Chapter 8 Application of Integrals | class 12th | Important Question for Maths

Class 12 Maths Chapter 8 Important Extra Questions Applications of the Integrals

Applications of the Integrals Important Extra Questions Very Short Answer Type

Question 1.
Find the area of region bounded by the curve y = x² and the line y = 4.
Answer:
322 sq. units.

Question 2.
Find the area bounded by the curve y = x³, x = 0 and the ordinates x = -2 and x = 1.
Answer:
174 sq. units.

Question 3.
Find the area bounded between parabolas y² = 4x and x² = 4y.
Answer:
163 sq. units.

Question 4.
Find the area enclosed between the curve y = cos x, 0 ≤ x ≤ π4 and the co-ordinate axes.
Answer:
12 sq. units.

Question 5.
Find the area between the x-axis curve y = cos x when 0≤ x < 2.
Answer:
4 sq. units

Question 6.
Find the ratio of the areas between the centre y = cos x and y = cos 2x and x-axis for x = 0 to
x = π3
Answer:
2:1.

Question 7.
Find the areas of the region:
{(x,y): x² + y² ≤ 1 ≤ x + 4}
Ans.
12 (π – 1) sq. units.

Applications of the Integrals Important Extra Questions Long Answer Type 2

Question 1.
Find the area enclosed by the circle:
x² + y² = a². (N.C.E.R.T.)
Solution:
The given circle is
x² + y² = a² ………….(1)
This is a circle whose centre is (0,0) and radius ‘a’.

Area of the circle=4 x (area of the region OABO, bounded by the curve, x-axis and ordinates x = 0, x = a)
[ ∵ Circle is symmetrical about both the axes]
= 4 ∫a0 ydx [Taking vertical strips] o
= 4∫a0a2−x2−−−−−−√dx
[ ∵ (1) ⇒ y = ± a2−x2−−−−−−√
But region OABO lies in 1st quadrant, ∴ y is + ve]

Question 2.
Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the liney = x and the circle x² + y² = 32. (C.B.S.E. 2018)
Solution:
We have :
y = x …(l)
and x² + y² = 32 …(2)
(1) is a st. line, passing through (0,0) and (2) is a circle with centre (0,0) and radius 4√2 units. Solving (1) and (2) :
Putting the value of y from (1) in (2), we get:
x² + x² = 32
2x² = 32
x² = 16
x = 4.
[∵ region lies in first quadrant]
Also y = 4
Thus the line (1) and the circle (2) meet each other at B (4,4), in the first quadrant.
Draw BM perp. to x – axis.

∴ Reqd. area = area of the region OMBO + area of the region BMAB …(3)
Now, area of the region OMBO

Again, area of the region BMAB

= 8π – (8 + 4π) = 4π – 8
∴ From (3),
Required area = 8 + (4π – 8) = 4π sq.units.

Question 3.
Find the area bounded by the curves y = √x , 2y + 3 = Y and Y-axis. (C.B.S.E. Sample Paper 2018-19)
Solution:
The given curves are
y = √x ………….(1)
and 2y + 3 = x …(2)
Solving (1) and (2), we get;
2y+3−−−−−√ = y
Squaring, 2y + 3 = y²
⇒ y²2 – 2y – 3 = 0
⇒ (y + 1)(y-3) = 0 ⇒ y = -1, 3
⇒ y = 3 [∵ y > 0]
Putting in (2),
x = 2(3) + 3 = 9.
Thus, (1) and (2) intersects at (9, 3).

Question 4.
Find the area of region:
{(x,y): x² + y² < 8, x² < 2y}. (C.B.S.E. Sample Paper 2018-19)
Solution:
The given curves are ;
x² + y² = 8 ………… (1)
x² = 2y ………… (2)

Solving (1) and (2):
8 – y²= 2y
⇒ y² + 2y – 8 = 0
⇒ (y + 4)(y – 2) = 0
= y = -4,2
⇒ y = 2. [∵ y > 0]
Putting in (2), x² = 4
⇒ x = -2 or 2.
Thus, (1) and (2) intersect at P(2, 2) and Q(-2, 2).

Question 5.
Using integration, find the area of the region enclosed between the two circles:
x² + y² = 1 and (x – 1)² + y² = 1. (C.B.S.E. 2019 C, C.B.S.E. 2019)
Solution:
The given circles are x² + y² =1 …(1)
and (x – 1)² + y² = 1
(1) is a circle with centre (0,0) and radius 1.
(2) is a circle with centre (1,0) and radius 1.
Solving (1) and (2):
(2)-(1) gives: -2x + 1 =0 ⇒ x = 12.
Putting in (1), 14 + y² = 1
y² = 34
⇒ y = ± 3√2
Thus, the circles intersect at A (12,3√2) and B(12,−3√2)

Reqd. area = 2 (shaded area)
= 2 (area (OAL) + area (ALC))

Question 6.
Using integration, find the area of the region: {(x, y); 9x² + 4y² ≤ 36,3x + 2y ≥ 6}. (C.B.S.E. 2019(C))
Solution:
We have: 9x² + 4y² = 36
x24+y29 = 1 …(1) ,
which is an upward ellipse
and 3x + 2y = 6 =» x2+y3 = 1 …(2),
which is a st. line.
Reqd. area is the shaded area, as shown in the figure:

Question 7.
Using the method of integration, find the area of the region bounded by the lines:
3x – 2y + 1 = 0,2x + 3y – 21 = 0 and x – 5y + 9 = 0.   (A.I.C.B.S.E. 2019, C.B.S.E. 2012)
Solution:
Let the sides AB, BC and CA of ΔABC be:
3x – 2y + 1 = 0 …(1)
2x + 3y – 21 = 0 …(2)
and x – 5y + 9 = 0 …(3) respectively.
Solving (3) and (1), we get A as (1,2).
Solving (1) and (2), we get B as (3,5).
Solving (2) and (3), we get C as (6,3).

Now ar (ΔABC) = ar (trap ALMB) + ar (trap BMNC) – ar (trap ALNC)

= 6510 = 6.5sq . units.

Question 8.
Find the area lying above the x-axis and included between the circle x² + y² = Sir and the parabola y² = 4x. (N.C.E.R. I; C.B.S.E. 2019,19 C)
Solution:
The given circle is x² + y² – 8x = 0
i.e. (x-4)² + y² = 16 ….. (1)
It has centre (4,0) and radius 4 units.
The given parabola is y² = 4x ….(2)

Solving (1) and (2) :
x² – 8x + 4x = 0 =
⇒ x² – 4x = 0
⇒ x(x-4) = 0
⇒ x = 0,4.
When x = 0,y = 0.
When x = 4,y² = 16
⇒ y = ±4.

Thus (1) intersects (2) at O (0, 0) and P (4, 4) above the x-axis.
∴ Area of the region OC APQO
= Area of the region OCPQO + Area of the region C APC
= ∫40y1dx+∫84y2dx
where yv y2 are ordinates of points on (2) and (1) respectively.
= ∫404x−−√dx+∫8442−(x−4)2−−−−−−−−−−√dx
[∵ Thinking +ve values as region lies above x-axis]

[Putting x-4 = t in 2nd integral so that dx = dt. When x = 4, t = 0; when x – 8, t = 4]

Question 9.
Using integration, find the area of the region bounded by the parabola y² = 4x and the circle 4x² + 4y² = 9. (Outside Delhi 2019)
Answer:

To find the points of intersection of the curves.
y² = 4x …(1)
and 4x² + 4y² = 9 …(2)
From (1) and (2),
4x² + 16x = 9
⇒ 4x²+ 10x – 9 = 0.
Solving, x = 12

Question 10.
Using integration, find the area of the region bounded by the iiney = 3x +2, the x-axis and the ordinates x = -2 and x = 1. (Outside Delhi 2019)
Answer:
The region is as shown in the figure.

Question 11.
Using integration, find the area of the region:
{(x,y): x² + y² ≤ 1, x + y ≥ 1 x ≥ 0, y ≥ 0}
Answer:
We have: x² + y² = 1 …(1)
and x + y = 1 …(2)
Solving (1) and (2), x² + (1 – x)² = 1 ⇒
2x² – 2x = 0
2x(x-1) = 0
x = 0
x = 1.
or

Required area = Shaded area ACBDA
= ar(OACBO – ar(OADBO)

Question 12.
Find the area enclosed between the parabola 4y = 3x²and the straight line 3x – 2y +12 = 0.
(A.I.C.B.S.E. 2017)
Answer:
The given parabola is 4y – 3x²
i.e. x² = 4y3…(1),
which is an upward parabola.
The given line is 3x – 2y + 12 = 0 ……….. (2)
Solving (1) and (2) :
From(1), y = 3×24 …(3)
Putting in (2),
3x – 2(3×24) + 12 = 0
x – x24 + 4 = 0
⇒ x² – 2x – 8 =0
⇒ (x-4)(x + 2)-0
⇒ x = – 2, 4.
When x = -2, then from (3),
y = 34(4) = 3.
When x = 4, then from (3),
y = 34(16) = 12.
Thus parabola (1) and line (2) meet each other at A (-2, 3) and B (4,12).

Question 13.
Using integration, find the area of the region :
{(x, y : |x-1| ≤ y ≤ 5−x2−−−−−√} (C.B.S.E. 2010)
Or
Sketch the region bounded by the curves:
y= 5−x2−−−−−√ and y = |x-1|and find its area, using integration.
(A.I.C.B.S.E. 2015)
Solution:
The given curves are : x² + y² = 5

The reqd. region is shown as shaded in the following figure:

y =x-1 meets x² + y² = 5 at B(2,1).
y = 1-x meets x² + y² = 5 atC(-1,2)
y = x -1 and y = 1 -xmeet at A(1, 0).

Reqd. area = ar (MCBLM) – ar (CMAC) – ar (ALBA)

Question 14.
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, 73). (C.B.S.E. 2015)
Solution:
The given circle is
x² + y² = 4 ……. (1)
Diff. w.r.t. x,
2x + 2y dydx = 0
dydx = −xy

Slope of normal at P (1, √3) =√3 .
∴ The equation of the tangent at P is :

⇒ 0 = -x + 3 + 1
⇒ x = 4.
Thus T is (4,0).
The equation of the normal at P is :
y – √3= √3(x-l)
⇒ y = √3x.
This meets x-axis i.e. y = 0, where x = 0.
Thus O is (0,0).
Now ar (ΔOPT) = ar (OPL) + ar (PLT)