Class 12 Maths NCERT Solutions for Chapter 8 Application of Integrals Miscellaneous Exercise

Integrals Miscellaneous Exercise Solutions

1. Find the area under the given curves and given lines:
(i) y = x², x = 1, x = 2 and x-axis
(ii) y = x⁴, x = 1, x = 5 and x –axis

Solution

(i) The required area is represented by the shaded area ADCBA as

Area ADCBA = ∫₁² y dx
= ∫₁² x² dx

= 8/3 - 1/3
= 7/3 units

(ii) The required area is represented by the shaded area ADCBA as

Area ADCBA = ∫₁⁵ x⁴ dx

= 624.8 units

2. Find the area between the curves y = x and y = x².

Solution

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x², is A (1, 1)
We draw AC perpendicular to x - axis.

Area (OBAO) = Area (ΔOCA) - Area (OCABO) ...(1)

= 1/2 - 1/3
= 1/6 units

3. Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y = 1 and y = 4.

Solution

The area in the first quadrant bounded by y = 4x² , x = 0, y = 1 and y = 4 is represented by the shaded area ABCDA as

Area ADCBA = ∫₁⁴ y dx
= ∫₁⁴ √y/2 dx

= 7/3 units

4. Sketch the graph of y = |x + 3| and evaluate ∫_-6⁰ |x + 3| dx.

Solution

The given equation is y = |x + 3|
The corresponding values of x and y are given in the following table :

On plotting these points, we obtain the graph of y = |x + 3| as follows.

It is known that, (x + 3) ≤ 0 for -6 ≤ x ≤ -3 and (x + 3) ≥ 0 for -3 ≤ x ≤ 0

= 9

5. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.

Solution

The graph of y = sin x can be drawn as

∴ Required area = Area OABO + Area BCDB

= [ - cosπ + cos 0] + |- cos 2π + cos π|
= 1 + 1+ |(-1-1)|
= 2 + |-2|
= 2 + 2 = 4 units

6. Find the area enclosed between the parabola y² = 4ax and the line y = mx.

Solution

The area enclosed between the parabola, y² = 4ax and the line y = mx, is represented by the shaded area OABO as

The points of intersection of both the curves are (0, 0) and (4a/m², 4a/m)
We draw AC perpendicular to x - axis.

Area OABO = Area OCABO - Area (ΔOCA)

7. Find the area enclosed by the parabola 4y = 3x² and the line 2y = 3x + 12

Solution

The area enclosed between the parabola, 4y = 3x² , and the line, 2y = 3x + 12, is represented by the shaded area OBAO as

The points of intersection of the given curves are A(-2, 3) and (4, 12).
We draw AC and BD perpendicular to x - axis.

Area OBAO = Area CDBA - (Area ODBO + Area OACO)

= 45 - 18
= 27 units

8. Find the area of the smaller region bounded by the ellipse x² /9 + y² /4 and the line x/3 + y/2 = 1.

Solution

The area of the smaller region bounded by the ellipse, x²/9 + y²/4 = 1, and the line x/3 + y/2 = 1 , is represented by the shaded region BCAB as

Area BCAB = Area (OBCAO) - Area (OBAO)

9. Find the area of the smaller region bounded by the ellipse x² /a² + y² /b² = 1 and the line x/a + y/b = 1.

Solution

The area of the smaller region bounded by the ellipse, x² /a² + y² /b² = 1 and the line, x/a + y/b = 1, is represented by the the shaded region BCAB as

Area BCAB = Area (OBCAO) - Area (OBAO)

10. Find the area of the region enclosed by the parabola x² = y, the line y = x + 2 and x-axis.

Solution

The area of the region enclosed by the parabola, x² = y , the line, y = x + 2 and x - axis is represented by the shaded region OABCO as

The point of intersection of the parabola, x² = y, and the line, y = x + 2, is A(-1, 1).
Area OABCO = Area (BCA) + Area COAC

11. Using the method of integration find the area bounded by the curve |x| + |y| = 1.
[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and – x – y = 1].

Solution

The area bounded by the curve, |x| + |y| = 1, is represented by the shaded region ADCB as

The curve intersects the axes at points A(0, 1), B(1, 0), C(0, -1), and D(-1, 0).
It can be observed that the given curve is symmetrical about x - axis and y - axis.

∴Area ADCB = 4 × Area OBAO

= 4(1/2)
= 2 units.

12. Find the area bounded by curves {(x, y) : y ≥ x² and y = |x|}.

Solution

The area bounded by the curves, {(x, y) : y ≥ x² and y = |x|}, is represented by the shaded region as

It can be observed that the required area is symmetrical about y - axis.
Required area = 2[Area (OCAO) - Area (OCADO)]

13. Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).

Solution

The Vertices of ΔABC are A(2, 0), B(4, 5) and C(6, 3).

Equation of line segment AB is
y - 0 = [(5 - 0)/(4 - 2)] (x - 2)
⇒ 2y = 5x - 10
⇒ y = (5/2) (x - 2) ...(1)
Equation of line segment BC is
y - 5 = [(3 - 5)(6 - 4)] (x - 4)
⇒ 2y - 10 = -2x + 8
⇒ 2y = -2x + 18
⇒ y = -x + 9 ...(2)
Equation of line segment CA is

⇒ -4y + 12 = -3x + 18
⇒ 4y = 3x - 6
⇒ y = (3/4)(x - 2) ...(3)
Area (ΔABC) = Area (ABLA) + Area (BLMCB) - Area(ACMA)

= 13 - 6
= 7 units

14. Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

Solution

The given equations of lines are
2x + y= 4 ...(1)
3x -2 y = 6 ...(2)
And, x - 3y + 5 = 0 ...(3)

The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x - axis.
Area (ΔABC) = Area (ALMCA) - Area (ALB) - Area(CMB)

15. Find the area of the region {(x, y) : y² ≤ 4x, 4x² + 4y² ≤ 9}

Solution

The area bounded by the curves, {(x, y) : y² ≤ 4x, 4x² + 4y² ≤ 9}, is represented as

The points of intersection of both the curves are (1/2, √2) and (1/2, -√2).
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x - axis.
Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC

Put 2x = t

⇒ dx = dt/2
When x = 3/2, t = 3 and when x = 1/2, t = 1

Therefore, the required area is

units

16. Choose the correct answer Area bounded by the curve y = x³, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B. -15/4
C. 15/4
D. 17/4

Solution

Required area = A(DCOD) + A(OAB)

Since the area is positive
∴ A(DCOD) = 16/4

Hence, the required area of the region is 17/4 units

17. Choose the correct answer The area bounded by the curve y = x | x| ,, x-axis and the ordinates x = –1 and x = 1 is given by
[Hint: y = x² if x > 0 and y = –x² if x < 0]
A. 0
B. 1/3
C. 2/3
D. 4/3