RD Sharma Class 12 Ex 20.4 Solutions Chapter 20 Definite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter20
Exercise20.4
CategoryRD 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, sin(2π -x)=-sinx

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

if 

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
2I=2π

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

if 

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

if

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

So

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

if 

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

if 

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 = I=π/4

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
2I=2a
I=a

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
2I=2π/3
I=π/3

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}}

2

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:

if 

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
2I=a
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

II=\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

Letx = sin\theta ,dx=acos\theta d\theta

Now, x=0 ,\theta = 0 , thenx=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

thenx=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}
x=A(1+x^2)+(Bx+C)(1+x)

Equating coefficients, we get

A+B=0; A=-B
B+C=1;-2A=1
A+C=0;A=-C
\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,t=cos(x)dx; dt=-sinxdx

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
I=2π/3

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 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, letI=\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 ssinx= \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
2I=6
I=3

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
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 ,t=sin^2x

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

I=0

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
I=πlog2+4I1

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
I=π^2

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

Letf(\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
I=2[log(4+2+1)-log(1)]
I=2log(7)

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

=0

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

Solution:

I= \int\limits_{-1}^{1} | xcosπx |dx

Let,f(x)=| xcosπx |dx

f(-x)=|- xcos(-πx) |=|- xcos(πx) |=| xcosπx |=f(x)
\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

Puttan(\frac{x}2)=t thensec^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
2I=0
I=0

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. Iff(2a-x)=-f(x) , 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)
I=0

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,I=0

\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
\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)] , whereh = \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) = x− 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) = ex.

=> 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) = ex.

=> 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)

= ea (eb-a −1)

= eb − ea

Therefore, the value of\int_{a}^{b}e^xdx  as limit of sum is eb − ea.

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 + e2x.
=> 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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