NCERT Solutions For Class 12 Maths Chapter 8 Application of Integrals Ex 8.1
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| Textbook | NCERT |
| Board | CBSE |
| Category | NCERT Solutions |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 8 |
| Exercise | Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1 |
| Number of Questions Solved | 13 |
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1
NCERT TEXTBOOK EXERCISES
Ex 8.1 Class 12 Maths 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 curve y² = x is a parabola with vertex at origin.Axis of x is the line of symmetry, which is the axis of parabola. The area of the region bounded by the curve, x = 1, x=4 and the x-axis. Area LMQP
Ex 8.1 Class 12 Maths 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 given curve is y² = 9x, which is a parabola with vertex at (0, 0) and axis along x-axis. It is symmetrical about x-axis, as it contains only even powers of y. x = 2 and x = 4 are straight lines parallel toy-axis at a positive distance of 2 and 4 units from it respectively.
∴ Required area = Area ABCD
Ex 8.1 Class 12 Maths 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:
The given curve x² = 4y is a parabola with vertex at (0,0). Also since it contains only even powers of x,it is symmetrical about y-axis.y = 2 and y = 4 are straight lines parallel to x-axis at a positive distance of 2 and 4 from it respectively.
Ex 8.1 Class 12 Maths Question 4.
Find the area of the region bounded by the ellipse $\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$
Solution:
The equation of the ellipse is $\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$
The given ellipse is symmetrical about both axis as it contains only even powers of y and x.
Ex 8.1 Class 12 Maths Question 5.
Find the area of the region bounded by the ellipse $\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1$
Solution:
$\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1$ It is an ellipse with centre (0,0) and length of major axis = 2a = 2×3 = 6 and length of minor axis = 2b = 2 × 2 = 4.
Ex 8.1 Class 12 Maths Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x² + y² = 4.
Solution:
Consider the two equations x² + y² = 4 … (i)
and x = √3y i.e. y=1/√3x …(ii)
(1) x² + y² = 4 is a circle with centre O (0,0) and radius = 2.
Ex 8.1 Class 12 Maths Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line x=a/√2
Solution:
The equation of the given curve are
x² + y² = a² …(i) and x=a/√2 …(ii)
Clearly, (i) represent a circle and (ii) is the equation of a straight line parallel to y-axis at a
Ex 8.1 Class 12 Maths 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:
Graph of the curve x = y² is a parabola as given in the figure. Its vertex is O and axis is x-axis. QR is the ordinate along x = 4
Ex 8.1 Class 12 Maths 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)
Ex 8.1 Class 12 Maths Question 10.
Find the area bounded by the curve x² = 4y and the line x = 4y – 2
Solution:
Given curve is x² = 4y …(i)
which is an upward parabola with vertex at (0,0) and it is symmetrical about y-axis
Equation of the line is x = 4y – 2 …(ii)
Solving (i) and (ii) simultaneously, we get; (4y – 2)² = 4y
⇒ 16y² – 16y + 4 = 4y
⇒ 4y² – 5y + 1 = 0
Ex 8.1 Class 12 Maths Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Solution:
The curve y² = 4x is a parabola as shown in the figure. Axis of the parabola is x-axis. The area of the region bounded by the curve y² = 4x and the line x = 3 is
A = Area of region OPQ = 2 (Area of the region OLQ)
Choose the correct answer in the following Exercises 12 and 13:
Ex 8.1 Class 12 Maths 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) π/2
(c) π/3
(d) π/4
Solution:
(a) x² + y² = 4. It is a circle at the centre (0,0) and r=2
Ex 8.1 Class 12 Maths Question 13.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
(a) 2
(b) 9/4
(c) 9/3
(d) 9/2
Solution:
(b) y² = 4x is a parabola
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1 PDF
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