OP Malhotra Class 12 Maths Solutions Chapter 25 Application of Integrals Ex 25(b)

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

Question 1.
Find the area of the region enclosed by the following curves or lines.
(a) y = 2x, y = x ²
(b) y² = 8x, x² = 8y
(c) y² = x,x² = y
Answer:
(a) The eqns. of given curves are:
y = 2x
and y = x²
eqn. (2) represents an upward parabola with vertex (0,0)
and the line (2) and intersects parabola when 2x = x² =
=> x(x-2) = 0
=> x = 0, 2
When x = 0 ∴ from (1); y = 0
When x = 2 ∴ from (1); y = 4
i. e. points of intersection are (0, 0) and (2, 4).
∴ Required area of shaded region
= \(\int_0^2\) y of line (1) – y of parabola (2) d x
= \(\int_0^2\) 2 x d x – \(\int_0^2\) x² dx
= x²]⁰₂ – \(\frac{x^3}{3}\)₀ ²
= 4 – \(\frac{8}{3}\)
= \(\frac{4}{3}\) sq.units

(b) Given eqns. of parabolas are
x² = 8y …..(1)
and y² = 8x …(2)
both parabolas intersects when \((\frac{x^2}{8})^2\) = 8x
=> x⁴ – 512x = 0
=> x(x³ – 512) = 0
=> x = 0, 8
∴ from (1); y = 0, 8
Thus the points of intersection of both curves are (0, 0) and (8, 8). Divide the region into vertical strips with lower end on x2 = 8y and upper end on y2 = 8* and this rectangle move from x = 0 to x = 8

(c) The given eqns. of curves are;
y² = x
and x² = y
eqn. (1) represents a right handed parabola with vertex (0, 0). eqn. (2) represents an upward parabola with vertex (0, 0). Both curves intersects when x⁴ = x
=> x(x³ – 1) = 0
=> x = 0, 1
when x = 0 ∴ from (2) ; y = 0
when x = 1 ∴ from (2) ; y = 1
∴ both curves intersects at (0, 0) and (1,1) Divide the shaded region into vertical strips. Each vertical strip has lower end on curve (1) and upper end on curve (1).
∴ required area = \(\int_0^1\)(\(\sqrt{x}\) – x²) dx
= \(\frac{2}{3} x^{3 / 2}\) – \(\frac{x^3}{3}\)]⁰₁
= \(\frac{2}{3}\) – \(\frac{1}{3}\)
= \(\frac{1}{3}\) sq. units

Question 2.
Find the area of the figure bounded by the graphs of the functions y = x², y = 2x – x².
eqn. (1) represents an upward parabola.
Answer:
Given eqns of curves are y = x² ….. (1)
and y = 2x – x² ……. (2)
eqn. (1) represents an upward parabola.
With vertex at (0, 0).
eqn. (2) can be written as x² – 2x = -y
x² – 2x + l – l = -y
=> (x- l)² = – (y – 1)
Clearly represents a downward parabola with vertex at (1, 1).
Both parabolas (1) and (2) intersects when x² = 2x – x².
2x² – 2x = 0
=> 2x (x – 1) = 0
=> x = 0, 1
When x = 0 from(l) ; y = 0
When x = 1 from (1) ; y = 1
i.e. points of intersection are (0, 0) and (1,1). Divide the shaded region into vertical strips. Each vertical strip is having lower end on curve (1) and upper end on curve (2).
∴ required area = \(\int_0^1\) [-x² + (2x – x²)] dx
= \(\int_0^1\) (2 x – 2 x²) d x
= x² – \(\frac{2}{3}\) x³]₀ ¹
= 1 – \(\frac{2}{3}\)
= \(\frac{1}{3}\) sq. units

Question 3.
(i) Find the area bounded by the curve x² = 4y and the straight line x = 4y – 2.
(ii) Find the area cut off from the para- bola 4y = 3 × 2 by the line 2y = 3x + 12.
(iii) Find the area of the region included between the parabola y² = x and the x + y = 2.
Answer:
(i) Given curve x² = 4y be an upward parabola with vertex (0,0). The given line x = 4y – 2 meets x-axis at (- 2, 0) and y-axis at (0, \(\frac{1}{2})\)
Both curves and line intersects when (4y – 2)² = 4y
=> 16y² – 16y + 4 = 4y
=> 16y² – 20y + 4 = 0
=> 4y² – 5y + 1 = 0
=> (y-1)(4y-1) = 0
=> y = 1, \(\frac{1}{4}\)
When y = 1 ; x = 2 and
When y = \(\frac{1}{4}\)
∴ x = -1
Thus points of intersection are (2, 1) and
(-1, \(\frac{1}{4}\))

∴ required area = area of shaded region
= \(\int_{-1}^2\) \(\frac{x+2}{4}\) – \(\frac{x^2}{4}\) dx
= \(\frac{1}{4}\) \(\frac{x^2}{2}\) + 2x – \(\frac{x^3}{3}\)₁ ²
= \(\frac{1}{4}\) [2 + 4 – \(\frac{8}{3}\) – \(\frac{1}{2}\) + 2 – \(\frac{1}{3}\)
= \(\frac{9}{8}\) sq. units

(ii) The given curve be 4y = 3x²
=> x² = \( \frac{4}{3}\)y …….(1)
and eqn. of given line be 2y = 3x + 12 …(2)
Curve (1) represents on upward parabola with vertex at (0,0). Line (2) meets x-axis at (- 4,0) and (0, 6).
Thus line (2) meets parabola (1)
When x² = \(\frac{4}{3}\) \(\frac{3 x+12}{2}\)
=> x² = 2 (x + 4)
=> x² – 2x – 8 = 0
=> (x – 4) (x + 2) = 0
=> x= 2, -2
When x = 4 ∴ from (2) ; y = 12
When x = -2 from (2) ; y = 3
Divide the region into vertical strips with ever vertical strip is having lower end on curve (1) and upper end on curve (2).

∴ required shaded area = \(\int_{-2}^4\) \(\frac{3 x+12}{2}\) – \(\frac{3}{4}\) x² d x \(\frac{3}{2}\) \(\frac{(x+4)^2}{2}\) – \(\frac{1}{4}\) x³]_{3 4}
= \(\frac{3}{4}\) × 64 – \(\frac{1}{4}\) × 4³ – \(\frac{3}{4}\) × 2² + \(\frac{1}{4}\)(-2)³
= 48 – 16 – 3 – 2 = 27 sq. units

(iii) We want to find the area of the region included between given two curves
y² = x …(1) and
x + y = 2 ……. (2)
Now eqn. (1) is a parabola with axis x-axis and vertex (0, 0). eqn. (2) represents a line having intercepts on coordinate axes are (2, 0) and (0, 2). eqn. (1) and (2) intersects when y² = 2 – y
=> y² + y – 2= 0
=> y = 1,-2
When y= 1 => x = 2 – 1 = 1
When y = – 2 => x = 4
Thus the points of intersection are (1, 1) and (4, – 2).
We divide the shaded region into horizontal strips. Each horizontal strip is having left end on curve y2 = x and right end on line x + y = 2.
∴ the approximating rectangle was length |X¹ – x²| land width dy

∴ required area = \(\int_{-2}^1\) |x₁ – x₂| dy
= \(\int_{-2}^1\) [(2-y) – y²] d y
= \(\frac{(2-y)^2}{-2}\)]²₁ – \(\frac{y^3}{3}\)² ₁
= \(-\frac{1}{2}\)[1 – 16] – \(\frac{1}{3}\)[1-(-8)]
= + \(\frac{15}{2}\) – 3
= \(\frac{9}{2}\) sq. units

Question 4.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and .v = 3. Also, sketch the region bounded by these curves.
Answer:
Given curves are y = x² + 2 …(1)
y = x …(2)
and x = 0, x = 1
eqn. (1) represents a parabola with vertex (0, 2).
eqn. (1) and (2) intersects when x = x² + 2 => x² – x + 2 = 0
∴ x does not gives real values.
Thus the line y = x does not meet the parabola y = x² + 2.
Clearly x = 1 meets y = x² + 2 at (1,3).

Divide the shaded region into vertical strips. Each vertical strips has lower end on y = x and upper end on y = x² + 2.
length of approximating rectangle = |y² – y¹| and width dx.
Clearly it moves from x = 0 to x = 1
∴ required area = \(\int_0^1\) [x² + 2 – x] dx
= \(\frac{x^3}{3}\) + 2x – \(\frac{x^2}{2}\)⁰ ₁
= \(\frac{1}{3}\) + 2 – \(\frac{1}{2}\)
= \(\frac{2+12-3}{6}\)
= \(\frac{11}{6}\)sq.units

Question 5.
Find the point of intersection of the liney = 4x with the curve y = x³ If A is that point of intersection which lies in the first quadrant and O is the origin, calculate the area between line OA and the curve.
Answer:
The eqn. of given line be y = 4x …(1)
and eqn. of given curve be y = x³ …….(2)
both curves (1) and (2) intersects when 4x = x³
When x = 0 ∴ from (1); y – 0
When x = ± 2 ∴ from (1); y = ± 8
Since the point of intersection of given line and curve lies in the first quadrant.
∴ Coordinates of A are (2, 8).
Divide the shaded region into vertical strips. Each vertical strip is having lower end on curve (2) and upper end on line (1).

∴ required area of shaded region
= \(\int_0^2\) [4 x – x³] d x
= 2x³ – \(\frac{x^4}{4}\)₀ ²
= (8 – 4) = 4 sq. units

Question 6.
Using integration, find the area of the mangle, whose vertices are
(i) (-1,0), (1,3) and (3,2)
(ii) (2, 5), (4, 7) and (6,2)
Answer:
(i) LetA (-1,0),B (l,3)andC (3,2) are the vertices of ∆ABC
eqn. of line AB be given by, y – 0 = \(\frac{3-0}{1+1}\)(x + 1) => y = \(\frac{3}{2}\)(x + 1)
eqn. of line AB be given by, y – 0 = \(\frac{2-0}{4}\) => y = \(\frac{1}{2}\) (x – 1)
eqn. of line AB be given by, y – 3 = \(\frac{2-3}{3-1}\)(x – 1)
=> y – 3 = \(– \frac{1}{2}\) (x – 1)
=> = \(-\frac{x}{2}\) + \(\frac{7}{2}\)
∴ reqd. area = area of region ABDA + area of region BDCB
= \(\int_{-1}^1\) \(\frac{3}{2}\)(x + 1) – \(\frac{1}{2}\) (x + 1) d x + \(\int_1^3\)\(\frac{-x}{2}\) + \(\frac{7}{2}\) – \(\frac{1}{2}\) x – \(\frac{1}{2}\) dx
= \(\frac{(x+1)^2}{2}\)^-1+1 + \(\int_1^3\)-(x – 3) dx
= \(\frac{1}{2}\)[4 – 0] – \(\frac{1}{2}\)(x – 3)² ]^{1 3}
= 2 – \(\frac{1}{2}\)(0 – 4) = 4 sq. units.

(ii) Let the given vertices of ∆ABC are A (2, 5), B (4, 7) and C (6, 2)
eqn. of line AB be given by
y – 5 = \(\frac{2-7}{6-4}\)(x – 2)
=> y – 5 = x – 2
=> y = x + 3
eqn. of line BC be given by
y – 7 = \(\frac{2-7}{6-4}\)(x- 4)
=> y – 7 = –\(\frac{5}{2}\) (x- 4)
=> 2y – 14 = -5x + 20
=> 2y + 5x = 34

eqn. of line AC be given by
y – 5 = \(\frac{2-5}{6-2}\) (x – 2)
=> y – 5 = \(\frac{-3}{4}\) (x – 2)
=> 4y – 20 = -3x + 6
=> 4y + 3x = 26
∴ area of shaded ∆ABC = area of region AMNB + area of region BNLC area of region AMLC

Question 7.
Using integration, find the area of the region bounded by the triangle whose sides are y = 2x + l,y = 3x + l and x = 4.
Answer:
To find the area of triangular region bounded by curves given as under
y = 2x + 1 …..(1)
y = 3x + 1 …..(2)
and x = 4 …..(3)
On solving (1) and (2); we have 2x + 1 = 3x + 1
=> x = 0
∴ y = 1
Thus line (1) and (2) intersects at (0, 1). eqn. (1) and (3) intersects at (4, 13). and eqn. (1) and (3) intersect at (4, 9). required area = area of ∆ABC.
Divide the region into vertical strips. Each vertical strip has lower end on liney = 2x + 1 and upper end on line y = 3x + 1.
So length of approximating rectangle = |y₁ – y₂| and width = dx

and clearly it moves from 0 to 4.
∴ required area = \(\int_0^4\) |ysub>1 – ysub>1| dx
= \(\int_0^4\)[(3x + 1) – (2x + 1)][/latex] dx
[∵ y₁ ≥ y₂
= \(\frac{x^2}{2}\)⁰₄ = 8 sq. units

Question 8.
Find the area of the region enclosed between two circles x² + y² = 4 and (x- 2)² + y² = 4.
Answer:
The given circles are x² + y² = 4 …(1)
and (x – 2)² + y² = 4 …(2)
eqn. (1) represents a circle with centre at (0, 0) and radius 2 and eqn. (2) represents a circle with centre (2, 0) and radius 2.
Now eqn. (1) and (2) intersects
when (x – 2)² – x² = 0
i.e x² – 4x + 4 – x² = 0 i.e x = 1
∴ y = ±√3
∴ pts. of intersections are (1,±√3).
∴ Required area = 2 [area OACO + area ABCA]
= 2 \(\int_0^1\) \(\sqrt{4-(x-2)^2}\) d x + \(\int_1^2\) \(\sqrt{4-x^2}\) d x
[Region lies in Ist quadrant]

Question 9.
Sketch the region common to the circle x² + y² = 16 and the parabola X² = 6y. Also find the area of the region using integration.
Answer:
Given region be {(x, y): x² + y² ≥ 16 ; x² ≤ 6y}
given eqn. of circle be JC2 +y2 = 16 which represents a circle with centre (0, 0) and radius 4. eqn. of given parabola be x2 = 6y
which represents an upward parabola with vertex at (0, 0). both curves eqn. (1) and eqn. (2) intersects y² + 6y – 16 = 0
(y – 2) (y + 8) = 0
=> y = 2,-8
when y = 2 ∴ from (2); x = ±2√3
Thus both curves intersects at (±2√3, 2)
and y = -8 does not gives any real values of x required area = 2 x area of OABO

Question 10.
Find the area given by x + y ≤ 6, x² + y² <, 6y and y2 ≤ 8x.
Answer:
The eqns. of given curves are
x + y = 6 …(1)
x2² + y² = 6y …(2)
and y² = 8x

lines (1) meets coordinates axes at A (6, 0) and B (0, 6). eqn. (2) represents a circle with centre C (0, 3) and radius 3. eqn. (3) represents a right handed parabola with vertex at O (0, 0). line (1) and circle (2) intersects when (6 – y)² + y² = 6y
=> 2y² – 18y + 36 = 0
=> y² – 9y + 18 = 0
=> (y – 3) (y – 6) = 0
=> y = 3, 6
when y = 3 ∴ from (1) ; X = 3
when y = 6 ∴ from (1) ; X = 0
i.e. points of intersection are (2, 4) and (18, – 12)
Divide the region into horizontal strips.
∴ required area = area of region OFDO + area of region EFDE

Question 11.
Indicate the region bounded by the curve x² = y, y = x + 2 and X-axis and obtain the area enclosed by them.
Answer:
Given eqns. of curves are, x² = y …..(1)
and y = x + 2 …..(2)
eqn. (1) represents a upward parabola with vertex (0, 0).
The line (2) meets coordinate axes at (- 2, 0) and (0, 2).
eqn. (1) and (2) intersects when
x² – x – 2 = 0
(x – 2) (x + 1) = 0 => x = -l, 2
When x = – 1 ∴ from (2); y = 1
When x = 2 ∴ from (2); y = 4
∴ points of intersection are (- 1, 1) and (2, 4).
Divide the region into vertical strips with upper end on line and lower end on parabola.
The length of approximating rectangle = |y² – y¹| and width = dx
Clearly the rectangle moves from x = – 1 to x = 2.

∴ required area = \(\int_{-1}^2\) |y₂ – y₁| d x
= \(\int_{-1}^2\) (y₂ – y₁) d x
= \(\int_{-1}^2\) [x + 2 – x²] d x
= [\(-\frac{x^2}{2}\) + 2 x – \(\frac{x^3}{3}\)]² ₁
= [2 + 4 – \(\frac{8}{3}\) – \(\frac{1}{2}\) + 2 – \(\frac{1}{3}\)]
= 5 – \(\frac{1}{2}\)
= \(\frac{9}{2}\) sq. units

Question 12.
Using integration, find the area of the triangle formed by positive x-axis and tangent and normal to the circle x² + y² = 4 at (1, \(\sqrt{3}\)).
Answer:
Given eqn. of circle be x² + y² = 4 with centre (0, 0) and radius 2 . Diff. given eqn. w.r.t. x; we have
2 x + 2 y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(-\frac{x}{y}\)

∴ slope of tangent to given circle at (1, \(\sqrt{3}\))
= \(\frac{-1}{\sqrt{3}}\)
Thus, eqn. of tangent to given circle at (1, \(\sqrt{3}\)) is given by
y – \(\sqrt{3}\) = \(-\frac{1}{\sqrt{3}}\)(x – 1)
⇒ x + \(\sqrt{3}\) y = 4 …………..(1)
∴ slope of normal to given circle at (1, \(\sqrt{3}\))
= \(\sqrt{3}\)
∴ eqn. of normal to given circle at point (1, \(\sqrt{3}\)) be given by
y – \(\sqrt{3}\) = \(\sqrt{3}\)(x – 1)
⇒ \(\sqrt{3}\) x – y = 0
line (1) meets coordinate axes at (4, 0) and (0, \(\frac{4}{\sqrt{3}}\)).
∴ required area = area of region OACO + area of region ACBA
= \(\int_0^1\) \(\sqrt{3}\) x d x + \(\int_1^4\) \(\frac{4-x}{\sqrt{3}}\) dx
= \(\sqrt{3}\) \(\frac{x^2}{2}\)]¹ ₀ + \(\frac{1}{\sqrt{3}}\) \(\frac{(4-x)^2}{-2}\)]⁴ ₁
= \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2 \sqrt{3}}\)[0 – 9]
= \(\frac{\sqrt{3}}{2}\) + \(\frac{9}{2 \sqrt{3}}\)
= \(\frac{\sqrt{3}}{2}\) + \(\frac{3 \sqrt{3}}{2}\)
= \(\frac{4 \sqrt{3}}{2}\)
= 2 \(\sqrt{3}\) sq. units

examples:

Question 1.
Calculate the area of the figure bounded by the curvey = log x, the straight line x = 2 and the .Y-axis,
Answer:
Given eqn. of given curve y = log x
The curve meets .v-axis i.e. y = 0
∴ loge x – 0
=> x = e⁰ = 1
Thus required area = \(\int_1^2\) y dx = \(\int_1^2\) log x dx = \(\int_1^2\) log x . 1 dx
= logx – x]² ₁ – \(\int_1^2\) \(\frac{1}{x}\) xdx
= xlogx – x]² ₁
= (2 log 2 – 2) – (1 log 1 – 1) = 2 log 2 – 2 – 0 + 1
= 2 log 2 – 1
= log 2² – log e
= log \(\frac{4}{e}\) sq units

Question 2.
Find the area of the figure bounded by the graphs of the function y = x² and y = 2x – x².
Answer:
Given eqns. of curves are y = x² …(1)
and y = 2x – x² …(2)
eqn. (1) represents an upward parabola.
With vertex at (0, 0). eqn. (2) can be written as x² – 2x = -y
x² – 2x + 1 – 1 = – y
=> (x – 1)² = -(y – 1)
Clearly represents a downward parabola with vertex at (1,1).
Both parabolas (1) and (2) intersects when x² = 2x – x²
2x² – 2x = 0
=> 2x(x – 1) = 0 => x = 0, 1

When x = 0 .’. from(1); y = 0
When x = 1 from (1); y = 1
i.e. points of intersection are (0, 0) and (1, 1).
Divide the shaded region into vertical strips.
Each vertical strip is having lower end on curve (1) and upper end on curve (2).
∴ required area = \(\int_0^1\) [-x² + (2x – x²)] dx
= \(\int_0^1\) (2x – x²)dx
= x² – \(\frac{2}{3}\)x³]¹ ₀
= 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\) sq.units

Question 3.
Show that the area ihehwled between the .Y-axis and the curve a²y = x²(x + a) is \(\frac{a² }{12}\)
Answer:
Clearly the curve a²y = 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
= \(\frac{1}{a^2 }\) [ \(\frac{x^4}{4}\) + \(\frac{a x^3}{3}\)]^-a ₀
= \(\frac{1}{a^2}\) [0 + 0 – \(\frac{a^4}{4}\) + \(\frac{a^4}{3}\)
= \(\frac{a^2 }{12}\) sq. units

Question 4.
Sketch and shade the area of the region lying in the first quadrant and bounded by y = 9 x², x = 0, y = 1 and y = 4. Find the area of the shaded region.
Answer:
Given eqn. of curve be y = 9 x² which represents an upward parabola with vertex at origin (0, 0). Divided the shaded region into horizontal strips. Each horizontal strip is having left end on y-axis and right end on given curve.
∴ Required area = \(\int_1^4\) x d y
= \(\int_1^4\) \(\frac{\sqrt{y}}{3}\) d y
= \(\frac{1}{3}\) \(\frac{y^{3 / 2}}{3 / 2}\)]¹ ₄
= \(\frac{2}{9}\)[8 – 1]
= \(\frac{14}{9}\) sq. units

Question 5.
Calculate the area bounded by the curve y = x(2 – x) and the lines x = 0, y = 0, x = 2.
Answer:
The eqn. of given curve be y = x(2 – x)
⇒ y = 2x – x² = -(x² – 2 x + 1 – 1)
⇒ y = -(x – 1)² + 1
⇒ (x – 1)² = -(y – 1)
which represents a downward parabola with vertex (1, 1). eqn. (1) meets x-axis i.e. y = 0
⇒ x = 0, 2 i.e. at points (0, 0) and (2, 0)
divide the shaded region into vertical strips. Each vertical strip is having lower end on x-axis and upper end on given curve 1 .
∴ required area = \(\int_0^2 \) y d x
= \(\int_0^2 \) (2 x – x²) d x
= x² – \(\frac{x^3}{3}\)]²₀
= 4 – \(\frac{8}{3}\)
= \(\frac{4}{3}\) sq. units

Question 6.
Find the area enclosed by the curves y² = x and y² = 4 – 3 x.
Answer:
The equations of given curves are
y² = x
y² = 4 – 3 x
and
eqn. (1) represents a right handed parabola with vertex O(0, 0).
eqn. (2) can be written as; y² = -3(x – \(\frac{4}{3}\))
which represents a left handed parabola with vertex at (\(\frac{4}{3}\), 0).
Both parabolas (1) and (2) intersects when 4 – 3 x = x
⇒ x = 1
∴ from (1); y² = 1
⇒ y = 1
Thus their points of intersections are (1, ± 1).
required area = 2 [area of region OABO]
= 2\(\int_0^1\) \(\sqrt{x}\) d x + \(\int_1^{4 / 3}\) \(\sqrt{4-3 x}\) d x
= 2\(\frac{2 x^{3 / 2}}{3}\)}¹
₀ + \(\frac{2}{3}\) \(\frac{(4 – 3 x)^{3 / 2}}{-3}\)^4/3 ₁
= 2\(\frac{2}{3}\)(1 – 0) – \(\frac{2}{9}\)(0 – 1)
= 2\(\frac{2}{3}\) + \(\frac{2}{9}\)
= \(\frac{16}{9}\) sq. units

Question 7.
Draw a rough sketch of the curve y² + 1 = x, x<2. Find the area enclosed by the curve and the line x = 2.
Answer:
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
[∵ y ≥ 0 ∴ |y| = y]
∴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 \(\frac{(x-1)^{3 / 2}}{3 / 2}\)¹ ₂
= \(\frac{4}{3}\)[1 – 0] = \(\frac{4}{3}\)

Question 8.
Draw a rough sketch of the curve x² + y = 9 and find the area enclosed by the curve, the x axis and the lines x + 1 = 0 and x – 2 = 0.
Answer:
Given eqn. of curve be x² + y = 9
⇒ x² = -(y – 9)
which represents a downward parabola with vertex (0, 9).

The given lines x + 1 = 0 i.e. x = -1 and x – 2 = 0 i.e. x = 2
Further given curve meets x-axis at (± 3, 0)
∴ Required area
= \(\int_{-1}^2\) y d x
= \(\int_{-1}^2\) (9 – x²) d x
= 9 x – \(\frac{x^3}{3}\)]²₁
= (18 – \(\frac{8}{3}\)) – (-9 + \(\frac{1}{3}\))
= 18 – \(\frac{8}{3}\) + 9 – \(\frac{1}{3}\)
= 27 – 3 = 24 sq. units

Question 9.
Draw a rough sketch of the curve y = x² – 5 x + 6 and find the area bounded by the curve and the x-axis.
Answer:
Given eqn. of curve be y = x² – 5 x + 6
i.e. y = x² – 5 x + \(\frac{25}{4}\) – \(\frac{25}{4}\) + 6
⇒ y = (x – \(\frac{5}{2}\))² – \(\frac{1}{4}\)
⇒ t(x – \(\frac{5}{2}\))²
= y + \(\frac{1}{4}\)
which represents an upward parabola with vertex (\(\frac{5}{2}\),- \(\frac{1}{4}\)).

The given curve meets x-axis i.e. y = 0
∴ from (1); we have x² – 5 x + 6 = 0
⇒ x = 2,3 i.e. at points (2, 0) and (3, 0)
and given curve meets y-axis at (0, 0)
∴ required area = \(\int_2^3\) |y| d x
= \(\int_2^3\) – (x² – 5 x + 6) d x
= –[\(\frac{x^3}{3}\) – \(\frac{5 x^2}{2}\) + 6 x]² ₃
= [9 – \(\frac{45}{2}\) + 18 – \(\frac{8}{3}\) + 10 – 12]
= -[25 – \(\frac{45}{2}\) – \(\frac{8}{3}\)]
= –\(\frac{150-135-16}{6}\)
= \(\frac{1}{6}\) sq. units

Question 10.
Draw a rough sketch of the curves y = (x – 1)² and y = |x – 1|. Hence, find the area of the region bounded by these curves.
Answer:
Given eqns. of curves are ; y = (x – 1)²
and
y = |x – 1|
eqn. (1) represents an upward parabola with vertex (1,0) and this parabola meets y-axis at (0,1).
Now y = |x – 1| = {x-1 ; x ≥ 1 -(x – 1) ; x < 1}.
Clearly the lines y = x – 1 meets the coordinate axes at (1, 0) and (0, -1).
and The line y = -x + 1 meets coordinate axes at (1, 0) and (0, 1).
Clearly the line y = x – 1 and y = (x – 1)²intersects when x – 1 = (x – 1)²
⇒ (x – 1)(x – 2) = 0
⇒ x = 1, 2
i.e. y = 0, 1 ; i.e. at points (1, 0) and (2, 1)
The line y = -(x – 1) meets the curve y = (x – 1)² when -(x – 1) = (x – 1)²
⇒ x = 1, 0
i.e. y = 0, 1 ie. at points (0, 1) and (1, 0).

Clearly required area = area of shaded region
= 2 \(\int_1^2\) [(x – 1) – (x + 1)²] d x
= 2\(\frac{(x-1)^2}{2}\) – \(\frac{(x-1)^3}{3}\)]₁₂
= 2\(\frac{1}{2}\) – \(\frac{1}{3}\) – 0 + 0
= \(\frac{1}{3}\) sq. units

Question 11.
Find the area of the region bounded by the curve x = 4 y – y² and the y-axis.
Answer:
Given eqn. of curve be
⇒ x = -(y¹ – 4 y + 4 – 4)
⇒ x = -(y – 2)¹ + 4
⇒ (y – 2)¹ = -(x – 4)
which represents a lift handed parabola with vertex (4, 2). Clearly eqn. (1) passes through O(0, 0). eqn. (1) meets y-axis at x = 0
∴ 4 y – y¹ = 0
⇒ y(4 – y) = 0
⇒ y = 0, 4
i.e. at points (0, 0) and (0, 4)

Divide the shaded region into horizontal strips.
Each horizontal strip is having left end on y-axis and right end on given curve.
∴ Required area = \(\int_0^4\) x d y.
= \(\int_0^4\) (4 y – y¹) d y
= \(\frac{4 y^2}{2}\) – \(\frac{y^3}{3}\)]₀ ⁴
= 32 – \(\frac{64}{3}\)
= \(\frac{32}{3}\) sq. units

Question 12.
Find the area bounded by the curve y = 2 x – x² , and the line y = x.
Answer:
Given eqn. of curve be y = 2x – x²
and eqn. of given line be y = x
eqn. (1) can be written as y = -(x² – 2 x + 1 – 1)
⇒ y = -(x – 1)² + 1
⇒ (x – 1)² = -(y – 1)
which represents a downward parabola with vertex at (1, 1). Given curve (1) meets x-axis at y = 0
∴ 2 x – x^2=0
⇒ x = 0, 2
i.e. at points (0, 0) and (2, 0).
Line (2) meets parabola (1) when x = 2 x – x²
⇒ x² = x
⇒ x = 0, 1
∴ from (2); y = 0, 1
i.e. both curves intersects at (0, 0) and (1, 1).
Divide the shaded region into vertical strips. Each strip is having upper end on curve (1) and lower end on line (2).

∴ Required area = \(\int_0^1\) [2x – x² – x] d x .
= \(\int_0^1\)(x – x²) d x
= \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\)_{0 1}
= \(\frac{1}{2}\) – \(\frac{1}{3}\)
= \(\frac{1}{6}\) sq. units

Question 13.
Find the smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2.
Answer:
Given eqn. of circle be
x² + y² = 4
Now eqn. (1) represents a circle with centre (0, 0) and radius 2
and eqn. of given line be x + y = 2 meets coordinates axes at (2, 0) and (0, 2).
∴ required area = area of shaded region ABCA
Divide the region into vertical strips with lower end on line x + y = 2 and upper end on x² + y² = 4. Clearly the rectangle move from x = 0 to x = 2

∴ required area = \(\int_0^2\) \(\sqrt{4-x^2}\) – (2 – x) d x
= \(\frac{x \sqrt{4-x^2}}{2}\) + \(\frac{4}{2}\) sin^-1 \(\frac{x}{2}\)₀ ² – \(\frac{(2-x)^2}{-2}\)² ₀
= [0 + 2 sin^-1(1) – 0 – 0] + \(\frac{1}{2}\)[0 – 4]
= 2 × \(\frac{\pi}{2}\) – 2
= (π- 2) sq. units
∴ Ans. (b)

Question 14.
Find the area of the region bounded by the curves y = 6x – x² and y = x² – 2 x.
Answer:
The given eqns. of curves are
and
y = 6x – x²
y = x² – 2 x
eqn. (1) can be written as y = -(x² – 6 x + 9 – 9)
⇒ y = -(x – 3)² + 9
⇒ (x – 3)² = -(y – 9)
which represents a downward parabola with vertex at (3, 9) and meets x-axis at y = 0
∴ from (1); 6 x – x² = 0
⇒ x(6 – x) = 0 ⇒ x = 0, 6
i.e. curve (1) meets x-axis at (0, 0) and (6, 0) eqn. (2) can be written as;
y = x² – 2 x
⇒ y = x² – 2 x + 1 – 1
⇒ y = (x – 1)² – 1
⇒ (x – 1)² = y + 1
which represent an upward parabola with vertex (1, -1) and meets x-axis at 0 = x² – 2 x i.e. x = 0, 2
i.e. curve (2) meets x-axis at (0, 0) and (2, 0)
both parabolas intersects when

6 x – x² = x² – 2 x
⇒ 2 x² – 8 x = 0 arrow x = 0, 4
When x = 0 ∴ from (1); y = 0
When x = 4 ∴ from (2); y = 24 – 16 = 8
Thus both parabolas intersects at (0, 0) and (4, 8).
Divide the region into vertical strips. Each vertical strip is having upper end on curve (1) and lower end on parabola (2).
∴ Required area = \(\int_0^4\) [(6 x – x²) – (x² – 2 x) d x
= \(\int_0^4\) (8 x – 2 x²) d x
= \(\frac{8 x^2}{2}\) – \(\frac{2 x^3}{3}\)₀⁴
= 4 × 4² – \(\frac{2}{3}\) × 4³
= 64 – \(\frac{128}{3}\)
= \(\frac{64}{3}\) sq. units