These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1
Question 1.
Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4, and the x-axis.
Solution:
The required area is shaded in the figure and lies between the lines x = 1 and x = 4. Hence the limits of integration are 1 and 4.
Question 2.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and x-axis in the first quadrant
Solution:
The required area is shaded in the figure lies between the lines x = 2 and x = 4. Hence the limits of integration are 2 and 4.
Question 3.
Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Solution:
x² = 4y ∴ x = 2\(\sqrt{y}\)
The required area is shaded in the figure and lies between the lines y = 2 and y = 4. Hence the limits of integration are 2 and 4.
Question 4.
Find the area of the region bounded by the ellipse \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 }\) = 1
Solution:
i. Put y = o in \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 }\) = 1, we get \(\frac{x^{2}}{16}=1\)
i.e., x² = 16 ∴ x = ± 4
∴ The points of intersection with x-axis are (-4, 0) and (4, 0)
ii. The shaded area lies between x = 0 and x = 4.
∴ The limits of integration are 0 and 4.
The curve is symmetrical w.r.t both the axes
∴ Required area = 4 x area of the shaded region
\(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 }\) = 1 i.e., \(\frac{y^{2}}{9}=1\) – \(\frac{x^{2}}{16}=1\)
Question 5.
Find the area of the region bounded by the ellipse \(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } \) = 1
Solution:
The points of intersection of the curve and the x-axis is obtained by substituting y = 0 in the equation of the curve.
We get \(\frac{x^{2}}{4}=1\), x² = 4 ∴ x = ±2
The curve intersects the x-axis at (-2, 0) and (2,0).
\(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } \) = 1 ∴ \(\frac{y^{2}}{9}\) = 1 – \(\frac{x^{2}}{4}=1\)
= \(\frac{4-x^{2}}{4}\)
∴ y² = \(\frac { 9 }{ 4 }\)(4 – x²) ∴ y = \(\frac{3}{2} \sqrt{4-x^{2}}\)
The curve is symmetrical about both the axes.
Area of the ellipse = 4 x area of the shaded region
Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line x = \(\sqrt{3}\)y and the circle x² + y² = 4.
Solution:
The given equations are x = \(\sqrt{3}\)y … (1)
x² + y = 4 … (2)
To get the point of intersection of the line and the circle, solve (1) and (2)
From (1), we get y = \(\frac{x}{\sqrt{3}}\)
∴ (2) → x² + \(\left(\frac{x}{\sqrt{3}}\right)^{2}\) = 4 ⇒ x² + \(\frac{x^{2}}{3}\) = 4
\(\frac{4x^{2}}{3}\) = 4 ⇒ x² = 3 ⇒ x = ±\(\sqrt{3}\)
To get the point of intersection of the circle and x-axis puty = 0 in (2)
(2) → x² = 4 ∴x = ± 2
∴ The x coordinate of A and B are \(\sqrt{3}\) and 2 respectively.
Required area = Area of OAC + Area of ACB
Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line x = \(\frac { a }{ \sqrt { 2 } }\).
Solution:
x² + y² = a² is a circle which intersects the positive x axis at (a, 0).
x² + y² = a²
∴ y = \(\sqrt{a^{2}-x^{2}}\)
The required area is shaded in the figure and lies between x = \(\frac { a }{ \sqrt { 2 } }\) and x – a
∴ Limits of integration are \(\frac { a }{ \sqrt { 2 } }\) and a
Area of the shaded region = 2 x Area of the shaded region in the first quadrant
Question 8.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Solution:
The given curves are x = y² … (1)
x = 4 ….. (2) and x = a … (3)
Since the line x = a divides the area in two equal parts, we get \(2 \int_{0}^{a} y d x=2 \int_{a}^{4} y d x\)
Since the areas are symmetric w.r.t. x-axis
Question 9.
Find the area of the region bounded by the parabola y = x² and y = |x|.
Solution:
Clearly x² = y represents a parabola with vertex at (0, 0) positive direction of y-axis as its axis it opens upwards.
y = |x| i.e., y = x and y = – x represent two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.
The required region is the shaded region as shown in the figure. Since both the curve are symmetrical about y-axis.
So,
Required area = 2 (shaded area in the first quadrant)
Question 10.
Find the area bounded by the curve x² = 4y and the line x = 4y – 2
Solution:
The given curves are x² = 4y … (1)
and x – 4y – 2 … (2)
Substituting (2) in (1),
we get (4y – 2)² = 4y
⇒ 16y² – 16y + 4 = 4y
⇒ 16y² – 20y + 4 = 0
⇒ 4y² – 5y + 1 = 0
⇒ 4y² – 4y – y + 1 = 0
⇒ 4y(y – 1) – 1(y – 1) = 0
⇒ (y – 1)(4y – 1) = 0
⇒ y = 1, y = \(\frac { 1 }{ 4 }\)
When y = 1, x = 2
When y = \(\frac { 1 }{ 4 }\), x = – 1
∴ Points of intersection are (-1, \(\frac { 1 }{ 4 }\)) and (2, 1)
The required area is shaded in the figure. Required area = (Area under the line x = 4y – 2) – (Area under the parabola x² = 4y) :
Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Solution:
The given curves are y² = 4x … (1)
and x = 3
The shaded region is symmetric w.r.t. x-axis.
∴ Required area = 2 x Shaded area in the first quadrant
Question 12.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
(a) π
(b) \(\frac { \pi }{ 2 } \)
(c) \(\frac { \pi }{ 3 } \)
(d) \(\frac { \pi }{ 4 } \)
Solution:
(a) π
Question 13.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
(a) 2
(b) \(\frac { 9 }{ 4 }\)
(c) \(\frac { 9 }{ 3 }\)
(d) \(\frac { 9 }{ 2 }\)
Solution:
