RD Sharma Class 12 Ex 21.1 Solutions Chapter 21 Areas of Bounded Regions

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter21
Exercise21.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 21.1 Solutions Chapter 21 Areas of Bounded Regions

Question 1. Using integration, find the area of region bounded between the line x = 2 and parabola y2 = 8x

Solution:

Here,

Given equations are:

x = 2       ……..(1)

y2 = 8x     ……..(2)

Here,

Equation (1) represents a line parallel to y-axis and equation (2) represents a parabola with vertex at origin and x-axis, 

Here is the rough sketch

We have to find the area of shaded region. We sliced it in vertical rectangle width of rectangle = △x,

Length = (y – 0) = y

Area of rectangle = y△x

This rectangle can move horizontal from x = 0 to x = 2

Required area = Shaded region OCBO

= 2 (Shaded region OABO)

\displaystyle =2\int_0^2y\ dx\\ =2\times2\sqrt2\int_0^2\sqrt{x}\ dx\\ =4\sqrt2\left[\frac{2}{3}\times\sqrt{x}\right]_0^2\\ =4\sqrt2\left[\left(\frac{2}{3}\times2\sqrt2\right)-\left(\frac{2}{3}\right)\times0\sqrt0\right]\\ =4\sqrt2\left(\frac{4\sqrt2}{3}\right)

Required area = \frac{32}{3}  square units.

Question 2. Using integration, find the area of the region bounded by the line y – 1 = x, the x-axis and the ordinates x = -2 and x = 3.

Solution:

The find area of region bounded by x-axis the ordinates x = -2 and x = 3

y – 1 = x   ………….(1)

Equation (1) is a line that meets at axes at (0, 1) and (-1, 0)

Here, is the rough sketch

Required area is enclosed between the lines.

Required area = Region ABCA + Region ADEA

\displaystyle A=\int^3_{-1}y\ dx + \left|\int^{-1}_{-2}y\ dx\right|\\ =\int_{-1}^{3}(x+1)dx+\left|\int_{-2}^{-1}(x+1)dx\right|\\ =\left(\frac{x^2}{2}+x\right)_{-1}^3+\left|\left(\frac{x^2}{2}+x\right)_{-2}^{-1}\right|\\ =\left[\left(\frac{9}{3}+3\right)-\left(\frac{1}{2}-1\right)\right]+\left|\left(\frac{1}{2}-1(2-2)\right)\right|\\ =\left[\frac{15}{2}+\frac{1}{2}\right]+\left|-\frac{1}{2}\right|\\ =8+\frac{1}{2}\\ A=\frac{17}{2}\ square\ units.

Question 3. Find the area of the region bounded by the parabola y2 = 4ax and the line x = a

Solution:

Here we have to find the area of the region that is bounded by

x = a     ………(1)

and 

y2 = 4a    ………(2)

Equation (1) represents a line parallel to y-axis and equation (2) represent a parabola with vertex at origin and axis as x-axis.

Here, is the rough sketch

Here we have to find area between the region,

Thus, slice it in rectangles of 

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This assumed triangle can go from x = 0 to x = a.

Required area = Region OCBO

= 2 (Region OABO)

\displaystyle =2\int_0^a\sqrt{4ax}\ dx\\ =2\times2\sqrt{a}\int_0^a\sqrt{x}\ dx\\ =4\sqrt{a}\left(\frac{2}{3}x\sqrt{x}\right)_0^a\\ =4\sqrt{a}\left(\frac{2}{3}a\sqrt{a}\right)

Required area = \frac{8}{3}a^2  square units

Question 4. Find the area lying above the x-axis and under the parabola y = 4x – x2.

Solution:

We have to find here the bounded area by x-axis and parabola

y = 4x – x2

x2 – 4x +4 = -y + 4

(x – 2)2 = -(y – 4)     ……….(1)

Equation (1) represent a downward parabola with vertex (2, 4) and passing through (0, 0) and (0, 4).

Here is the rough sketch

Here the shaded region represents the required area.

We slice the region in approximation rectangles

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = a

Thus,

Required area = Region OABO

\displaystyle =\int_0^4(4x-x^2)\ dx\\ =\left(4\frac{x^2}{2}-\frac{x^3}{3}\right)_0^4\\ =\left(\frac{4\times16}{2}-\frac{64}{3}\right)-(0-0)\\ =\frac{64}{6}

Required area = \frac{32}{3}  square units

Question 5. Draw a rough sketch to indicate the region bounded between the curve y2 = 4x and the line x = 3. Also, find the area of this region.

Solution:

We have to find area bounded by

y2 = 4x   ……..(1)

and

x = 3   ………..(2)

Equation (1) represents a parabola with vertex at origin and axis as x-axis and equation (2) represents a line parallel to y-axis

Here is the rough sketch

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

This approximation rectangle slide from x = 0 to x = 3

Required area = Region OCBO

= 2(Region OABO)

\displaystyle =2\int_0^3y\ dx\\ =2\int_0^3\sqrt{4x}\ dx\\ =4\int_0^3\sqrt{x}\ dx\\ =4\left(\frac{2}{3}x\sqrt{x}\right)_0^3\\ =\frac{8}{3}\times3\sqrt3

Required area = 8\sqrt3  square units

Question 6. Make a rough sketch of the graph of the function y = 4 – x2, 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x-axis, and the line x = 0 and x = 2.

Solution:

Here, we will find the area enclosed by

y = 4 – x2

x2 = -(y – 4)     ……….(1)

x = 0    ………(2)

x = 2     ……….(3)

Equation (1) represents a downward parabola with vertex at (0, 4) and passing through (2, 0), (-2, 0). Equation (2) represents y-axis and equation (3) represents a line parallel to y-axis.

Here’s a rough sketch

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

length = y – 0 = y

This approximation rectangle slide from x = 0 to x = 2

Required area = Region OABO

\displaystyle =\int_0^2(4-x^2)\ dx\\ =\left(4x-\frac{x^3}{3}\right)_0^2\\ =\left[4(2)-\frac{(2)^3}{3}\right]-[0]\\ =\left[\frac{24-8}{3}\right]

Required area = \frac{16}{3}  square units.

Question 7. Sketch the graph of y=\sqrt{x+1}   in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the line x = 0 and x = 4.

Solution:

Here, we will find the area enclosed by x-axis and

y=\sqrt{x+1}

y2 = x + 1  …….(1)

x = 0   ………(2)

x = 4    ………(3)

Equation (1) represents a parabola with vertex at (-1, 0) and passing through (0, 1) and (0, -1). Equation (2) represents y-axis and equation (3) represents a line parallel to y-axis passing through (4, 0).

Thus, here is the rough sketch;

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = 4

Required area = Region OECDO

\displaystyle =\int_0^4y\ dx\\ =\int_0^4\sqrt{x+1}\ dx\\ =\left(\frac{2}{3}(x+1)\sqrt{x+1}\right)_0^4\\ =\frac{2}{3}[((4+1)\sqrt{4+1})-((0+1)\sqrt{0+1})]

Required area = \frac{2}{3}[5\sqrt5-1]  square units or \frac{2}{3}\left[5^\frac{3}{2}-1\right]     square units.

Question 8. Find the area under the curve y=\sqrt{6x+4}   above x-axis from x = 0 to x = 2. Draw a sketch of the curve also.

Solution:

Here, we will find the area enclosed by x-axis

x = 0,

x = 2    ………(1)

y2 = 6x + 4     ……….(2)

Equation (1) represents y-axis and a line parallel to y-axis passing through (2, 0). Equation (2) represents a parabola with vertex at \left(-\frac{2}{3},\ 0\right)     and passes through the points (0, 2) , ( 0, -2).

Thus, here is the rough sketch;

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = 2,

Required area = Region OABCO

\displaystyle =\int_0^2\sqrt{6x+4}\ dx\\ =\left(\frac{2}{3}\frac{(6x+4)\sqrt{6x+4}}{6}\right)_0^2\\ =\frac{1}{9}[((12+4)\sqrt{12+4})-((0+4)\sqrt{0+4})]\\ =\frac{1}{9}[16\sqrt{16}-4\sqrt{4}]\\ =\frac{1}{9}(64-8)

Required area =  \frac{56}{9}  square units.

Question 9. Draw the rough sketch of y2 + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.

Solution:

Here, we will find the area enclosed by x-axis

y2 = x + 1     ……….(1)

and

x = 2  ……………(2)

Equation (1) is a parabola with vertex at (0, 1) and axis as x-axis.

Equation (2) represents a line parallel to y-axis passing through  (2, 0)

Thus, here is the rough sketch;

Shaded region represents the required area that we have sliced in the form of a rectangle of

Width = △x

Length = y – 0 = y

Area of rectangle = y△x

This approximation rectangle slide from x = 0 to x = 2,

Required area = Region ABCA

= 2(Region AOCA)

\displaystyle =2\int_1^2y\ dx\\ =2\int_1^2\sqrt{x-1}\ dx\\ =2\left(\frac{2}{3}(x-1)\times\sqrt{x-1}^2\right)_1^2\\ =\frac{4}{3}[((2-1)\sqrt{2-1})-((1-1)\sqrt{1-1})]\\ =\frac{4}{3}(1-0)

Required area = \frac{4}{3}  square units.

Question 10. Draw a rough sketch of the graph of the curve \frac{x^2}{4}+\frac{y^2}{9}=1   and evaluate the area of the region under the curve and above the x-axis

Solution:

Here, we can observe that ellipse is symmetrical about x-axis.

Area bounded by ellipse = \displaystyle =2\int_0^2 y\ dx\\ = 2\int_0^23\sqrt{1-\frac{x^2}{4}}\ dx\\ =3\int\sqrt{4-x^2}\ dx\\ =3\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}sin^{-1}\frac{x}{2}\right]_0^2\\ =3[1(0)+2sin^{-1}-0-2sin^{-1}(0)]\\ =3[\pi]\\ =3\pi\ sq.\ units

Question 11. Sketch the region {(x, y) : 9x2 + 4y2 = 36} and find the area enclosed by it, using integration.

Solution:

9×2 + 4y2 = 36

\frac{x^2}{4}+\frac{y^2}{9}=1\\ y=\pm\sqrt{\frac{36-9x^2}{4}}

Area of Sector OABCO = \displaystyle \int_0^2\sqrt{\frac{36-9x^2}{4}}\ dx\\ =\frac{3}{2}\int_0^2\sqrt{4-x^2}\ dx\\ =\frac{3}{2}\left[\frac{x\sqrt{4-x^2}}{2}+\frac{2^2}{2}sin^{-1}\left(\frac{x}{2}\right)\right]_0^2\\ =\frac{3}{2}\left[\frac{2\sqrt{4-2^2}}{2}+\frac{2^2}{2}sin^{-1}\left(\frac{2}{2}\right)\right]-\frac{3}{2}\left[\frac{0\sqrt{4-0^2}}{2}+\frac{2^2}{2}sin^{-1}\left(\frac{0}{2}\right)\right]\\ =\frac{3}{2}\times2\times\frac{\pi}{2}-0\\ =\frac{3\pi}{2}\ sq.\ units

Area of the whole figure = 4 x area of DOABCO

=4\times\frac{3\pi}{2}\\ =6\pi\ sq.\ units.

Question 12. Draw a rough sketch of the graph of the function y=2\sqrt{1-x^2} , x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis

Solution:

Here, we have to find the area enclosed between the curve and x-axis.

y=2\sqrt{1-x^2},\ x\ ∈\ [0,\ 1 ]\\ \Rightarrow y^2+4x^2=4,\ x\ ∈\ [0,\ 1]\\ \Rightarrow \frac{x^2}{1}+\frac{y^2}{4},\ x\ ∈\ [0,\ 1]

Equation (1) represents an ellipse with centre at origin and passes through (±1, 0) and (0, ±2) and x ∈ [0, 1] as represented by region between y-axis and line x = 1.

Here, is the rough sketch.

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = 1,

Thus,

Required area = Region OAPBO

\displaystyle =\int_0^1y\ dx\\ =\int_0^12\sqrt{1-x^2}\ dx\\ =2\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}sin^{-1}(x)\right]_0^1\\ =2\left[\left(\frac{1}{2}\sqrt{1-x^2}+\frac{1}{2}sin^{-1}(1)\right)-(0+0)\right]\\ =2\left[0+\frac{1}{2}\times\frac{\pi}{2}\right]

Required area = \frac{\pi}{2}   square units

Question 13. Determine the area under the curve y=\sqrt{a^2-x^2}  included between the line x = 0 and x = 8.

Solution:

Here,

We have to find area under the curve

y =\sqrt{a^2-x^2}

x2 + y2 = a  ………..(1)

between x = 0    ………(2)

x = a    ………..(3)

Equation (1) represents a circle with Centre (0, 0) and passes axes at (0, ±a), (±a, 0).

Equation (2) represents y-axis and 

Equation x = a represents  a line parallel to y-axis passing through (a, 0)

Here, is the rough sketch,

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = a,

Thus,

Required area = Region OAPBO

\displaystyle =\int_0^ay\ dx\\ =\int_0^a\sqrt{a^2-x^2}\ dx\\ =\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}\right]_0^a\\ =\left[\left(\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}sin^{-1}(1)\right)-(0)\right]\\ =2\left[0+\frac{a^2}{2}\times\frac{\pi}{2}\right]

Required area = \frac{\pi}{4}a^2   square units

Question 14. Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8

Solution:

Here,

We have to find area bounded by x-axis

2y + 5x = 7   ………(1)

x = 2      ……..(2)

x = 8    ………(3)

Equation (1)  represents line passing through \left(-\frac{7}{5},\ 0\right)   and \left(0,\ \frac{7}{2}\right)   equation.

Equation (2), (3) shows line parallel to y-axis passing through (2, 0), (8, 0) respectively.

Here, is the rough sketch;

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 2 to x = 8,

Thus,

Required area = Region ABCDA

\displaystyle =\int_2^8\left(\frac{5x+7}{2}\right)\\ =\frac{1}{2}\left(\frac{5x^2}{2}+7x\right)_2^8\\ =\frac{1}{2}\left[\left(\frac{5(8)^2}{2}+7(8)\right)-\left(\frac{5(2)^2}{2}+7(2)\right)\right]\\ =\frac{1}{2}[(160+56)-(10+14)]\\ =\frac{192}{2}

Required area = 96 square units

Question 15. Using definite integrals, find the area of the circle x2 + y2 = a2

Solution:

Here, we have to find the area of circle,

x2 + y2 = a2

Equation (1) represents a circle with centre (0, 0) and radius a, Thus is meets the axes (±a, 0), (0, ±a).

Here, is the rough sketch;

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = a,

Thus,

Required area = Region ABCDA

= 4 ( Region ABOA)

\displaystyle =4\int_0^ay\ dx\\ =4\int_0^a\sqrt{a^2-x^2}\ dx\\ =4\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}\right]_0^a\\ =4\left[\left(\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}sin^{-1}(1)\right)-(0+0)\right]\\ =4\left[0+\frac{a^2}{2}\times\frac{\pi}{2}\right]\\ =4\left(\frac{a^2\pi}{4}\right)

Question 16. Using the integration, find the area of the region bounded by the following curves, after making a rough sketch: y = 1 + |x + 1|, x = -2, x = 3, y = 

Solution:

Here, we have to find the area enclosed by;

x = -2,

x = 3,

y = 0 and

y = 1 + |x + 1|

⇒ y = 1 + x + 1, if x + 1 0

⇒ y = 2 + x    ……….(1), if x ≥ -1

and

⇒ y = 1 – (x + 1), if x + 1 < 0

⇒ y = 1 – x – 1, if x < -1

⇒ y = -x    ………(2), if x < -1

Thus,

Equation (1) is a straight line that passes through (0, 2) and (-1 , 1).

Equation (2) is a line passing through (-1, 1) and (-2, 2) and it is enclosed by line x = 2 and x = 3 which are lines parallel to y-axis and pass through (2, 0) and (3, 0) respectively y = 0 is x-axis 

Here is the rough sketch

Shaded region represents the required area.

Thus,

Required area = Region (ABECDFA)

Required area = (Region ABEFA + Region ECDFE)    ……..(1)

Region ECDFE

We slice it into approximation rectangle of 

Width = △x

Length = y1

Area of rectangle = y1△x

The approx rectangles slide from x = -2 to x = -1,

Region ABEFA

We slice it into approximation rectangle of 

Width = △x

Length = y2

Area of rectangle = y2△x

The approx rectangles slide from x = -1 to x = 3,

Required area = \displaystyle \int_{-2}^{-1}y_1\ dx+\int_{-1}^3y_2\ dx\\ =-\left[\frac{x^2}{2}\right]^{-1}_{-2}+\left[\frac{x^2}{2}+2x\right]_{-1}^3\\ =-\left[\frac{1}{2}+\frac{4}{2}\right]+\left[\left(\frac{9}{2}+6\right)-\left(\frac{1}{2}-2\right)\right]\\ =\frac{3}{2}+\left(\frac{21}{2}+\frac{3}{2}\right)\\ =\frac{27}{2}

Required area = \frac{27}{2}   square units

Question 17. Sketch the graph y = |x -5|. Evaluate \displaystyle \int_0^1|x-5|\ dx . What does the value of the integral represent on the graph?

Solution:

Here, is the sketch of the given graph: 

y = |x – 5|

Hence,

Required area = \displaystyle =\int_0^1y\ dx\\ =\int_0^1|x-5|\ dx\\ =\int_0^1-(x-5)\ dx\\ =\left[\frac{-x^2}{2}+5x\right]_0^1\\ =\left[-\frac{1}{2}+5\right]\\ =\frac{9}{2}\ sq.\ units

Thus, 

The given integral represents the area bounded by the curves that are,

x = 0,

y = 0,

x = 1

and

y = -(x – 5).

Question 18. Sketch the graph of y = |x + 3| and evaluate \displaystyle \int_{-6}^0|x+3|\ dx. What does this integral represents on the graph?

Solution:

Here,

The given equation is y = |x + 3|

The corresponding values of x and y are given in the following table.

x-6-5-4-3-2-10
y3210123

Thus, 

After plotting these points,

We will get the graph of y = |x + 3|

It is shown as;

It is known that (x + 3) ≤ 0 for -6 ≤ x ≤  -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0

Therefore,

\displaystyle \int_{-6}^0|(x+3)|\ dx=-\int_{-6}^{-3}(x + 3)\ dx\\ =-\left[\frac{x^2}{2}+3x\right]_{-6}^{-3}+\left[\frac{x^2}{2}+3x\right]_{-3}^0\\ =\left[\left(\frac{(-3)^2}{2}+3(-3)\right)-\left(\frac{(-6)^2}{2}+3(-6)\right)\right]+\left[0-\frac{(-3)^2}{2}+3(-3)\right]\\ =-\left[-\frac{9}{2}\right]-\left[-\frac{9}{2}\right]\\ =9

Question 19. Sketch the graph y = |x + 1|. Evaluate \displaystyle \int_{4}^2|(x+1)|\ dx. What does the value of this integral represent on this graph?

Solution:

Here,

Given:

y = |x + 1|=\begin{cases} x+1,\ if\ x+1\ge 0\\ -(x+1),\ if\ x+1<0 \end{cases}\\ y=\begin{cases} (x+1),\ if\ x\ge -1\\ -x-1,\ if\ x<-1 \end{cases}

y = x + 1   …………(1)

and

y = -x – 1   ……….(2)

Equation (1) represents a line which meets axes at (0, 1).

Equation (2) represents a line passing through (0, -1) and (-1, 0)

Here is the rough sketch

\displaystyle \int_{4}^2|(x+1)|\ dx=-\int_{-4}^{-1}(x + 1)\ dx\ +\int_{-1}^2(x+1)\ dx\\ =-\left[\frac{x^2}{2}+x\right]_{-4}^{-1}+\left[\frac{x^2}{2}+x\right]_{-1}^2\\ =-\left[\left(\frac{1}{2}-1\right)-\left(\frac{16}{2}-4\right)\right]+\left[\left(\frac{4}{2}+2\right)-\left(\frac{1}{2}-1\right)\right]\\ =-\left[\left(-\frac{1}{2}-4\right)\right]+\left[4+\frac{1}{2}\right]\\ =\frac{9}{2}+\frac{9}{2}\\ =\frac{18}{2}

Required area = 9 square units.

Question 20. Find the area of the region bounded by the curve xy – 3x – 2y – 10 = 0, x-axis and the lines x = 3, x = 4.

Solution:

Here,

We have to find the area bounded by

x axis,

x = 3,

x = 4

and

xy – 3x -2y – 10 = 0

⇒ y(x – 2) = 3x + 10

⇒ y=\frac{3x+10}{x-2}

Here, is the rough sketch

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 3 to x = 4,

Required area = Region ABCDA

\displaystyle =\int_3^4y\ dx\\ =\int_3^4\left(\frac{3x+10}{x-2}\right)\ dx\\ =\int_3^4\left(3+\frac{16}{x-2}\right)\ dx\\ =(3x)_3^4+16(log|x-2|)_3^4\\ =(12-9)+16(log2-log1)

Required area = (3 + 16 log2) square units.

Question 21. Draw a rough sketch of the curve y=\frac{π}{2}+2sin^2x  and find the area between x-axis, the curve and the ordinates x = 0, x = π

Solution:

Here, we have to find the bounded by 

y=\frac{\pi}{2}+2sin^2x

x-axis, x = 0 and x = π

Here is the table for values of y=\frac{\pi}{2}+2sin^2x

\frac{\pi}{6}

Here is the rough sketch,

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABCDO

\displaystyle =\int_0^π y\ dx\\ =\int_0^π\left(\frac{π}{2}+2sin^2x\right)\ dx\\ =\int_0^π\left(\frac{π}{2}+1-cos\ 2x\right)\ dx\\ =\left[\frac{π}{2}x+x-\frac{sin\ 2x}{2}\right]_0^π\\ =\left[\left(\frac{π^2}{2}+π-\frac{sin\ 2x}{2}\right)-(0)\right]\\ =\frac{π^2}{2}+π

Required area = \frac{π}{2}(π+2)      square units

Question 22. Draw a rough sketch of the curve y=\frac{x}{π}+2sin^2x  and find the area between the x-axis, the curve and the ordinates x = 0, x = π.

Solution:

Here, we have the area between y-axis,

x = 0,

x = π

and

y=\frac{x}{π}+2\ sin^2x\ \ \ \ \ ......(1)

Thus, the table for equation (1) is

\frac{π}{6}

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABOA

\displaystyle =\int_0^π y\ dx\\ =\int_0^π\left(\frac{π}{2}+2sin^2x\right)\ dx\\ =\int_0^π\left(\frac{π}{2}+1-cos\ 2x\right)\ dx\\ =\left[\frac{π}{2x}x+x-\frac{sin\ 2x}{2}\right]_0^π\\ =\left[\left(\frac{π^2}{2x}+π-0\right)-(0)\right]\\

Required area = \frac{3\pi}{2} square units

Question 23. Find the area bounded by the curve y = cos x between x = 0 and x = 2π

Solution:

Here from the figure we can see that

The required area = area of the region OABO + area of the region BCDB + area of the region DEFD

Therefore,

The required area = \displaystyle \int_0^{\frac{\pi}{2}}cos\ x\ dx+\left|\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}cos\ x\ dx\right|+\int_{\frac{3\pi}{2}}^{2\pi}cos\ x\ dx\\ =[sin\ x]_0^{\frac{\pi}{2}}+\left|[sin\ x]^{\frac{3\pi}{2}}_{\frac{\pi}{2}}\right|+[sin\ x]^{2\pi}_{\frac{3\pi}{2}}\\ =\left[sin\frac{\pi}{2}-sin0\right]+\left|sin\frac{3\pi}{2}-sin\frac{\pi}{2}\right|+\left[sin\ 2x-sin\frac{3\pi}{2}\right]\\ =1+2+1\\ =4\ sq.\ units

Question 24. Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = \frac{π}{3}  are the ratio 2:3.

Solution:

We have to find the area under the curve

y = sin x    ……..(1)

and

y = sin 2x     …………(2)

Between x = 0 and x = \frac{\pi}{3}

0\ \ \ \ \ \ \frac{\pi}{6}\ \ \ \ \ \ \frac{\pi}{4}\ \ \ \ \ \ \ \frac{\pi}{3}\\ 0\ \ \ \ \  0.5\ \ \ \  0.7\ \ \ \ 0.8

Here is the rough sketch 

Area under curve y = sin 2x

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y1

Area of rectangle = y1△x

The approx rectangles slide from x = 0 to x = \frac{\pi}{3}     ,

Thus,

Required area = Region OPACO

\displaystyle A_1=\int_0^{\frac{\pi}{3}}y_1\ dx\\ =\int_0^{\frac{\pi}{3}}sin\ 2x\ dx\\ =\left[\frac{-cos\ 2x}{2}\right]_0^{\frac{\pi}{3}}\\ =-\left[-\frac{1}{4}-\frac{1}{2}\right]\\ A_1=\frac{3}{4}\ sq.\ units

We slice it into approximation rectangle of 

Width = △x

Length = y2

Area of rectangle = y2△x

The approx rectangles slide from x = 0 to x = \frac{\pi}{3},

Thus,

Required area = Region OQACO

\displaystyle =\int_0^{\frac{\pi}{3}}y_2\ dx\\ =\int_0^{\frac{\pi}{3}}sin\ x\ dx\\ =\left[{-cos\ x}\right]_0^{\frac{\pi}{3}}\\ =-\left[\frac{1}{2}-1\right]\\ A_2=\frac{1}{2}\ sq.\ units

Thus,

A_2:A_1=\frac{1}{2}:\frac{3}{4}\\ A_2:A_1=2:3

Question 25. Compare the area under the curves y = cos2x and y = sin2x between x = 0 and x = π

Solution:

Here to compare area under curves

y = cos2x

and

y = sin2x

Between x = 0 and x = π

This is the table for y = cos2x and y = sin2x

\frac{\pi}{6}\ \ \ \ \ \ \ \ \ \frac{\pi}{4}\\ 0.75\ \ \ \ 0.5

Area of region enclosed by

y = cos2x and axis

A1 = Region OABO + Region BCDB

= 2(Region BCDB)

\displaystyle =2\int_{\frac{\pi}{2}}^\pi cos^2x\ dx\\ =2\int_{\frac{\pi}{2}}^\pi\left(\frac{1-cos\ 2x}{2}\right)\ dx\\ =\left[x-\frac{sin\ 2x}{2}\right]^\pi_{\frac{\pi}{2}}\\ =\left[(x-0)-\left(\frac{\pi}{2}-0\right)\right]\\ =\pi-\frac{\pi}{2}\\ A_1=\frac{\pi}{2}\ sq.\ units\ \ \ \ \ \ .....(1)

Area of region enclosed by y = sin2x and axis

A2 = Region OEDO

\displaystyle =\int_{0}^\pi sin^2x\ dx\\ =\int_{0}^\pi\left(\frac{1-cos\ 2x}{2}\right)\ dx\\ =\frac{1}{2}\left[x-\frac{sin\ 2x}{2}\right]^\pi_{0}\\ =\frac{1}{2}[(x-0)-(0)]\\ A_2=\frac{\pi}{2}\ sq.\ units\ \ \ \ \ \ .....(2)

From equation (1) and (2),

A1 = A2

Thus,

Area enclosed by y = cos2x = Area enclosed by y = sin2x.

Question 26. Find the area bounded by the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  and the ordinates x = 0 and x = ae, where, b2 = a2(1 – e 2) and e < 1.

Solution:

Thus, the required area in the figure below of the region BOB’RFSB is enclosed by the ellipse and the lines x = 0 and x = ae

Here is the area of the region BOB’RFSB

\displaystyle =2\int_0^{ae}y\ dx\\ =2\frac{b}{a}\int_0^{ae}\sqrt{a^2-x^2}\ dx\\ =\frac{2b}{a}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}\right]_0^{ae}\\ =\frac{2b}{2a}\left[ae\sqrt{a^2-a^2e^2}+a^2sin^{-1}e\right]\\ =ab\left[e\sqrt{1-e^2}+sin^{-1}e\right]

Question 27. Find the area of the minor segment of the circle x2 + y2 = a2 cut-off by the line x = \frac{a}{2}  .

Solution:

Area of the mirror segment of the circle

\displaystyle =2\int_{\frac{a}{2}}^a\sqrt{a^2-x^2}\ dx\\ =2\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{2}\right]_{\frac{a}{2}^a}\\ =2\left[\frac{a}{2}(0)+\frac{a^2}{2}sin^{-1}\left(\frac{a}{2}\right)-\frac{a}{4}\sqrt{a^2-\frac{a^2}{4}}-\frac{a^2}{2}sin^{-1}\frac{a}{4}\right]\\ =2\left[\frac{a^2}{2}sin^{-1}\left(\frac{a}{2}\right)-\frac{a}{4}\sqrt{a^2-\frac{a^2}{4}}-\frac{a^2}{2}sin^{-1}\frac{a}{4}\right]\\ =\frac{a^2}{12}[4\pi-3\sqrt3]\ sq.\ units

Question 28. Find the area of the region bounded by the curve x = at, y = 2at between the ordinates corresponding t = 1 and t = 2.

Solution:

Area of the bounded region

\displaystyle =2\int_1^2y\ \frac{dx}{dt}\ dt\\ =2\int_1^2(2at)(2at)\ dt\\ =8a^2\int_1^2t^2\ dt\\ =8a^2\left[\frac{t^3}{3}\right]_1^2\\ =8a^2\left[\frac{8}{3}-\frac{1}{3}\right]\\ =\frac{56a^2}{3}\ sq.\ units

Question 29. Find the area enclosed by the curve x = 3 cos t, y = 2 sin t.

Solution:

Area of the bounded region

\displaystyle =4\int_0^{\frac{\pi}{2}}2sin\ t\ dt\\ =-8[cos\ t]_0^{\frac{\pi}{2}}

= -8 [0 – 1]

= 8 square units 

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