Application of Integrals Class 12 Maths Important Questions Chapter 8

Question 10.
Using integration, find the area of region bounded by the triangle whose vertices are (- 2, 1), (0, 4) and (2, 3). (Delhi 2017)
Answer:
4 sq units

Question 11.
Find the area bounded by the circle x² + y² = 16 and the line √3y = x in the first quadrant, using integration. (Delhi 2017)
Answer:
Given equations of circle is x² + y² = 16 and x = √3y.
⇒ y = 13√x represents a line through the origin.
The line y = 13√x intersects the circle
∴ x² + x23 = 16
⇒ 3x2+x23 = 16
⇒ 4x² = 48 ⇒ x² = 12 ⇒ x = ±2√3
when x = 2√3, then y = 23√3√ = 2

Required area (shaded region in first quadrant)
= (Area under the line y = 13√x from x = 0 to 2√3) + (Area under the circle from x = 2√3 to x = 4)

Question 12.
Using the method of integration, find the area of the ∆ABC, coordinates of whose vertices are A (4, 1), B(6, 6) and C (8, 4). (All India 2017)
Answer:
7 sq units

Question 13.
Find the area enclosed between the parabola 4y = 3x² and the straight line 3x – 2y + 12 = 0. (All India 2017)
Answer:
Given parabola is 4y = 3x² …… (i)
represent an upward parabola with vertex (0, 0) and equation of line is 2y = 3x + 12 …… (ii)

From Eqs. (j) and (ii), we get
2(3x + 12) = 3x²
⇒ 3x² – 6x – 24 = 0
⇒ x² – 2x – 8 = 0
⇒ (x – 4) (x + 2) = 0
⇒ x = 4, – 2
when x = 4, then y = 3×4+122 = 12 [from Eq. (ii)]
when x = – 2, then y = 3×(−2)+122 = 3
Thus, intersection points are (- 2, 3) and (4, 12).

Question 14.
Using integration, find the area of the region
{(x, y): x² + y² ≤ 2ax, y² ≥ ax;x, y ≥ 0}. (Delhi 2016)
Answer:
Given region is
{(x, y): x² + y² ≤ 2ax, y² ≥ ox; x, y ≥ 0}
Now, we have x² + y² ≤ 2ax
⇒ x² + y² – 2ax ≤ 0
⇒ x² – 2ax + a² + y² < a²
[adding a² both sides of inequality]
⇒ (x – a)² + y² < a²
which is the interior of the circle having centre (a, 0) and radius a.
Also, we have y²
which is the exterior of the parabola having vertex (0, 0) and axis is X-axis.
Now, let us find the intersection point of circle
(x – a)² + y² = a² and parabola y² = ax.
On substituting y² = ax in the given circle, we get
(x – a)² + ax = a²
⇒ x² + a² – 2ax + ax = a²
⇒ x² – ax = 0
⇒ x(x – a) = 0
⇒ x = 0, a
When x = 0, then
y² = 0 ⇒ y = 0
When x = a, then
y² = a² ⇒ y = ± a
So, the points of intersection are 0(0, 0), A (a, a) and B(a, -a). (1)
Now, draw the graph of given curve as shown below:

Clearly, the required area of region will lie in first
quadrant as x, y ≥ 0.
∴ Required area = Area of shaded region

Question 15.
Using integration, find the area of the triangular region whose vertices are (2, – 2), (4, 3) and (1, 2). (All India 2016)
Answer:
132 sq units

Question 16.
Using integration, find the area of the region bounded by the curves y = 4−x2−−−−−√, x² + y² – 4x = 0 and the x-axis. (Foreign 2016)
Answer:
Given equations of curves are
y = 4−x2−−−−−√ ……. (i)
and x² + y² – 4x = 0
Consider the curve
y = 4−x2−−−−−√ = y² = 4 – x²
⇒ x² + y² = 4 which represents a circle with centre (0, 0) and radius 2 units.
Now, consider the curve x² + y² – 4x = 0
⇒ (x – 2)² + y² = 4, which also represents a circle with centre (2, 0) and radius 2 units.
Now, let us sketch the graph of given curves and find their points of intersection.

On substituting the value of y from Eq. (i) in Eq. (ii), we get
x² + (4 – x²) – 4x = 0
⇒ 4 – 4x = 0
⇒ x = 1
On substituting x = 1 in Eq. (1). we get y = √3
Thus, the point of intersection is (1, √3).
Clearly, required area = Area of shaded region OABO

Question 17.
Using integration, find the area of the region in the first quadrant enclosed by the Y-axis, the line y = x and the circle x² + y² = 32. (Delhi 2015C)
Answer:
Given, equation of circle is x² + y² = 32
and equation of line is
y = x
Consider x² + y² = 32
⇒ x² + y² = (4√2)²
So, given circle has centre (0, 0) and radius 4√2 units.
Now, let us sketch the graph of given curves and find their points of intersection

On substituting y = x in Eq. (i), we get
2x² = 32 ⇒ x² = 16 ⇒ x = ± 4
Thus, the points of intersection are (4, 4) and (-4, -4).
So, given line and the circle intersect in the first quadrant at point A(4, 4) and the circle cut the y-axis at point B (0, 4√2).
Let us draw AM perpendicular to y-axis.
Clearly, required area
= Area of shaded region OABO

Question 18.
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, √3). (Delhi 2015)
Answer:
First, differentiate the given curve w.r.t. x and determine the value of dydx at (1, √3).
Find the equations of tangent and normal at
point(1, √3) by using formula y – y₁ = dydx(x – x₁) and y – y₁ = – 1dy/dx(x – x₁)
Further, plot the above lines on a graph paper and find the area by using integration.

Given equation of circle is
x² + y² = 4
On differentiating both sides of Eq. (i) w.r.t. x,
we get
2x + 2ydydx = 0

On putting y = 0 in Eq. (ii), we get
x + 0 = 4 ⇒ x = 4
∴ the tangent line x + √3y = 4 cuts the X-axis at A(4, 0).
∴ Required area = Area of shaded region OAB

Question 19.
Using integration, find the area of the region bounded by the line x – y + 2 = 0, the curve x = √y and Y-axis. (Foreign 2015)
Answer:
Given curves are
x – y + 2 = 0 ….. (i)
and x = √y ….. (ii)
Consider x = √y ⇒ x² = y, which represents the parabola whose vertex is (0, 0) and axis is Y-axis.
Now, the point of intersection of Eqs.(i) and (ii)
is given by x = x+2−−−−√
⇒ x² = x + 2
⇒ x² – x – 2 = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = – 1, 2
But x = – 1 does not satisfy the Eq. (ii).
∴ x = 2
Now, putting x = 2 in Eq. (ii), we get
2 = √y ⇒ y = 4
Hence, the point of intersection is (2, 4).
But actual equation of given parabola is x = √y, it means a semi-parabola which is on right side of Y – axis.
The graph of given curves are shown below:

Question 20.
Find the area of the region {(x, y): y² ≤ 4x, 4x² + 4y² ≤ 9}, using method of integration. (All India 2015C, 2013, 2008C: Delhi 2008C)
Answer:
First, find the intersection points of given curves and then draw a rough diagram to represent the required area. If it is symmetrical about X-axis or Y-axis, then we first find area of only one portion , from them and then required area is twice of that area.

Given curves are
y² = 4x …… (i)
and 4x² + 4y² = 9
⇒ x² + y² = 94 …… (ii)
Eq. (i) represents a parabola having vertex (0, 0) and axis is X-axis and Eq. (ii) represents a circle having centre (0, 0) and radius 32.
On substituting y² = 4x in Eq. (ii), we get
x² + 4x = 94
⇒ 4x² + 18x – 2x – 9 = 0
⇒ 2x(2x + 9) – 1 (2x + 9) = 0
⇒ (2x + 9) (2x – 1) = 0
⇒ x = 12,−92
On putting x = 12 in Eq. (i), we get
y = ±√2
At x = –92, y have imaginary values.
So, intersection points are p(12,2–√) and p(12,−2–√).
Now, the shaded region represents the required region as shown below:

∴ Required area = 2[Area of the region ORPO + Area of the region RAPR]

Question 21.
Using integration, find the area of the region in the first quadrant enclosed by the X-axis, the line y = x and the circle x² + y² = 18. (All India 2014C)
Answer:
9π4 sq units

Question 22.
Using integration, find the area of the region bounded by the curves y = |x + 1| + 1, x = – 3, x = 3 and y = 0. (Delhi 2014C)
Answer:
Given curves are

Eq. (ii) represents the line parallel to Y-axis and passes through the point (- 3, 0).
Eq. (iii) represents the line parallel to Y-axis and passes through the point (3, 0).
Eq. (iv) represent X-axis.
Now, Eqs. (i), (ii) (iii) and (iv) can be represented in graph as shown below:

Clearly, required area

Hence, the required area is 16 sq units.

Question 23.
Using integration, find the area of ∆ PQR, coordinates of whose vertices are P(2, 0), Q(4, 5) and R (6, 3). (All India 2014C)
Answer:
7 sq units

Question 24.
Using integration, find the area of the region bounded by the triangle whose vertices are (- 1, 2), (1, 5) and (3, 4). (All India 2014)
Answer:
4 sq units

Question 25.
Find the area of the smaller region bounded by the ellipse x29+y24 = 1 and the line x3+y2 = 1. (Foreign 2014)
Or
Using integration, find the area of the following region. (Delhi 2010)
{(x,y):x29+y24≤1≤x3+y2}
Answer:
Given equation of ellipse is
x29+y24 = 1
and equation of line is x3+y2 = 1
For the points of intersection of ellipse and line, put the value of x from Eq. (ii) in Eq. (i), we get
(1−y2)2+y24 = 1
⇒ 1 + y24 – y + y24 = 1
⇒ y² – 2y = 0
⇒ y = 0, 2
When y – 0, then x = 3 and point is A(3, 0).
When y = 2, then x = 0 and point is B{0, 2).

Question 26.
Using integration, find the area of the region bounded by the lines 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0. (All India 2014C; Foreign 2011; Delhi 2009)
Answer:
Given lines are
2x + y = 4 ……… (i)
3x – 2y = 6 ……… (ii)
and x – 3y = – 5 ……… (iii)
Clearly, the line 2x + y = 4 passes through the points (2, 0) and (0, 4), the line 3x – 2y = 6 passes through the points (2, 0) and (0, – 3) and the line x – 3y = – 5 passes through the points (-5, 0) and (0, 53).
Now, the region bounded by these lines is shown below:

On solving Eqs. (i) and (ii),
we get x = 2 and y = 0
So, lines 2x + y = 4 and 3x – 2y = 6 meet at the point C(2,0).
Again, solving Eqs. (ii) and (iii), we get
x = 4 and y = 3
So, lines 3x – 2y = 6 and x – 3y = – 5 meet at the point B(4, 3).
On solving Eqs. (iii) and (i), we get
x = 1 and y = 2
So, lines 2x + y = 4 and x – 3y = – 5 meet at the point A (1, 2).
Now, required area of ∆ABC
= Area of region ABNMA – (Area of ∆AMC + Area of ∆BCN)

Question 27.
Using integration, find the area bounded by the curve x² = 4y and the line x = 4y – 2. (Delhi 2014C, 2013, 2010)
Answer:
Given curves are
x² = 4y ……… (i)
and x = 4y – 2 …….. (ii)
Eq. (i) represents a parabola with vertex at origin and axis along positive direction of Y-axis. Eq. (ii) represents a straight line which meets the coordinate axes at (- 2, 0) and (0, 12) respectively. To find the points of intersection of the given parabola and the line, we solve Eqs. (i) and (ii), simultaneously.
On substituting x = 4y – 2 in Eq. (i), we get
(4y – 2)² = 4y
⇒ 16y² + 4 – 16y = 4y
⇒ 16y² – 20y + 4 = o
⇒ 4y² – 5y + 1 = 0
⇒ (4y – 1) (y – 1) = 0
⇒ y = 1, 14
On putting the values of y in Eq. (ii), we get x = 2, – 1
So, the points of intersection of the given parabola and the line are (2, 1) and (- 1, 1/4).

The region whose area is to be found out is shaded in figure.
∴ Required area, A is given by

Question 28.
Using integration, find the area of the region bounded by the curves y = x² and y = x. (Delhi 2013C)
Answer:
16 sq units

Question 29.
Find the area of the region {(x, y):y² ≤ 6ax and x² + y² ≤ 16a²}, using method of integration. (All India 2013)
Answer:
(43√3a² + 16a2π3) sq units.

Question 30.
Find the area of the region bounded by the parabola y = x2 and the line y = |x|. (All India 2013)
Or
Find the area of the region given by {(x, y): x² ≤ y ≤ |x|}. (Delhi 2011C; All India 2009, 2008C)
Answer:
Given curves x² = y ……. (i)
and y = |x| …… (ii)
From Eqs. (i) and (ii), we get
x² = |x|

Case I When x ≤ 0
Then, x² = – x
⇒ x(x + 1) = 0
∴ x = 0, – 1
On putting the values of x in Eq. (i), we get
y = 0, 1

Case II When x ≥ 0
Then, x² = x ⇒ x(x – 1) = 0
∴ x = 0, 1
On putting the values of x in Eq. (j), we get
y = 0, 1
So, both curves cut each other at points
A (-1, 1), 0(0, 0) and B(1, 1).
The graphs of given curves is shown below,clearly the shaded region is symmetrical about the Y-axis.

Now, area of region OPBO

Hence, required area = 2 × Area of region OPBO
[∵ region is symmetrical about Y-axis]
= 2 ×16 = 13 sq unit

Question 31.
Using integration, find the area of the circle x² + y² = 16, which is exterior to the parabola y² = 6x. (All India 2012C)
Answer:
Given, equation of circle is x² + y² = 16 ….. (i)
and equation of parabola is y² = 6x ……. (ii)
Clearly, the given circle has centre (0, 0) and radius 4 units and the given parabola has vertex (0, 0) and axis parallel to X-axis.
Now, let us sketch the graph of given curves and find their points of intersection.

On substituting y² = 6x in Eq. (i), we get
x² + 6x -16 = 0
⇒ (x + 8)(x – 2) = 0
⇒ x = – 8 or x = 2
Clearly, from Eq. (ii), when x = – 8, then y² = – 48, which is not possible. So, x ≠ – 8
∴ x = 2
Now, on substituting x = 2in Eq. (ii), we get
y² = 12 ⇒ y = ± 2√3
Thus, the points of intersection are (2, – 2√3) and (2, 2√3).
Clearly, required area = Area of shaded region
= Area of circle – Area of region OABCO
= π(4)² – 2(Area of region OBCO)

Question 32.
Using method of integration, find the area of region bounded by lines 3x – 2y + 1 = 0, 2x + 3y – 21 = 0 and x – 5y + 9 = 0. (Delhi 2012)
Answer:
132 sq units

Question 33.
Using integration, find the area of the region bounded by the lines 3x – y – 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0. (Delhi 2012)
Answer:
10 sq units

Question 34.
Using integration, find the area of the region bounded by the lines 5x – 2y – 10 = 0, x + y – 9 = 0 and 2x – 5y – 4 = 0. (Delhi 2012)
Answer:
212 sq units

Question 35.
Find the area of the region
{(x, y): x² + y² ≤ 4, x + y ≥ 2}. (All India 2012).
Answer:
Given region is {(x, y): x² + y² ≤ 4, x + y ≥ 2}.
The above region has a circle with equation
x² + y² = 4 … (i)
whose centre is (0, 0) and radius is 2, and line with equation
x + y = 2 …(ii)
Point of intersection is calculated as follows
x² + y² = 4
⇒ x² + (2 – x)² = 4 [from Eq. (ii)]
⇒ x² + 4 + x² – 4x = 4
⇒ 2x² – 4x = 0
⇒ 2x (x – 2) = 0
⇒ x = 0 or 2
When x = 0, then y = 2 – 0 = 2 and when x = 2, then y = 2 – 2 = 0
So, points of intersection are (0, 2) and (2, 0).
On drawing the graph, we get the shaded region as shown below:

Question 36.
Find the area of the region
{(x, y):(x² + y²) ≤ 1 ≤ x + y}. (All India 2011; Delhi 2010C)
Answer:
(π4−12) sq units

Question 37.
Sketch the graph of y = | x + 3 | and evaluate the area under the curve y = |x + 3| above X-axis and between x = – 6 to x = 0. (All India 2011 )
Answer:
First, we sketch the graph of
y = |x + 3|

So, we have y = x + 3 for x ≥ – 3 and y = – x – 3 for x < – 3
A sketch of y = |x + 3| is shown below:

Here, y = x + 3 is the straight line which cuts X and Y-axes at (- 3, 0) and (0, 3) respectively.
Thus, y = x + 3 for x ≥ – 3 represents the part of line which lies on the right side of x = – 3.
Similarly, y = – x – 3, x < – 3 represents the part of line y = – x – 3, which lies on left side of x = – 3.
Clearly, required area
= Area of region ABPA + Area of region PCOP

Question 38.
Using integration, find the area of the region {(x, y): x² + y² ≤ 16, x² ≤ 6y} (Delhi 2010C)
Answer:
Given region is {(x, y): x² + y² ≤ 16, x² ≤ 6y}
Above region has a circle x2 + y2 = 16 whose centre is (0, 0) and radius 4 and a parabola whose vertex is (0, 0) and axis along y-axis.
First, let us sketch the region, as shown below:

For finding the points of intersection of two curves, we have
x² + y² = 16
and x² = 6y
On putting x² = 6y from Eq. (ii) in Eq. (i), we get
y² + 6y -16 = 0
⇒ y² + 8y – 2y -16 = 0
⇒ y(y + 8) – 2(y + 8) = 0
⇒ (y – 2) (y + 8) = 0
⇒ y = 2 or – 8
When y = 2, then from Eq. (ii), we get
x = ± √12 = ± 2√3
and when y = – 8, then from Eq. (ii), we get
x² = – 48 which is not possible.
So, y = – 8 is rejected.
Thus, the two curves meet at points C(2√3, 2) and D(- 2√3, 2).
Now, required area
= Area of shaded region OCBDO
= 2 [ Area of region OACO + Area of region ABCA]

Question 39.
Find the area of circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y. (All India 2010)
Answer:
2√6+94 sin^-1(22√3)sq units

Hint:

Question 40.
Using integration, find the area of the following region.
{(x, y):|x – 1| ≤ y ≤ 5−x2−−−−−√} (Delhi 2010)
Answer:
First, write the given curves separately, i.e. y = |x – 1| and y = 5−x2−−−−−√.
Then, sketch all the above defined functions and find the required area.
Given region is {(x, y): |x – 1| ≤ y ≤ 5−x2−−−−−√}
Above region has two equations
y = |x – 1| and y = 5−x2−−−−−√

Also, other curve is y = 5−x2−−−−−√
On squaring both sides, we get
y² = 5 – x²
⇒ x² + y² = 5
which represents equation of circle with centre (0, 0) and radius, r = √5.
But the actual equation of curve is y = V5- x2 which represents a semi-circle whose centre is (0, 0) and radius r = √5.
On drawing the rough sketch, we get the following graph:

For finding the points of intersection of the curves, we have
y = 1 – x … (i)
y = x – 1 …… (ii)
and y = 5−x2−−−−−√ ….. (iii)
On putting y = 1 – x from Eq. (i) in Eq. (iii), we get
(1 – x) = 5−x2−−−−−√
⇒ x² + (1 – x)² = 5
⇒ x² + 1 + x² – 2x = 5
⇒ 2x² – 2x – 4 = 0
⇒ x² – x – 2 = 0
⇒ x² – 2x + x – 2 = 0
⇒ x(x – 2) + 1(x – 2) = 0
⇒ (x + 1) (x – 2) = 0
∴ x = – 1 or 2
Now, when x = -1, then
y = 5−x2−−−−−√ = 5−1−−−−√ = √4
⇒ y = 2
and when x = 2, then y = 5−x2−−−−−√ = 5−4−−−−√ = 1
⇒ y = 1
So, points of intersection of Eqs. (i) and (iii) are (-1, 2) and (2, 1).
Similarly, on solving Eq. (ii) and Eq. (iii), we get
x = – 1 or 2
From Eq. (iii), at x = – 1, y = 2and at x = 2, y = 1.
Hence, the two curves intersect at (-1, 2) and (2, 1).
Now, required area