Practicing S Chand Class 12 Maths Solutions Chapter 25 Application of Integrals Ex 25(a) is the ultimate need for students who intend to score good marks in examinations.

## S Chand Class 12 ICSE Maths Solutions Chapter 25 Application of Integrals Ex 25(a)

Question 1.
Using integration, find the area of the region bounded by
(i) the line 3 y = 2 x + 4, the x-axis and the lines x = 1 and x = 3.
(ii) 2 y = -x + 6, the x-axis and the lines x = 2 and x = 4.
(i) The eqn. of given line be
3 y = 2 x + 4 ……………..(1)
eqn. (1) meets x-axis at A(-2, 0) and y-axis at B (0, $$\frac{4}{3}$$).
∴ required area = $$\int_1^3$$ y d x
= $$\int_1^3$$ $$\frac{2 x+4}{3}$$ d x = $$\frac{1}{3}$$ $$\frac{(2 x+4)^2}{2 \times 2}]_1^3$$
= $$\frac{1}{12}$$ [(10)2 – 62]
= $$\frac{1}{12}$$(100 – 36)
= $$\frac{64}{12}$$ = $$\frac{16}{3}$$ sq. units

(ii) The eqn. of given line be
2y = -x + 6
it meets coordinate axes at A(6, 0) and B(0, 3).
∴ Required area = $$\int_2^4$$ y d x
= $$\int_2^4$$ $$\frac{-x+6}{2}$$ d x
= $$\frac{1}{2}$$ $$\frac{(6-x)^2}{-2}]_2^4$$
= $$-\frac{1}{4}$$ [4 – 16]
= 3 sq. units

Question 2.
(i) Using integration, find the area of the region bounded between the line x = 2 and the parabola y2 = 8 x.
(ii) Find the area of the region bounded by the parabola y2 = 12 x and its latus rectum.
(iii) Find the area enclosed between the co-ordinate axes and the curve y2 = 4 a(x + λ) in the second quadrant.
(i) Clearly the figure is symmetrical about x-axis.
∴ area enclosed by the parabola
= 2 × area enclosed by parabola
and x = 2 in Ist quadrant
Divide the region in first quadrant into vertical strips each vertical strip has its lower end on x-axis and upper and on the parabola y = $$\sqrt{8 x}$$. So the approximating rectangle has length |y| and width d x. This rectangle can move between x = 0 and x = 2.

[∵ y ≥ 0 ∴|y|=y]
∴ required area = 2 × $$\int_0^2$$|y| d y
= 2 $$\int_0^2$$ y d y
= 2 $$\int_0^2$$ √ 8 √ x d x
= 2 × 2 $$\sqrt{2}$$ [$$\frac{2}{3}$$ x3/2 ]2 0
= $$\frac{8 \sqrt{2}}{3}$$ [23/2 – 0]
= $$\frac{32}{3}$$ sq. units

(ii) Given eqn. of parabola be y2 = 12 x represents a right handed parabola with vertex at (0,0). Here 4 a = 12
⇒ a = 3
∴ required area = 2 × area of region OABO
= 2 × $$\int_0^3$$ $$\sqrt{12 x}$$ d x
= 2 × 2 $$\sqrt{3}$$ $$\int_0^3$$ x1/2 d x
= 4 $$\sqrt{3}$$ $$\frac{x^{3 / 2}}{3 / 2}]_0^3$$
= $$\frac{8 \sqrt{3}}{3}$$[3 $$\sqrt{3}$$ – 0]
= 24 sq. units

(iii) The given eqn. of parabola be
y2 = 4 a(x + λ)
represents a right handed parabola with vertex (-λ, 0) it meets y-axis at A(0, 2 $$\sqrt{a \lambda}$$)
∴ required area = $$\int_{-\lambda}^0 2 \sqrt{a}$$ $$\sqrt{x+\lambda}$$ d x
[∵ y > 0 in 2nd quad ]
= 2 $$\sqrt{a}$$ $$\frac{(x+\lambda)^{3 / 2}}{3 / 2}]_{-\lambda}^0$$
= $$\frac{4 \sqrt{a}}{3}$$ λ3/2

Question 3.
Draw the rough sketch of the curve y = $$\sqrt{3 x + 4 x}$$ and find the area under the curve, above the x-axis and between x = 0 and x = 4.
The eqn. of given curve y = $$\sqrt{3 x+4}$$
The table of values is given as under :

 x 0 4 2 y 2 4 √10

Plot the points (0, 2) ;(4, 4) and (2, $$\sqrt{10}$$) on graph paper and join then by free hand curve.
∴ Required area of region OABCO
= $$\int_0^4$$ $$\sqrt{3 x+4}$$ d x
= $$\frac{(3 x+4)^{3 / 2}}{\frac{3}{2} \cdot 3}]_0^4$$
= $$\frac{2}{9}[(16)^{3 / 2}-(4)^{3 / 2}]$$
= $$\frac{2}{9}$$[64 – 8]
= $$\frac{112}{9}$$ sq. units

Question 4.
Sketch the graph of y = |x + 1|. Evaluate $$\int_{-4}^2$$|x + 1| d x. What does the value of the integral represent on the graph?
For Graph ; For x < -1; y = f(x) = -(x + 1) For x = -1 ; y = 0 For x > -1 ; y = x + 1
∴ $$\int_{-4}^2$$|x + 1| d x
= $$\int_{-4}^{-1}$$ – (x + 1) d x + $$\int_{-1}^2$$ (x + 1) d x
∴ I = $$-\frac{(x+1)^2}{2}]_{-4}^1$$ + $$\frac{(x+1)^2}{2}]_{-1}^2$$
= $$-\frac{1}{2}$$(0 – 9) + $$\frac{1}{2}$$(9 – 0)
= $$\frac{9}{2}$$ + $$\frac{9}{2}$$ = 9

When x = -4 < -1 ∴ y = -(-4 + 1) = 3 i.e. point becomes (-4, 3). When x = 2 > -1
∴ y = 2 + 1 = 3 i.e. point becomes (2, 3).
Further, area of ∆ ABC = $$|\frac{1}{2} \times(-3) \times(3)|$$ = $$\frac{9}{2}$$ sq. units
and area of ∆CDE = $$|\frac{1}{2} \times 3 \times 3|$$ = $$\frac{9}{2}$$ sq. units
∴ $$\int_{-4}^2$$|x + 1| d x
= sum of areas of two ∆ from -4 to -1 and -1 to 2 .

Question 5.
Find the area bounded by the curve y = x2 and the line y = 16. [Fig. 25.41]

(i) The given curve y = x2
and the line y = 16 intersects when 16 = x2
⇒ x = ± 4
∴ Region R = {(x, y) ; x2 ≤ y ≤ 10 ; 0 ≤ x ≤ 4}
∴ Required area = $$\int_0^4$$[16 – x2 ] d x
= 16 x – $$\frac{x^3}{3}]_0^4$$
= 64 – $$\frac{64}{3}$$
= $$\frac{128}{3}$$ sq. units

(ii) required area = $$\int_1^4$$ x d y
= $$\int_1^4$$ $$\frac{\sqrt{y}}{2}$$ d y
= $$\frac{1}{2}$$ $$\frac{y^{3 / 2}}{3 / 2}]_1^4$$
= $$\frac{1}{3}$$[43/2 – 1] = $$\frac{7}{3}$$ sq. units

Question 6.
Sketch the region lying in the first quadrant and bounded by y = 4 x2, x = 0, y = 1 and y = 4. Find the area of the region, using integration.

The given curve is y = 4 x2 be an upward parabola with vertex at (0, 0) and is symmetrical about y-axis.
∴ required area = area of region OABCO
= $$\int_1^4$$ x d y [Taking Horizontal strips]

= $$\int_1^4$$ $$\frac{\sqrt{y}}{2}$$ d y
[∵ y = 4 x2 ⇒ x = ± $$\frac{\sqrt{y}}{2}$$ as region lies in 1st quadrant ∴ x > 0 i.e. x = $$\frac{\sqrt{y}}{2}$$]
= $$=\frac{1}{2}$$ $$\frac{y^{3 / 2}}{3 / 2}]_1^4$$
= $$\frac{1}{3}$$[8 – 1]
= $$\frac{7}{3}$$ square units

Question 7.
Find the gradients of the curve y = x2(2 – x) at the points (0, 0) and (2, 0) and sketch the part of the curve of which both x and y are positive. Find the area between this part of the curve and the axis.
Given eqn. of curve y = x2(2 – x)
∴ $$\frac{d y}{d x}$$ = 4 x – 3 x2
Thus gradient of curve at (0, 0) = $$\frac{d y}{d x}$$(0,0) = 0
and gradient of curve at (2, 0) = $$\frac{d y}{d x}$$(2,0)
= 8 – 12 = -4

The table of values is given as under:

 x 0 1 2 $$\frac{3}{2}$$ y 0 1 0 $$\frac{9}{8}$$

by plotting all these points on graph paper and join them by free hand curve.
∴ required area of region OBAO = $$\int_0^2$$ x2(2 – x) d x
= $$\frac{2 x^3}{3}$$ – $$\frac{x^4}{4}]_0^2$$
= $$\frac{16}{3}$$ – 4
= $$\frac{4}{3}$$ sq. units

Question 8.
Find the area included between the curve y = 9 – x2 , the X-axis and the lines x = -2 and x = ± 2.
The given curve
y = 9 – x2
x2 = 9 – y = -(y – 9)
⇒ represents a downward parabola with vertex (0, 9).
∴ required area of region P A Q P
= $$\int_{-2}^2$$ y d x
= $$\int_{-2}^2$$(9 – x2) d x
= 9 x – $$\frac{x^3}{3}]_{-2}^2$$
= (18 – $$\frac{8}{3}$$) – (-18 + $$\frac{8}{3}$$)
= 36 – $$\frac{16}{3}$$
= $$\frac{108-16}{3}$$
= $$\frac{92}{3}$$ sq. units

Question 9.
Find the area bounded by the curve y2 = 4 x, the line y = 3, and the y-axis.
The given curve y2 = 4 x represents a right handed parabola with vertex at origin O(0, 0). Now the line y = 3 meets the given curve y2 = 4 x when 9 = 4 x
i.e. x = $$\frac{9}{4}$$ i.e. at point ($$\frac{9}{4}$$, 3).
∴ R = {(x, y) ; 0 ≤ x ≤ $$\frac{y^2}{4}$$ ; 0 ≤ y ≤3}
Thus required area = $$\int_0^3$$ $$\frac{y^2}{4}$$ d y
= $$\frac{1}{4}$$ $$\frac{y^3}{3}]_0^3$$
= $$\frac{1}{12}$$ × 27
= $$\frac{9}{4}$$ sq. units

Aliter: Divide the region into vertical strips with upper end on line y = 3
and. low erend on given curve y2 = 4 × and 0 ≤ x ≤ $$\frac{9}{4}$$
∴ required area = $$\int_0^{9 / 4}$$[3 – 2 $$\sqrt{x}$$] dx = 3x – $$\frac{2 x^{3 / 2}}{3 / 2}]_0^{9 / 4}$$
= 3($$\frac{9}{4}$$ – 0) – $$\frac{4}{3}$$[$$\frac{9}{4})^{3 / 2}$$ – 0]
= $$\frac{27}{4}$$ – $$\frac{4}{3}$$ × $$\frac{27}{8}$$
= $$\frac{27}{4}$$ – $$\frac{9}{2}$$
= $$\frac{27-18}{4}$$ = $$\frac{9}{4}$$ sq. units

Question 10.
Find the area of the region bounded by y = -1, y = 2, x = y and x = 0.
Given lines are
y = -1 ……………….(1)
y = 2 ……………….(2)
x = y ……………….(3)
x = 0 …………………(4)
and
Since y = -1 and y = 2 are the lines are parallel to x-axis and x = 0 represents y-axis.
The line y = x passes through origin and making a slope of 45° with positive direction of x-axis.
Divide the region into horizontal strips as shown in shaded portion.

R1 = {(x, y); 0 ≤ x ≤ y ; 0 ≤ y ≤ 2}
R2 = {(x, y) ; y ≤ x ≤ 0 ;-1 ≤ y ≤ 0}
∴ Required area = $$\int_{-1}^0$$|x| d y + $$\int_0^2$$ x d y
= $$\int_{-1}^0$$ – y d y + $$\int_0^2$$ y d y
= $$-\frac{y^2}{2}]_{-1}^0$$ + $$\frac{y^2}{2}]_0^2$$
= $$-\frac{1}{2}$$(0 – 1) + $$\frac{1}{2}$$ (4 – 0)
= $$\frac{1}{2}$$ + 2 = $$\frac{5}{2}$$ sq. units

Question 11.
Find the area bounded by the parabola y2 = 2 x and the ordinates x = 1 and x = 4.
The given parabola y2 = 2 x be a right handed parabola with vertex (0, 0).
Divide the region into vertical strips with lower end on x-axis and upper end on given curve
y2 = 2 x .
∴ required area = 2 $$\int_1^4$$ y d x
= 2 $$\int_1^4$$ $$\sqrt{2}$$ $$\sqrt{x}$$ d x
= 2 $$\sqrt{2}$$ $$\frac{x^{3 / 2}}{3 / 2}]_1^4$$
= $$\frac{4 \sqrt{2}}{3}$$[43/2 – 1]
= $$\frac{4 \sqrt{2}}{3}$$[8 – 1]
= $$\frac{28 \sqrt{2}}{3}$$ sq.units

Question 12.
Find the area bounded by the curve y2 = 4 a2 (x – 1) and the lines x = 1 and y = 4 a.
Given eqn. of curve be y2 = 4 a2 (x – 1) which represents a right handed parabola with vertex (1, 0).
The line x = 1 meets given parabola at A(1, 0).
The line y = 4 a meets given parabola when 16 a2 = 4 a2 (x – 1)
⇒ x = 5 i.e. at B(5, 4 a).
∴ R = {(x, y) ; 4 a ≤ y ≤ 2 a $$\sqrt{x-1}$$ ; 1 ≤ x ≤ 5}
Thus required area = $$\int_1^5$$ [4 a – 2 a $$\sqrt{x-1}$$] d x
= 4 a x – $$\frac{2 a(x-1)^{3 / 2}}{3 / 2}]_1^5$$
= 4 a(5 – 1) – $$\frac{4 a}{3}$$ [43/2 – 0]
= 16 a – $$\frac{32 a}{3}$$
= $$\frac{16 a}{3}$$ sq.units

Question 13.
Evaluate the area of the region bounded by the curve y = 2 $$\sqrt{1-x^2}$$, and the X-axis, after drawings rough sketch of the same.
Given equation of curve be y = 2 $$\sqrt{1-x^2}$$
⇒ y2 = 4 (1 – x2)
⇒ $$\frac{y^2}{4}$$ + $$\frac{x^2}{1}$$ = 1
So it represents an ellipse with major axis on y-axis and minor axis as x-axis. Clearly it contains even power of x and y so the given curve is symmetrical about both coordinate axis.

∴ area of whole ellipse = 4 × area of ellipse in first quadrant and coordinate axes.
Now, we divide the region in first quadrant into vertical strips each vertical strip has lower and on x-axis and upper end on ellipse. So the approximating rectangle has length |y| and width d x and clearly it move from x = 0 and x = 1.
∴ area of whole ellipse = 4 × $$\int_0^1$$ |y| d x
= 4 $$\int_0^1$$ y d x [∵ y ≥ 0]
= 4 $$\int_0^1$$ 2 $$\sqrt{1-x^2}$$ d x
= 8 $$\frac{x \sqrt{1-x^2}}{2}$$ + $$\frac{1}{2}$$ sin-1 x]1 0
= 8[0 + $$\frac{1}{2}$$ sin-1 (1) – 0 – 0]
= 8 × $$\frac{1}{2}$$ × $$\frac{\pi}{2}$$
= 2 π square units
Here required area = area of region in first quadrant = $$\frac{1}{4}$$ × 2π = $$\frac{\pi}{2}$$ square units

Question 14.
Make a rough sketch of the curve 3 y = (2 x + 1)(2 – x), and calculate its gradient at the point where it meets the axis of y. Find the area bounded by the curve and the axis of x.
Given eqn. of curve be
3y = (2 x + 1)(2 – x) ……………….(1)
The table of values is given as under :

The given eqn. (1) can be written as,
3y = -2 x2 + 3 x + 2
⇒ 3y = -2[x2 – $$\frac{3}{2}$$ x + $$\frac{9}{16}$$ – $$\frac{9}{16}$$ – 1]
⇒ 3y = -2[(x – $$\frac{3}{4}$$)2 – $$\frac{25}{16}$$]
⇒ (x – $$\frac{3}{4}$$)2
= $$\frac{25}{16}$$ – $$\frac{3}{2}$$
y = $$-\frac{3}{2}$$(y – $$\frac{25}{16}$$ × $$\frac{2}{3}$$)
⇒ (x – $$\frac{3}{4}$$)2
= $$-\frac{3}{2}$$ (y – $$\frac{25}{24}$$)
which represents a downward parabola with vertex ($$\frac{3}{4}$$, $$\frac{25}{24}$$).
Diff. (1) both sides w.r.t. x, we have
3 $$\frac{d y}{d x}$$ = -4 x + 3
⇒ $$\frac{d y}{d x}$$ = $$\frac{-4 x+3}{3}$$
curve given by eqn. (1) meets y-axis i.e. x = 0
∴ y = $$\frac{2}{3}$$ i.e. at point (0, $$\frac{2}{3}$$).
∴ gradient of the curve at (0, $$\frac{2}{3}$$)
= $$\frac{d y}{d x}$$ (0, $$\frac{2}{3}$$)
= $$-\frac{4 × 0 + 3}{3}$$ = 1
Given curve meets x-axis at y = 0 i.e. (2 x + 1)(2 – x) = 0
⇒ x = -\frac{1}{2}, 2
∴ required area = $$\frac{1}{3}$$ $$\int_{-1 / 2}^2$$ (2 x + 1)(2 – x) d x
= $$\frac{1}{3}$$ $$\int_{-1 / 2}^2$$ (- 2 x2 + 3 x + 2) d x
= $$\frac{1}{3}$$[$$-\frac{2}{3}$$ x3 + $$\frac{3 x^2}{2}$$ + 2 x]-1/2 2
= $$\frac{1}{3}$$[($$-\frac{16}{3}$$ + 6 + 4) – ($$\frac{1}{12}$$ + $$\frac{3}{8}$$ – 1)]
= $$\frac{1}{3}$$ [$$\frac{14}{3}$$ – $$\frac{1}{12}$$ – $$\frac{3}{8}$$ + 1]
= $$\frac{125}{72}$$ sq. units

Question 15.
Given alongside figure shows a sketch of the curve y = (x – 1)(4 – x). If P is the point (2, 2), verify that the tangent at P passes through the origin O as shown. Calculate the area O A P enclosed between the tangent, the curve and the axis of x.

The eqn. of given curve be
y = (x – 1)(4 – x)
∴ $$\frac{d y}{d x}$$ = (x – 1)(-1) + (4 – x) 1
= -x + 1 + 4 – x = 5 – 2 x

∴ slope of tangent to given curve (1) at point (2, 2)
= $$\frac{d y}{d x}$$ {(2,2)} = 5 – 4 = 1
Thus eqn. of tangent through the point (2, 2) be given by
y – 2 = 1(x – 2)
⇒ y = x
Clearly eqn. (2) passes through origin O(0, 0).
∴ area of shaded region OAPO
= area of ∆ OPQ – area of region APQA
= $$\frac{1}{2}$$ × 2 × 2 – $$\int_1^2$$ (x – 1)(4 – x) d x
[since curve (1) meets x-axis i.e. y = 0 ∴ x = 1, x = 4 ]
= 2 – $$\int_1^2$$ (-x2 + 5 x – 4) d x
= 2 + $$\int_1^2$$ (x2 – 5 x + 4) d x
= 2 + $$\frac{x^3}{3}$$ – $$\frac{5 x^2}{2}$$ + 4 x]21
= 2 + $$\frac{8}{3}$$ – 10 + 8 – $$\frac{1}{3}$$ + $$\frac{5}{2}$$ – 4]
= $$\frac{8}{3}$$ – $$\frac{1}{3}$$ + $$\frac{5}{2}$$ – 4
= $$\frac{7}{3}$$ – $$\frac{3}{2}$$
= $$\frac{5}{6}$$ sq. units

Question 16.
The curve y = a x2 + b x + c passes through the points (1, 0),(2, 0) and its gradient at the point (2, 0) is 2 . Find the numerical value of the area included between the curve and the axis of x.
The eqn. of given curve y = a x2 + b x + c eqn. (1) passes through the point (1, 0) and (2, 0).
∴ 0 = a + b + c
0 = 4 a + 2 b + c
Diff. eqn. (1) both sides w.r.t. x; ; we have
$$\frac{d y}{d x}$$ = 2 a x + b
∴ gradient of curve at point (2, 0) = 2
⇒ ( $$\frac{d y}{d x}$$) (2,0) = 2
⇒ 4 a + b = 2
eqn. (3) -2 x eqn. (4) ; we have
4 a + 2 b + c – 8 a – 2 b + 4 = 0
⇒ -4 a + c = -4
⇒ 4 a – c = 4
eqn. (4) – eqn. (2) ; we have
3a – c = 2
eqn. (5) – eqn. (6) gives; a = 2
∴ from (6) ; c = 6 – 2 = 4
∴ from (4); b = 2 – 4 a = 2 – 8 = -6
Thus from (1); we have
y = 2 x2 – 6 x + 4
eqn. (7) meets x-axis when y = 0
∴ 2 x2 – 6 x + 4 = 0
⇒ (x – 1)(2 x – 4) = 0
⇒ x = 1,2
∴ required area = $$\int_1^2$$ y d x
= $$\int_1^2$$ (2 x2 – 6 x + 4) d x
= 2 $$\int_1^2$$ (x2 – 3 x + 2) d x
= 2$$\frac{x^3}{3}$$ – $$\frac{3 x^2}{2}$$ + 2 x]2 1
= |2$$\frac{8}{3}$$ – 6 + 4 – $$\frac{1}{3}$$ + $$\frac{3}{2}$$ – 2]|
= 2[$$\frac{7}{3}$$ + $$\frac{3}{2}$$ – 4]
= |2$$\frac{14+9-24}{6}$$|
= $$\frac{1}{3}$$ sq. units

Question 17.
The line y = 2 x meets the curve y2 = 4 x at the point O (the origin) and P, and P N is perpendicular to the y-axis. Prove that the area between the curve and O P is one-half the area enclosed by the lines O N, N P and the curve.
eqn. of given line be
y = 2 x

y2 = 4 x
and eqn. of given curve be
eqn. (2) represents a right handed parabola with vertex at (0, 0) and line (1) meets curve (2) when 4 x2 = 4 x
⇒ 4 x(x – 1) = 0
⇒ x = 0,1
∴ from (1); y = 0, 2
Thus the point of intersection are O(0, 0) and P(1, 2).
∴ area betwen the curve and OP
A1 = $$\int_0^1$$ [2 $$\sqrt{x}$$ – 2 x] d x
= $$\frac{2 x^{3 / 2}}{3 / 2}-x^2$$]_0^1
= $$\frac{4}{3}$$ – 1
= $$\frac{1}{3}$$ sq. units
A2 = area of region ONPO
= area of ∆ ONP – area between curve and OP
= $$\frac{1}{2}$$ × 2 × 1 – $$\frac{1}{3}$$
= $$\frac{2}{3}$$ sq. units
∴ ⇒ $$\frac{\mathrm{A}_1}{\mathrm{~A}_2}$$
= $$\frac{1}{2}$$
⇒ area between the curve and OP is one-half the area enclosed by the lines ON, NP and the curve.

Question 18.
Calculate the areas of the two parts into which the area enclosed by the x-axis and the curve y=18 x-3 x^2 is divided by the line x=4.
Given eqn. of curve be
y = 18 x – 3 x2
⇒ y = -3(x2 – 6 x + 9 – 9)
⇒ y = -3(x – 3)2 + 27
⇒ 3(x – 3)2 = 27 – y
⇒ (x – 3)2
= $$-\frac{1}{3}$$(y – 27)
which represents a downward parabola with vertex (3, 27). Curve (1) meets x-axis at y = 0
∴18 x – 3 x2 =0
⇒ 3 x(6 – x) = 0
⇒ x = 0, 6
i.e. at points (0, 0) and Q(6, 0)
The line x = 4 meets curve (1)
When y = 72 – 48 = 24 i.e. at P(4, 24)

A1 = area of shaded region OAPRO
= $$\int_0^4$$ y d x
= $$\int_0^4$$ (18 x – 3 x2 ) d x
= 9 x2 – x3]0 4
= 144 – 64
= 80 sq. units
A2 = area of shaded region PQRP
= $$\int_4^6$$ (18 x – 3 x2) d x
= 9 x2 – x3]46
= (9 × 36 – 216) – (144 – 64) = 108 – 80
= 28 sq. units

Question 19.
Draw the rough sketch of y2 + 1 = x, x < 2 and find the area enclosed by the curve and the line x = 2.
Given equation of curve be y2 + 1 = x
⇒ y2 = x – 1 Clearly it represents a parabola (right handed) with vertex (1, 0) and does not meeting y-axis at any point.
Further the given curve meets the line x = 2 at y2 + 1 = 2
⇒ y = ± 1 i.e. at points (2, + 1) and (2, – 1).

Clearly the given curve is symmetrical about x-axis.
∴ required area = 2 × area of region enclosed by parabola and line x = 2 in first quadrant.
Divide this region R into vertical strips with lower end on x axis and upper end on y = $$\sqrt{x-1}$$ and corresponding rectangle move from x = 1 and x = 2
∴ required area = 2 $$\int_1^2$$ |y| d y = 2 $$\int_1^2$$ y d x
= 2 $$\int_1^2$$ $$\sqrt{x-1}$$ d x
= 2 $$\frac{(x-1)^{3 / 2}}{3 / 2}$$]21
= $$\frac{4}{3}$$[1 – 0]
= $$\frac{4}{3}$$ square units.
[∵ y ≥ 0 ∴ |y| = y]

Question 20.
Find the area bounded by the curve y = 4 – x2 and the line y = 0 and y = 3.

We want to find the area of region bounded by curves given below:
and
y = 4 – x2
y = 0, y = 3
eqn. (1) represents a parabola (downward) with vertex (0,4). y = 4 – x2 meets y = 3 when 3 = 4 – x2
⇒ x2 = 1
⇒ x = ± 1
∴ points of intersection are ( ± 1,3).
∴ required area = 2 × area of region OABCO
Divide the region into small horizontal strips. Each strip has left end on y-axis and right end on curve y = 4 – x2.
∴ required area = 2 $$\int_0^3$$ x d y
= 2 $$\int_0^3$$ $$\sqrt{4-y}$$ d y
= 2 $$\frac{(4-y)^{3 / 2}}{-3 / 2}$$]0 3
= $$-\frac{4}{3}$$[1 – 8]
= $$\frac{28}{3}$$ sq. units

Question 21.
Make a rough sketch of the following curves. Also, find the area enclosed between the curves and the axes.
(i) y = cos x, 0 ≤ x ≤ $$\frac{\pi}{2}$$
(ii) y = cos 2 x, 0 ≤ x ≤ $$\frac{\pi}{4}$$
(iii) y = sin x, 0 ≤ x ≤ $$\frac{\pi}{2}$$
(iv) y = cos ^2 x, 0 ≤ x ≤ $$\frac{\pi}{2}$$
(i) Given eqn. of curve be y = cos x ; 0 ≤ x ≤ $$\frac{\pi}{2}$$
Table of values is given as under:

 x 0 $$\frac{\pi}{2}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ y 1 $$\frac{1}{2}$$ 0 0

∴ required area = $$\int_0^{\pi / 2}$$ y d x
= $$\int_0^{\pi / 2}$$ cos x d x
= sin x]π/2 0
= sin $$\frac{\pi}{2}$$ – sin 0
= 1 – 0 = 1 sq. units

(ii) Given eqn. of curve be
y = cos 2 x ; 0 ≤ x ≤ $$\frac{\pi}{4}$$

 x 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ y 1 $$\frac{1}{2}$$ 0

∴ required area = $$\int_0^{\pi / 4}$$ y d x
= $$\int_0^{\pi / 4}$$ cos 2 x d x
= $$\frac{\sin 2 x}{2}$$]π/4 0
= $$\frac{1}{2}$$ [sin $$\frac{\pi}{2}$$ – 0]
= $$\frac{1}{2}$$ sq. units

(iii) Equation of given curve be
y = sin x ; 0 ≤ x ≤ $$\frac{\pi}{2}$$

 x 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ y 0 $$\frac{1}{2}$$ $$\frac{1}{\sqrt{2}}$$ $$\frac{\sqrt{3}}{2}$$ 1

∴ required area = $$\int_0^{\pi / 2}$$ y dx = $$\int_0^{\pi / 2}$$ sin x d x
= -cos x]π/2 0
= -[0 – 1]
= 1 sq. units

(iv) Given curve is y = cos2 x
For rough sketch we construct a table of values as under :

 x 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ y 0 $$\frac{3}{4}$$ $$\frac{1}{2}$$ $$\frac{1}{4}$$ 1

∴ Required area = $$\int_0^{\pi / 2}$$ y d x [using vertical strips]
= $$\int_0^{\pi / 2}$$ cos2 x d x
= $$\frac{1}{2}$$ $$\int_0^{\pi / 2}$$ (1 + cos 2 x) d x
= $$\frac{1}{2}$$ [x + $$\frac{\sin 2 x}{2}$$]π/2 0
= $$\frac{\pi}{4}$$ sq. units

Question 22.
Draw a rough sketch of the curve y = cos2 x in [0, π] and find the area enclosed by the the lines x = 0, x = π and the x-axis.
Given eqn. of curve be y = cos2 x in [0, π]

 x 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ $$\frac{2\pi}{3}$$ π y 1 $$\frac{3}{4}$$ $$\frac{1}{2}$$ $$\frac{1}{4}$$ 0 $$\frac{1}{4}$$ 1

∴ Required area = $$\int_0^{\pi / 2}$$ cos2 x + $$\int_{\pi / 2}^\pi$$ cos2 x d x
= $$\int_0^\pi$$ cos2 x d x
= $$\int_0^\pi$$ $$\frac{1+\cos 2 x}{2}$$ d x
= $$\frac{1}{2}$$ [x + $$\frac{\sin 2 x}{2}$$]π 0
= $$\frac{1}{2}$$[π + 0 – 0 – 0]
= $$\frac{\pi}{2}$$ sq. units

Question 23.
Draw a rough sketch of the curve y = $$\frac{1}{2}$$π + 2 sin2 x and find the area between the x-axis, the curve and the ordinates x = 0 and x = π
Given eqn. of curve be y = $$\frac{\pi}{2}$$ + 2 sin2 x

 x 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ $$\frac{2\pi}{3}$$ $$\frac{3\pi}{4}$$ $$\frac{5\pi}{6}$$ π y $$\frac{\pi}{2}$$ $$\frac{\pi}{2}$$+$$\frac{1}{2}$$ $$\frac{\pi}{2}$$+1 $$\frac{\pi}{2}$$+$$\frac{3}{2}$$ 3.57 3.07 2.57 2.07 1.57

The rough sketch for given curve is given as under.
Divide the region into vertical strips. Each vertical strip has lower end on x-axis and upper end on given curve.
Thus the corresponding rectangle has length |y| and width d x and this rectangle move from x = 0 and x = π.
∴ Required area = $$\int_0^\pi$$ |y| d x = $$\int_0^\pi$$ y d x
[∵ y ≥ 0 ∴ |y| = y]
= $$\int_0^\pi$$ $$\frac{\pi}{2}$$ + 2 sin2 x] d x
= $$\frac{\pi}{2}$$ x]π 0 + $$\int_0^\pi$$ 2[ $$\frac{1-\cos 2 x}{2}$$] d x
= $$\frac{\pi}{2}$$(π – 0) + [x – $$\frac{\sin 2 x}{2}$$]π 0
= $$\frac{\pi^2}{2}$$ + [π – 0 – 0 – 0]
= $$\frac{\pi^2}{2}$$ + π
= $$\frac{\pi}{2}$$ (π + 2) sq. units

Question 24.
Show that the area included between the x-axis and the curve a2 y = x2 (x + a) is $$\frac{a^2}{12}$$
Clearly the curve a2 y = x2 (x + a)
meets x-axis i.e. y = 0, where x2 (x + a) = 0
⇒ x = 0,-a
∴ curve meets x-axis at (0,0) and (-a, 0).
∴ required area
= $$\int_{-a}^0$$ y d x
= $$\int_{-a}^0$$ $$\frac{1}{a^2}$$ (x3 + a x2) d x
= $$\frac{1}{a^2}$$ $$\frac{x^4}{4}$$ + $$\frac{a x^3}{3}$$]0 a
= $$\frac{1}{a^2}$$ [0 + 0 – $$\frac{a^4}{4}$$ + $$\frac{a^4}{3}$$]
= $$\frac{a^2}{12}$$ sq. units

Question 24.
Show that the area included between the x-axis and the curve a2 y = x2 (x + a) is $$\frac{a^2}{12}$$
Clearly the curve a2 y = x2 (x + a)
meets x-axis i.e. y = 0, where x2 (x + a) = 0
⇒ x = 0,-a
∴ curve meets x-axis at (0,0) and (-a, 0).
∴ required area
= $$\int_{-a}^0$$ y d x
= $$\int_{-a}^0$$ $$\frac{1}{a^2}$$ (x3 + a x2) d x
= $$\frac{1}{a^2}$$ $$\frac{x^4}{4}$$ + $$\frac{a x^3}{3}$$]0 a
= $$\frac{1}{a^2}$$ [0 + 0 – $$\frac{a^4}{4}$$ + $$\frac{a^4}{3}$$]
= $$\frac{a^2}{12}$$ sq. units

Question 25.
(i) Calculate the area bounded by the curve y = x2 – 1, the x-axis and the line y = 8.
(ii). Calculate the approximate increase in this area, if the line y = 8 is changed to y = 8.01.
(i) The given curve y = x2 – 1
⇒ x2 = y + 1
it represents an upward parabola with vertex (0, -1) and meets x-axis at y = 0.
∴ from (1), x2
= 1 ⇒ x = ± 1
i.e. at points ( ± 1, 0).
The line y = 8 meets the curve (1) when x2 = 9
⇒ x = ± 3 i.e. at points ( ± 3, 8).
∴ Required area = 2 $$\int_0^8$$ x d y
= 2 $$\int_0^8$$ $$\sqrt{y+1}$$ d y
= 2 $$\frac{(y+1)^{3 / 2}}{3 / 2}$$]0 8
= $$\frac{4}{3}$$ [93/2 – 1]
= $$\frac{4}{3}$$ × 26
= $$\frac{104}{3}$$ sq. units

(ii) Thus required area = 2 $$\int_8^{8.01}$$ x d y
= 2 $$\int_8^{8.01}$$ $$\sqrt{y+1}$$ d y
= 2 $$\frac{(y+1)^{3 / 2}}{3 / 2}$$]8 0.88
= $$\frac{4}{3}$$[(9.01)3/2 – 93/2 ]
= $$\frac{4}{3}$$[27.0450 – 27]
= 0.06 sq. units

Question 26.
Find the area of the region bounded by y = -1, y = 2, x = y3 and x = 0.
The given curve be x = y3 and lines are y = -1 ; y = 2 and x = 0
The line y = 2 intersects the curve x = y3 at (8, 2)
and y = -1 meets curve at the point (-1, -1).
∴ Required area = $$\int_0^2$$ x d y + $$\int_{-1}^0$$ x d y
= $$\int_0^2 y^3$$ d y + $$\int_{-1}^0$$ (0 – y3) d y
= $$\frac{y^4}{4}$$]_0^2 – $$\frac{y^4}{4}$$]-1 2
= 4 – $$\frac{1}{4}$$ [0 – 1]
= 4 + $$\frac{1}{4}$$
= $$\frac{19}{4}$$ sq. units

Question 27.
Find the area bounded by x = a t2 , y = 2 a t between the ordinates corresponding to t = 1 and t = 2.
∴ Required area = 2 $$\int_a^{4 a}$$|y| d x
= 2 $$\int_a^{4 a}$$ 2 $$\sqrt{a x}$$ d x [∵ y ≥ 0]
= 4 $$\sqrt{a}$$ $$\frac{x^{3 / 2}}{3 / 2}$$]a 4a
= $$\frac{8 \sqrt{a}}{3}$$ [(4 a)3/2 – a3/2 ]
= $$\frac{8 \sqrt{a}}{3}$$[8 – 1] a3/2
= $$\frac{56}{3}$$ a2 sq. units