that

(i) $\int\limits_{-a}^{a} f(x^2)dx= 2\int\limits_{0}^{a} f(x^2)dx$

Solution:

we have , $I=\int\limits_{-a}^{a} f(x^2)dx$

clearly f(x²) is an even function .

So, $\int\limits_{-a}^{a} f(t)= 2\int\limits_{0}^{a} f(t)dx$

$I=2\int\limits_{0}^{a} f(x^2)dx$

(ii) $\int\limits_{-a}^{a}x f(x^2)dx=0$

Solution:

$I=\int\limits_{-a}^{a}x f(x^2)dx$

clearly , xf(x²) is odd function .

So,

$\therefore \int\limits_{-a}^{a}x f(x^2)dx=0$

Question 43. If f(x) is a continuous function defined on [0,2a] . Then, prove that

$\int\limits_{0}^{2a} f(x)dx =\int\limits_{0}^{a} {f(x)+f(2a-x)}dx$

Solution:

We have from LHS,

$I=\int\limits_{0}^{2a} f(x)dx =\int\limits_{0}^{a} f(x)dx+ \int\limits_{a}^{2a}f(x)dx$

$\therefore \int\limits_0^{2a}f(x)dx= - \int\limits_a^{0}f(2a-t)dt$

$\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}f(2a-x)dx$

substituting $\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}f(2a-x)dx$

we get,

$\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}f(x)dx+ \int\limits_0^{a}f(2a-x)dx$

$\int\limits_0^{2a}f(x)dx= \int\limits_0^{a}\{f(x)+f(2a-x)\}dx$

Question 44. If f(a+b-x) = f(x), then prove that

$\int\limits_{a}^{b} f(x)dx= (\frac{a+b}{2})\int\limits_{a}^{b} f(x)dx$

Solution:

$I=\int\limits_{a}^b xf(x)dx$

$I=\int\limits_{a}^{b}(a+b-x) f(a+b-x)dx$

$I=\int\limits_a^bf(x)dx$ ——————[ Given that f(a+b-x) = f(x) ]

$I=\int\limits_a^b (a+b) f(x)dx-\int\limits_a^b xf(x)dx$

$I=\int\limits_a^b (a+b)f(x)dx - I$

$2I=\int\limits_a^b (a+b)f(x)dx$

$I=\frac{a+b}{2}\int\limits_a^bf(x)dx$

Question 45. If f(x) is a continuous function defined on [-a,a], then prove that

$\int\limits_{-a}^{a} f(x)dx = \int\limits_{0}^{a} f(x)+f(-x)dx$

Solution:

we have , $I=\int\limits_{-a}^{a} f(x)dx = \int\limits_{-a}^{0} f(x)+ \int\limits_{0}^{a}f(-x)dx$

Let, x=-t, then dx=-dt

x=-a ⇒ t=a

x=0 ⇒ t=0

$\therefore \int\limits_{-a}^a f(x)dx= \int\limits_{-a}^0f(-t)(-dt) =\int\limits_{-a}^0-f(-t)dt$

$\int\limits_{-a}^a f(x)dx=\int\limits_{0}^a f(-t)(dt)$

$\int\limits_{-a}^0 f(x)dx=\int\limits_{0}^a f(-x)dx$

$\therefore \int\limits_{-a}^a f(x)dx=\int\limits_{0}^a f(-x)dx +\int\limits_{0}^a f(x)dx$

$\int\limits_{-a}^a f(x)dx = \int\limits_{0}^a \{f(-x)+f(x)\}dx$

Hence, Proved.

Question 46. Prove that: $\int\limits_{0}^{π} xf(sinx)dx = \frac{π}{2} \int\limits_{0}^{π}f(sinx)dx$

Solution:

$I= \int\limits_{0}^{π} xf(sinx)dx$

$I=\int\limits_{0}^{π}(π- x)f(sin(π-x))dx$

$I=\int\limits_0^π (π -x) f(sinx)dx$

$2I=\int\limits_0^π π f(sinx)dx$

$I=\fracπ2 \int\limits_0^π f(sinx)dx$

Evaluate the following definite integrals as limits of sums:

Question 1. $\int_{0}^{3}(x+4)dx$

Solution:

We have,
I = $\int_{0}^{3}(x+4)dx$
We know, $\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 3 and f(x) = x + 4.
=> h = 3/n
=> nh = 3
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[4+(h+4)+(2h+4)+...+(n-1)h+4]$
= $\lim_{h\to0}h[4n+h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[4n+h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{3}{n}[4n+\frac{3}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}\left[12+\frac{9}{2}(1-\frac{1}{n})\right]$
= 12 + $\frac{9}{2}$
= $\frac{33}{2}$
Therefore, the value of $\int_{0}^{3}(x+4)dx$ as limit of sum is $\frac{33}{2}$ .

Question 2. $\int_{0}^{2}(x+3)dx$

Solution:

We have,
I = $\int_{0}^{2}(x+3)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 2 and f(x) = x + 3.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[3+(h+3)+(2h+3)+...+(n-1)h+3]$
= $\lim_{h\to0}h[3n+h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[3n+h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[3n+\frac{2}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[6+2(1-\frac{1}{n})]$
= 6 + 2
= 8
Therefore, the value of $\int_{0}^{2}(x+3)dx$ as limit of sum is 8.

Question 3. $\int_{1}^{3}(3x-2)dx$

Solution:

We have,
I = $\int_{1}^{3}(3x-2)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 1, b = 3 and f(x) = 3x − 2.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[1+[3(1+h)-2]+[3(1+2h)-2]+...+[3(1+(n-1)h)-2]]$
= $\lim_{h\to0}h[n+3h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[n+3h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[n+\frac{6}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[2+6(1-\frac{1}{n})]$
= 2 + 6
= 8
Therefore, the value of $\int_{1}^{3}(3x-2)dx$ as limit of sum is 8.

Question 4. $\int_{-1}^{1}(x+3)dx$

Solution:

We have,
I = $\int_{-1}^{1}(x+3)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = −1, b = 1 and f(x) = x + 3.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(-1)+f(-1+h)+f(-1+2h)+...+f(-1+(n-1)h)]$
= $\lim_{h\to0}h[2+(2+h)+(2+2h)+...+((n-1)h+2)]$
= $\lim_{h\to0}h[2n+h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[2n+h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[2n+\frac{2}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[4+2(1-\frac{1}{n})]$
= 4 + 2
= 6
Therefore, the value of $\int_{-1}^{1}(x+3)dx$ as limit of sum is 6.

Question 5. $\int_{0}^{5}(x+1)dx$

Solution:

We have,
I = $\int_{0}^{5}(x+1)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 5 and f(x) = x + 1.
=> h = 5/n
=> nh = 5
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[1+(h+1)+(2h+1)+...+((n-1)h+1)]$
= $\lim_{h\to0}h[n+h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[n+h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{5}{n}[n+\frac{5}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[5+\frac{25}{2}(1-\frac{1}{n})]$
= 5 + $\frac{25}{2}$
= $\frac{35}{2}$
Therefore, the value of $\int_{0}^{5}(x+1)dx$ as limit of sum is $\frac{35}{2}$ .

Question 6. $\int_{1}^{3}(2x+3)dx$

Solution:

We have,
I = $\int_{1}^{3}(2x+3)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where $h = \frac{b-a}{n}$
Here a = 1, b = 3 and f(x) = 2x + 3.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[2+3+[2(1+h)+3]+[2(1+2h)+3]+...+[2(1+(n-1)h+3)]]$
= $\lim_{h\to0}h[5+(5+2h)+(5+4h)+...+(5+2(n-1)h)]$
= $\lim_{h\to0}h[5n+2h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[5n+2h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[5n+\frac{4}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[10+4(1-\frac{1}{n})]$
= 10 + 4
= 14
Therefore, the value of $\int_{1}^{3}(2x+3)dx$ as limit of sum is 14.

Question 7. $\int_{3}^{5}(2-x)dx$

Solution:

We have,
I = $\int_{3}^{5}(2-x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 3, b = 5 and f(x) = 2 − x.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(3)+f(3+h)+f(3+2h)+...+f(3+(n-1)h)]$
= $\lim_{h\to0}h[(2-3)+(2-(3+h))+f(2-(3+2h))+...+f(2-(3+(n-1)h))]$
= $\lim_{h\to0}h[-1+(-1-h)+(-1-2h)+...+(-1-(n-1)h)]$
= $\lim_{h\to0}h[-n-h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[-n-h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[-n-\frac{2}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[-2-2(1-\frac{1}{n})]$
= –2 – 2
= –4
Therefore, the value of $\int_{3}^{5}(2-x)dx$ as limit of sum is –4.

Question 8. $\int_{0}^{2}(x^2+1)dx$

Solution:

We have,
I = $\int_{0}^{2}(x^2+1)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 2 and f(x) = x² + 1.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[1+(h^2+1)+((2h)^2+1)+...+(((n-1)h)^2+1)]$
= $\lim_{h\to0}h[n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[n+h^2\frac{n(n-1)(2n-1)}{6}]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[n+\frac{4}{n^2}\frac{n(n-1)(2n-1)}{6}]$
= $\lim_{n\to\infty}[2+\frac{4n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= $2+\frac{4×2}{3}$
= $2+\frac{8}{3}$
= $\frac{14}{3}$
Therefore, the value of $\int_{0}^{2}(x^2+1)dx$ as limit of sum is $\frac{14}{3}$ .

Question 9. $\int_{1}^{2}x^2dx$

Solution:

We have,
I = $\int_{1}^{2}x^2dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 1, b = 2 and f(x) = x2.
=> h = 1/n
=> nh = 1
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[1+(1+h)^2+(1+2h)^2+...+(1+(n-1)h)^2]$
= $\lim_{h\to0}h[1+(1+2h+h^2)+(1+4h+4h^2)+...+(1+2(n-1)h+(n-1)^2h^2)]$
= $\lim_{h\to0}h[n+2h(1+2+3+...+(n-1))+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[n+2h(\frac{n(n-1)}{2})+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{1}{n}[n+\frac{2}{n}(\frac{n(n-1)}{2})+\frac{1}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[1+(1-\frac{1}{n})+\frac{n^3}{6n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= 1 + 1 + $\frac{2}{6}$
= 1 + 1 + $\frac{1}{3}$
= $\frac{7}{3}$
Therefore, the value of $\int_{1}^{2}x^2dx$ as limit of sum is $\frac{7}{3}$ .

Question 10. $\int_{2}^{3}(2x^2+1)dx$

Solution:

We have,
I = $\int_{2}^{3}(2x^2+1)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 2, b = 3 and f(x) = 2x² + 1.
=> h = 1/n
=> nh = 1
So, we get,
I = $\lim_{h\to0}h[f(2)+f(2+h)+f(2+2h)+...+f(2+(n-1)h)]$
= $\lim_{h\to0}h[9+[2(2+h)^2+1]+[2(2+2h)^2+1]+...+[2(2+(n-1)h)^2+1]]$
= $\lim_{h\to0}h[9n+8h(1+2+3+...+(n-1))+2h^2(1^2+2^2+3^2+...(n-1)^2)]$
= $\lim_{h\to0}h[9n+8h(\frac{n(n-1)}{2})+2h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{1}{n}[9n+\frac{8}{n}(\frac{n(n-1)}{2})+\frac{2}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[9+\frac{4n^2}{n^2}(1-\frac{1}{n})+\frac{n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= 9 + 4 + $\frac{2}{3}$
= $\frac{41}{3}$
Therefore, the value of $\int_{2}^{3}(2x^2+1)dx$ as limit of sum is $\frac{41}{3}$ .

Question 11. $\int_{1}^{2}(x^2-1)dx$

Solution:

We have,
I = $\int_{1}^{2}(x^2-1)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 1, b = 2 and f(x) = x² − 1.
=> h = 1/n
=> nh = 1
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[(1^2-1)+[(1+h)^2-1]+[(1+2h)^2-1]+...+[(1+(n-1)h)^2-1)]$
= $\lim_{h\to0}h[0+2h(1+2+3+...+(n-1))+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[2h(\frac{n(n-1)}{2})+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{1}{n}[\frac{2}{n}(\frac{n(n-1)}{2})+\frac{1}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[\frac{n^2}{n^2}(1-\frac{1}{n})+\frac{n^3}{6n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= 1 + $\frac{2}{6}$
= 1 + $\frac{1}{3}$
= $\frac{4}{3}$
Therefore, the value of $\int_{1}^{2}(x^2-1)dx$ as limit of sum is $\frac{4}{3}$ .

Evaluate the following definite integrals as limits of sums:

Question 12. $\int_{0}^{2}(x^2+4)dx$

Solution:

We have,
I = $\int_{0}^{2}(x^2+4)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 2 and f(x) = x² + 4.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[4+(h^2+4)+((2h)^2+4)+...+(((n-1)h)^2+4)]$
= $\lim_{h\to0}h[4n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[4n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[4n+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[4n+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[8+\frac{4n^3}{3n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= 8 + $\frac{4(2)}{3}$
= 8 + $\frac{8}{3}$
= $\frac{32}{3}$
Therefore, the value of $\int_{0}^{2}(x^2+4)dx$ as limit of sum is $\frac{32}{3}$ .

Question 13. $\int_{1}^{4}(x^2-x)dx$

Solution:

We have,
I = $\int_{1}^{4}(x^2-x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 1, b = 4 and f(x) = x²− x.
=> h = 3/n
=> nh = 3
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[(1^2-1)+[(1+h)^2-(1+h)]+[(1+2h)^2-(1+2h)]+...+[(1+(n-1)h)^2-(1+(n-1)h)]]$
= $\lim_{h\to0}h[0+[h+h^2]+[2h+(2h)^2]+...+[(n-1)h+((n-1)h)^2]]$
= $\lim_{h\to0}h[h(1+2+3+...+(n-1))+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[h(\frac{n(n-1)}{2})+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{3}{n}[\frac{3}{n}(\frac{n(n-1)}{2})+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[\frac{9n^2}{2n^2}(1-\frac{1}{n})+\frac{3n^3}{2n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= $\frac{9}{2}+3$
= $\frac{15}{2}$
Therefore, the value of $\int_{1}^{4}(x^2-x)dx$ as limit of sum is $\frac{15}{2}$ .

Question 14. $\int_{0}^{1}(3x^2+5x)dx$

Solution:

We have,
I = $\int_{0}^{1}(3x^2+5x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 1 and f(x) = 3x² + 5x.
=> h = 1/n
=> nh = 1
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[0+[3h^2+5h]+[3(2h)^2+5(2h)]+...+[3((n-1)h)^2+5((n-1)h)]$
= $\lim_{h\to0}h[3h^2(1^2+2^2+3^2+...+(n-1)^2)+5h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[3h^2(\frac{n(n-1)(2n-1)}{6})+5h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{1}{n}[\frac{3}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{5}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[\frac{n^3}{2n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{5n^2}{2n^2}(1-\frac{1}{n})]$
= 1 + $\frac{5}{2}$
= $\frac{7}{2}$
Therefore, the value of $\int_{0}^{1}(3x^2+5x)dx$ as limit of sum is $\frac{7}{2}$ .

Question 15. $\int_{0}^{2}e^xdx$

Solution:

We have,
I = $\int_{0}^{2}e^xdx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 2 and f(x) = e^x.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[1+e^h+e^{2h}+...+e^{(n-1)h}]$
= $\lim_{h\to0}h[\frac{(e^h)^n-1}{e^h-1}]$
= $\lim_{h\to0}h[\frac{e^{nh}-1}{e^h-1}]$
= $\lim_{h\to0}h[\frac{e^{2}-1}{e^h-1}]$
= $\lim_{h\to0}\left(\frac{e^{2}-1}{\frac{e^h-1}{h}}\right)$
= e² − 1
Therefore, the value of $\int_{0}^{2}e^xdx$ as limit of sum is e² − 1.

Question 16. $\int_{a}^{b}e^xdx$

Solution:

We have,
I = $\int_{a}^{b}e^xdx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = a, b = b and f(x) = e^x.
=> h = $\frac{b-a}{n}$
=> nh = b − a
So, we get,
I = $\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$
= $\lim_{h\to0}h[e^a+e^{a+h}+e^{a+2h}+...+e^{a+(n-1)h}]$
= $\lim_{h\to0}he^a[1+e^{h}+e^{2h}+...+e^{(n-1)h}]$
= $\lim_{h\to0}he^a[\frac{(e^h)^n-1}{e^h-1}]$
= $\lim_{h\to0}he^a[\frac{e^{nh}-1}{e^h-1}]$
= $\lim_{h\to0}he^a[\frac{e^{b-a}-1}{e^h-1}]$
= $\lim_{h\to0}e^a\left(\frac{e^{b-a}-1}{\frac{e^h-1}{h}}\right)$
= e^a (e^b-a−1)
= e^b − e^a
Therefore, the value of $\int_{a}^{b}e^xdx$ as limit of sum is e^b − e^a.

Question 17. $\int_{a}^{b}cosxdx$

Solution:

We have,
I = $\int_{a}^{b}cosxdx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = a, b = b and f(x) = cos x.
=> h = $\frac{b-a}{n}$
=> nh = b − a
So, we get,
I = $\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$
= $\lim_{h\to0}h[cosa+cos(a+h)+cos(a+2h)+...+cos(a+(n-1)h)]$
= $\lim_{h\to0}h\left[\frac{cos(a+(n-1)\frac{h}{2})sin\frac{nh}{2}}{sin\frac{h}{2}}\right]$
= $\lim_{h\to0}h\left[\frac{cos(a+\frac{nh}{2}-\frac{h}{2})sin\frac{nh}{2}}{sin\frac{h}{2}}\right]$
= $\lim_{h\to0}h\left[\frac{cos(a+\frac{b-a}{2}-\frac{h}{2})sin\frac{b-a}{2}}{sin\frac{h}{2}}\right]$
= $\lim_{h\to0}\left[\frac{\frac{h}{2}}{sin\frac{h}{2}}×2cos(a+\frac{b-a}{2}-\frac{h}{2})(sin\frac{b-a}{2})\right]$
= $\lim_{h\to0}\left[2cos(a+\frac{b-a}{2})sin(\frac{b-a}{2})\right]$
= $2cos(\frac{a+b}{2})sin(\frac{b-a}{2})$
= sin b − sin a
Therefore, the value of $\int_{a}^{b}cosxdx$ as limit of sum is sin b − sin a.

Question 18. $\int_{0}^{\frac{\pi}{2}}sinxdx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}sinxdx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = $\frac{\pi}{2}$ and f(x) = sin x.
=> h = $\frac{\pi}{2n}$
=> nh = $\frac{2}{\pi}$
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[sin0+sinh+sin2h+...+sin(n-1)h]$
= $\lim_{h\to0}h\frac{sin(\frac{(n-1)h}{2})sin(\frac{nh}{2})}{sin\frac{h}{2}}$
= $\lim_{h\to0}h\frac{sin(\frac{nh}{2}-\frac{h}{2})sin(\frac{nh}{2})}{sin\frac{h}{2}}$
= $\lim_{h\to0}h\frac{sin(\frac{\pi}{4}-\frac{h}{2})sin(\frac{\pi}{4})}{sin\frac{h}{2}}$
= $2(\frac{1}{2})$
= 1
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}sinxdx$ as limit of sum is 1.

Question 19. $\int_{0}^{\frac{\pi}{2}}cosxdx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}cosxdx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = $\frac{\pi}{2}$ and f(x) = cos x.
=> h = $\frac{\pi}{2n}$
=> nh = $\frac{2}{\pi}$
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[cos0+cosh+cos2h+...+cos(n-1)h]$
= $\lim_{h\to0}h[\frac{cos(\frac{nh}{2}-\frac{h}{2})cos\frac{nh}{2}}{cos\frac{h}{2}}]$
= $\lim_{h\to0}h[\frac{cos(\frac{\pi}{4}-\frac{h}{2})cos\frac{\pi}{4}}{cos\frac{h}{2}}]$
= $2(\frac{1}{2})$
= 1
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}cosxdx$ as limit of sum is 1.

Question 20. $\int_{1}^{4}(3x^2+2x)dx$

Solution:

We have,
I = $\int_{1}^{4}(3x^2+2x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a =1, b = 4 and f(x) = 3x² + 2x.
=> h = 3/n
=> nh = 3
So, we get,
I = $\lim_{h\to0}h[f(0)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[(3+2)+[3(1+h)^2+2(1+h)]+[3(1+2h)^2+2(1+2h)]+...+[3(1+(n-1)h)^2+2(1+(n-1)h)]]$
= $\lim_{h\to0}h[5n+8h(1+2+3...+(n-1))+3h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[5n+8h(\frac{n(n-1)}{2})+3h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{3}{n}[5n+\frac{24}{n}(\frac{n(n-1)}{2})+\frac{27}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[15+\frac{36n^2}{n^2}(1-\frac{1}{n})+\frac{27n^3}{2n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= 15 + 36 + 27
= 78
Therefore, the value of $\int_{1}^{4}(3x^2+2x)dx$ as limit of sum is 78.

Question 21. $\int_{0}^{2}(3x^2-2)dx$

Solution:

We have,
I = $\int_{0}^{2}(3x^2-2)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a =0, b = 2 and f(x) = 3x² − 2.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[-2+[3h^2-2]+[3(2h)^2-2]+...+[3((n-1)h)^2-2]]$
= $\lim_{h\to0}h[-2n+3h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[-2n+3h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[-2n+\frac{12}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[-4+\frac{4n^3}{n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= −4 + 8
= 4
Therefore, the value of $\int_{0}^{2}(3x^2-2)dx$ as limit of sum is 4.

Question 22. $\int_{0}^{2}(x^2+2)dx$

Solution:

We have,
I = $\int_{0}^{2}(x^2+2)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a =0, b = 2 and f(x) = x² + 2.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[2+(h^2+2)+[((2h)^2+2)]+...+[((n-1)h)^2+2]]$
= $\lim_{h\to0}h[2n+h^2(1^2+2^2+3^2+...+(n-1)^2)]$
= $\lim_{h\to0}h[2n+h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[2+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[4+\frac{4n^3}{n^3}(1-\frac{1}{n})(1-\frac{2}{n})]$
= 4 + $\frac{4(2)}{3}$
= 4 + $\frac{8}{3}$
= $\frac{20}{3}$
Therefore, the value of $\int_{0}^{2}(x^2+2)dx$ as limit of sum is $\frac{20}{3}$ .
Evaluate the following definite integrals as limits of sums:
Question 23. $\int_{0}^{4}(x+e^{2x})dx$
Solution:

We have,
I = $\int_{0}^{4}(x+e^{2x})dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 4 and f(x) = x + e^2x.
=> h = 4/n
=> nh = 4
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[1+[h+e^{2h}]+[2h+e^{4h}]+...+[(n-1)h+e^{2(n-1)h}]]$
= $\lim_{h\to0}h[h(1+2+3+...+(n-1))+(1+e^{2h}+e^{4h}+...e^{2(n-1)h})]$
= $\lim_{h\to0}h[h(\frac{n(n-1)}{2})+(\frac{(e^{2h})^2-1}{e^{2h}-1})]$
= $\lim_{h\to0}h[h(\frac{n(n-1)}{2})+(\frac{e^{2nh}-1}{e^{2h}-1})]$
= $\lim_{h\to0}h^2\left[(\frac{n(n-1)}{2})+\left(\frac{e^{8}-1}{2(\frac{e^{2h}-1}{h})}\right)\right]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}[\frac{16}{n^2}(\frac{n(n-1)}{2})+(\frac{e^{8}-1}{2})]$
= $\lim_{n\to\infty}[\frac{8n^2}{n^2}(1-\frac{1}{n})+(\frac{e^{8}-1}{2})]$
= $8+(\frac{e^{8}-1}{2})$
= $\frac{15+e^{8}}{2}$
Therefore, the value of $\int_{0}^{4}(x+e^{2x})dx$ as limit of sum is $\frac{15+e^{8}}{2}$ .
Question 24. $\int_{0}^{2}(x^2+x)dx$
Solution:
We have,
I = $\int_{0}^{2}(x^2+x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 2 and f(x) = x² + x.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[0+(h^2+h)+[(2h)^2+2h]+...+[(n-1)h)^2+(n-1)h]]$
= $\lim_{h\to0}h[h^2(1^2+2^2+3^2+...(n-1)^2)+h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[h^2(\frac{n(n-1)(2n-1)}{6})+h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{2}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[\frac{4n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{2n^2}{n^2}(1-\frac{1}{n})]$
= $\frac{8}{3}+2$
= $\frac{14}{3}$
Therefore, the value of $\int_{0}^{2}(x^2+x)dx$ as limit of sum is $\frac{14}{3}$ .
Question 25. $\int_{0}^{2}(x^2+2x+1)dx$
Solution:
We have,
I = $\int_{0}^{2}(x^2+2x+1)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 2 and f(x) = x² + 2x + 1.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[1+(h^2+2h+1)+[(2h)^2+2(2h)+1]+...[(n-1)^2+2(n-1)+1]$
= $\lim_{h\to0}h[n+h^2(1^2+2^2+3^2+...(n-1)^2)+2h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[n+h^2(\frac{n(n-1)(2n-1)}{6})+2h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[n+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{4}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[2+\frac{4n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{4n^2}{n^2}(1-\frac{1}{n})]$
= $2+\frac{8}{3}+4$
= $\frac{26}{3}$
Therefore, the value of $\int_{0}^{2}(x^2+2x+1)dx$ as limit of sum is $\frac{26}{3}$ .
Question 26. $\int_{0}^{3}(2x^2+3x+5)dx$
Solution:
We have,
I = $\int_{0}^{3}(2x^2+3x+5)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 3 and f(x) = 2x² + 3x + 5.
=> h = 3/n
=> nh = 3
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[5+(2h^2+3h+5)+[2(2h)^2+3(2h)+5]+...+[2(n-1)^2h^2+3((n-1)h)+5]]$
= $\lim_{h\to0}h[5n+2h^2(1^2+2^2+3^2+...+(n-1)^2)+3h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[5n+2h^2(\frac{n(n-1)(2n-1)}{6})+3h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{3}{n}[5n+\frac{18}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{9}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[15+\frac{9n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{27n^2}{2n^2}(1-\frac{1}{n})]$
= 15 + 18 + $\frac{27}{2}$
= $\frac{93}{2}$
Therefore, the value of $\int_{0}^{3}(2x^2+3x+5)dx$ as limit of sum is $\frac{93}{2}$ .
Question 27. $\int_{a}^{b}xdx$
Solution:
We have,
I = $\int_{a}^{b}xdx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = a, b = b and f(x) = x.
=> h = $\frac{b-a}{n}$
=> nh = b − a
So, we get,
I = $\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$
= $\lim_{h\to0}h[a+(a+h)+(a+2h)+...+(a+(n-1)h)]$
= $\lim_{h\to0}h[na+h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[na+h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{b-a}{n}[na+\frac{b-a}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}(b-a)[a+\frac{b-a}{n}(\frac{n-1}{2})]$
= $\lim_{n\to\infty}(b-a)[a+(b-a)(\frac{1-\frac{1}{n}}{2})]$
= $(b-a)[a+\frac{b-a}{2}]$
= $\frac{(b-a)(b+a)}{2}$
= $\frac{b^2-a^2}{2}$
Therefore, the value of $\int_{a}^{b}xdx$ as limit of sum is $\frac{b^2-a^2}{2}$ .
Question 28. $\int_{0}^{5}(x+1)dx$
Solution:
We have,
I = $\int_{0}^{5}(x+1)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 5 and f(x) = x + 1.
=> h =5/n
=> nh = 5
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[1+(h+1)+(2h+1)+...+((n-1)h+1)]$
= $\lim_{h\to0}h[n+h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[n+h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{5}{n}[n+\frac{5}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[5+\frac{25n^2}{2n^2}(1-\frac{1}{n})]$
= 5 + $\frac{25}{2}$
= $\frac{35}{2}$
Therefore, the value of $\int_{0}^{5}(x+1)dx$ as limit of sum is $\frac{35}{2}$ .
Question 29. $\int_{2}^{3}x^2dx$
Solution:
We have,
I = $\int_{2}^{3}x^2dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 2, b = 3 and f(x) = x².
=> h = 1/n
=> nh = 1
So, we get,
I = $\lim_{h\to0}h[f(2)+f(2+h)+f(2+2h)+...+f(2+(n-1)h)]$
= $\lim_{h\to0}h[4+(2+h)^2+(2+2h)^2+...+(2+(n-1)h)^2]$
= $\lim_{h\to0}h[4+[2^2+2.h+h^2]+[2^2+2.2h+(2h)^2]+...]$
= $\lim_{h\to0}h[4n+h^2(1^2+2^2+...+(n-1)^2)+4h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[4n+h^2(\frac{n(n-1)(2n-1)}{6})+4h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{1}{n}[4n+\frac{1}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{4}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[4+\frac{n^3}{6n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{2n^2}{n^2}(1-\frac{1}{n})]$
= $4+\frac{2}{6}+2$
= $4+\frac{1}{3}+2$
= $\frac{19}{3}$
Therefore, the value of $\int_{2}^{3}x^2dx$ as limit of sum is $\frac{19}{3}$ .
Question 30. $\int_{1}^{3}(x^2+x)dx$
Solution:
We have,
I = $\int_{1}^{3}(x^2+x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 1, b = 3 and f(x) = x² + x.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[2+[(1+h)^2+(1+h)]+[(1+2h)^2+(1+2h)]+...]$
= $\lim_{h\to0}h[2n+h^2(1^2+2^2+3^2+...+(n-1)^2)+3h(1+2+3+...+(n-1))]$
= $\lim_{h\to0}h[2n+h^2(\frac{n(n-1)(2n-1)}{6})+3h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[2n+\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[4+\frac{4n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})+\frac{6n^2}{n^2}(1-\frac{1}{n})]$
= $4+\frac{8}{3}+6$
= $\frac{38}{3}$
Therefore, the value of $\int_{1}^{3}(x^2+x)dx$ as limit of sum is $\frac{38}{3}$ .
Question 31. $\int_{0}^{2}(x^2-x)dx$
Solution:
We have,
I = $\int_{0}^{2}(x^2-x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 0, b = 2 and f(x) = x² − x.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(0)+f(h)+f(2h)+...+f((n-1)h)]$
= $\lim_{h\to0}h[0+[h^2-h]+[(2h)^2-2h]+...]$
= $\lim_{h\to0}h[h^2(1^2+2^2+...+(n-1)^2)-h(1+2+...+(n-1))]$
= $\lim_{h\to0}h[h^2(\frac{n(n-1)(2n-1)}{6})-h(\frac{n(n-1)}{2})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[\frac{4}{n^2}(\frac{n(n-1)(2n-1)}{6})-\frac{2}{n}(\frac{n(n-1)}{2})]$
= $\lim_{n\to\infty}[\frac{4n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})-\frac{2n^2}{n^2}(1-\frac{1}{n})]$
= $\frac{8}{3}-2$
= $\frac{2}{3}$
Therefore, the value of $\int_{0}^{2}(x^2-x)dx$ as limit of sum is $\frac{2}{3}$ .
Question 32. $\int_{1}^{3}(2x^2+5x)dx$
Solution:
We have,
I = $\int_{1}^{3}(2x^2+5x)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 1, b = 3 and f(x) = 2x² + 5x.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[7+[2(1+h)^2+5(1+h)]+[2(1+2h)^2+5(1+2h)]+...]$
= $\lim_{h\to0}h[7n+9h(1+2+3+...+(n-1))+2h^2(1^2+2^2+3^2+...(n-1)^2)]$
= $\lim_{h\to0}h[7n+9h(\frac{n(n+1)}{2})+2h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[7n+\frac{18}{n}(\frac{n(n-1)}{2})+\frac{8}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[14+\frac{18n^2}{n^2}(1-\frac{1}{n})+\frac{8n^3}{3n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= 14 + 18 + $\frac{16}{3}$
= $\frac{112}{3}$
Therefore, the value of $\int_{1}^{3}(2x^2+5x)dx$ as limit of sum is $\frac{112}{3}$ .
Question 33. $\int_{1}^{3}(3x^2+1)dx$
Solution:
We have,
I = $\int_{1}^{3}(3x^2+1)dx$
We know,
$\int_{a}^{b}f(x)dx=\lim_{h\to0}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$ , where h = $\frac{b-a}{n}$
Here a = 1, b = 3 and f(x) = 3x² + 1.
=> h = 2/n
=> nh = 2
So, we get,
I = $\lim_{h\to0}h[f(1)+f(1+h)+f(1+2h)+...+f(1+(n-1)h)]$
= $\lim_{h\to0}h[4+[3(1+h)^2+1]+[3(1+2h)^2+1]+...]$
= $\lim_{h\to0}h[4n+6h(1+2+3+...(n-1))+3h^2(1^2+2^2+3^2+...(n-1)^2)]$
= $\lim_{h\to0}h[4n+6h(\frac{n(n-1)}{2})+3h^2(\frac{n(n-1)(2n-1)}{6})]$
Now if h −> 0, then n −> ∞. So, we have,
= $\lim_{n\to\infty}\frac{2}{n}[4n+\frac{12}{n}(\frac{n(n-1)}{2})+\frac{12}{n^2}(\frac{n(n-1)(2n-1)}{6})]$
= $\lim_{n\to\infty}[8+\frac{12n^2}{n^2}(1-\frac{1}{n})+\frac{4n^3}{n^3}(1-\frac{1}{n})(2-\frac{1}{n})]$
= 8 + 12 + 8
= 28
Therefore, the value of $\int_{1}^{3}(3x^2+1)dx$ as limit of sum is 28.

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RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals

Evaluate of each of the following integrals (1-16):

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16. If f(a+b-x)=f(x), then prove that

Evaluate of each of the following integrals (1-46):

Question 1.

Question 2. ,

Question 3. ,

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Question 21.

Question 22.

Question 23.

Question 24.

Question 25.

Question 26.

Question 27.

Question 28.

Question 29.

Question 30.

Question 31.

Question 32.

Question 33.

Question 34.

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40. If f is an integrable function such that f(2a-x)=f(x), then prove that

Question 41. If, prove that

Question 42. If f is an integrable function, show that

(i)

(ii)

Question 43. If f(x) is a continuous function defined on [0,2a] . Then, prove that

Question 44. If f(a+b-x) = f(x), then prove that

Question 45. If f(x) is a continuous function defined on [-a,a], then prove that

Question 46. Prove that:

Evaluate the following definite integrals as limits of sums:

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Evaluate the following definite integrals as limits of sums:

Question 1. $\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx$

Question 2. $\int\limits_0^{2π} log (secx+tanx)dx$

Question 3. $\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx$

Question 4. $\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx$

Question 5. $\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx$

Question 6. $\int\limits_{-a}^a \frac{1}{1+a^x}dx,a>0″ height=”60″ width=”190″><span class=$

Question 7. $\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx$

Question 8. $\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx$

Question 9. $\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx$

Question 10. $\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx,n\in N,n\geq 2$

Question 11. $\int\limits_0^{π/2} (2logcosx-logsinx2x)dx$

Question 12. $\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx$

Question 13. $\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx$

Question 14. $\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx$

Question 15. $\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx$

Question 16. If f(a+b-x)=f(x), then prove that $\int\limits_a^b xf(x)dx=\frac{a+b}{2} \int\limits_a^bf(x)dx$

Question 1. $\int\limits_0^{\frac{π}{2}} \frac{1}{1+tanx}dx$

Question 2. $\int\limits_0^{\frac{π}{2}} \frac{1}{1+cotx}dx$ ,

Question 3. $\int\limits_0^{\frac{π}{2}} \frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx$ ,

Question 4. $\int\limits_0^{\frac{π}{2}} \frac{sin^{3/2}x}{sin^{3/2}x+cos^{3/2}x}dx$

Question 5. $\int\limits_0^{\frac{π}{2}} \frac{sin^nx}{sin^nx+cos^nx}dx$

Question 6. $\int\limits_0^{\frac{π}{2}} \frac{1}{1+\sqrt{tanx}}dx$

Question 7. $\int\limits_0^{a} \frac{1}{x+\sqrt{a^2-x^2}}dx$

Question 8. $\int\limits_0^{\infty} \frac{logx}{1+x^2}dx$

Question 9. $\int\limits_0^{1} \frac{log(1+x)}{1+x^2}dx$

Question 10. $\int\limits_0^{\infty} \frac{x}{(1+x)(1+x^2)}dx$

Question 11. $\int\limits_0^{π} \frac{xtanx}{secx cosecx}dx$

Question 12. $\int\limits_0^{π}{xsinxcos^4x}dx$

Question 13. $\int\limits_0^{π}{xsin^3xdx}$

Question 14. $\int\limits_0^{π} {xlogsinx}dx$

Question 15. $\int\limits_0^{π} \frac{xsinx}{1+sinx}dx$

Question 16. $\int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx, 0<\alpha <π$

Question 17. $\int\limits_0^{π} xcos^2xdx$

Question 18. $\int\limits_{π/6}^{π/3} \frac{1}{1+cot^{3/2}x}dx$

Question 19. $\int\limits_{0}^{π/2} \frac{tan^7x}{tan^7x+cot^7x}dx$

Question 20. $\int\limits_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x} +\sqrt{10-x}}dx$

Question 21. $\int\limits_{0}^{π} {xsinxcos^2x}dx$

Question 22. $\int\limits_{0}^{π/2} \frac{xsinxcosx}{sin^4x+cos^4x}dx$

Question 23. $\int\limits_{-π/2}^{π/2} sin^3xdx$

Question 24. $\int\limits_{-π/2}^{π/2} sin^4xdx$

Question 25. $\int\limits_{-1}^{1} log(\frac{2-x}{2+x})dx$

Question 26. $\int\limits_{-π/4}^{π/4} sin^2xdx$

Question 27. $\int\limits_{0}^{π} log(1-cosx)dx$

Question 28. $\int\limits_{-π/4}^{π/4} log(\frac {2-sinx}{2+sinx})dx$

Question 29. $\int\limits_{-π}^{π} \frac{2x(1+sinx)}{1+cos^2x}dx$

Question 30. $\int\limits_{-a}^{a} log(\frac{a-sin\theta}{a+sin\theta})d\theta ,a>0″ height=”60″ width=”261″><span class=$

Question 31. $\int\limits_{-2}^{2} \frac{3x^3+2| x |+1}{x^2+| x |+1}dx$

Question 32. $\int\limits_{-3π/2}^{-π/2} sin^2(3π+x)+(π+x)^3)dx$

Question 33. $\int\limits_{0}^{2} x\sqrt{2-x} dx$

Question 34. $\int\limits_{0}^{1} log(\frac{1}{x}-1)dx$

Question 35. $\int\limits_{-1}^{1} | xcosπx |dx$

Question 36. $\int\limits_{0}^{π} (\frac{x}{1+sin^2x}+cos^7x) dx$

Question 37. $\int\limits_{0}^{π} (\frac{x}{1+cos\alpha sinx}) dx$

Question 38. $\int\limits_{0}^{2π} (sin^{100}xcos^{101}x) dx$

Question 39. $\int\limits_{0}^{π/2} \frac{(asinx+bcosx)}{sinx+cosx} dx$

Question 40. If f is an integrable function such that f(2a-x)=f(x), then prove that $\int\limits_{0}^{2a} f(x)dx= 2\int\limits_{0}^{a} f(x)dx$

Question 41. If, prove that $\int\limits_{0}^{2a} f(x)dx =0$

(i) $\int\limits_{-a}^{a} f(x^2)dx= 2\int\limits_{0}^{a} f(x^2)dx$

(ii) $\int\limits_{-a}^{a}x f(x^2)dx=0$

Question 46. Prove that: $\int\limits_{0}^{π} xf(sinx)dx = \frac{π}{2} \int\limits_{0}^{π}f(sinx)dx$

Question 1. $\int_{0}^{3}(x+4)dx$

Question 2. $\int_{0}^{2}(x+3)dx$

Question 3. $\int_{1}^{3}(3x-2)dx$

Question 4. $\int_{-1}^{1}(x+3)dx$

Question 5. $\int_{0}^{5}(x+1)dx$

Question 6. $\int_{1}^{3}(2x+3)dx$

Question 7. $\int_{3}^{5}(2-x)dx$

Question 8. $\int_{0}^{2}(x^2+1)dx$

Question 9. $\int_{1}^{2}x^2dx$

Question 10. $\int_{2}^{3}(2x^2+1)dx$

Question 11. $\int_{1}^{2}(x^2-1)dx$

Question 12. $\int_{0}^{2}(x^2+4)dx$

Question 13. $\int_{1}^{4}(x^2-x)dx$

Question 14. $\int_{0}^{1}(3x^2+5x)dx$

Question 15. $\int_{0}^{2}e^xdx$

Question 16. $\int_{a}^{b}e^xdx$

Question 17. $\int_{a}^{b}cosxdx$

Question 18. $\int_{0}^{\frac{\pi}{2}}sinxdx$

Question 19. $\int_{0}^{\frac{\pi}{2}}cosxdx$

Question 20. $\int_{1}^{4}(3x^2+2x)dx$