# 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 = x^{2}, x = 1, x = 2 and x-axis(ii) y = x^{4}, x = 1, x = 5 and x –axis**

**Solution**

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

Area ADCBA = ∫_{1}^{2} y dx

= ∫_{1}^{2} x^{2} dx

= 8/3 - 1/3

= 7/3 units

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

Area ADCBA = ∫_{1}^{5} x^{4} dx

= 624.8 units

**2. Find the area between the curves y = x and y = x^{2}.**

**Solution**

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x

^{2}, is A (1, 1)

We draw AC perpendicular to x - axis.

Area (OBAO) = Area (Î”OCA) - Area (OCABO)

= 1/2 - 1/3

= 1/6 units

**...(1)**= 1/2 - 1/3

= 1/6 units

**3. Find the area of the region lying in the first quadrant and bounded by**

*y*= 4*x*^{2},*x*= 0,*y*= 1 and*y*= 4.**Solution**

The area in the first quadrant bounded by y = 4x

^{2}, x = 0, y = 1 and y = 4 is represented by the shaded area ABCDA asArea ADCBA = ∫

_{1}

^{4}y dx

= ∫

_{1}

^{4}√y/2 dx

= 7/3 units

**4. Sketch the graph of y = |x + 3| and evaluate ∫**

_{-6}^{0}|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

∴ Required area = Area OABO + Area BCDB

= [ - cosÏ€ + cos 0] + |- cos 2Ï€ + cos Ï€|

= 1 + 1+ |(-1-1)|

= 2 + |-2|

= 2 + 2 = 4 units

The points of intersection of both the curves are (0, 0) and (4a/m

We draw AC perpendicular to x - axis.

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 BCAB = Area (OBCAO) - Area (OBAO)

The point of intersection of the parabola, x

Area OABCO = Area (BCA) + Area COAC

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.

It can be observed that the required area is symmetrical about y - axis.

Required area = 2[Area (OCAO) - Area (OCADO)]

Equation of line segment AB is

y - 0 = [(5 - 0)/(4 - 2)] (x - 2)

⇒ 2y = 5x - 10

⇒ y = (5/2) (x - 2)

Equation of line segment BC is

y - 5 = [(3 - 5)(6 - 4)] (x - 4)

⇒ 2y - 10 = -2x + 8

⇒ 2y = -2x + 18

⇒ y = -x + 9

Equation of line segment CA is

⇒ -4y + 12 = -3x + 18

⇒ 4y = 3x - 6

⇒ y = (3/4)(x - 2)

Area (Î”ABC) = Area (ABLA) + Area (BLMCB) - Area(ACMA)

= 13 - 6

= 7 units

The area of the region bounded by the lines is the area of Î”ABC. AL and CM are the perpendiculars on x - axis.

Area (Î”ABC) = Area (ALMCA) - Area (ALB) - Area(CMB)

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

Therefore, the required area is units

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

= 1/3

∴ A(OAB) = 1/3

Thus, the total area = 1/3 + 1/3 = 2/3 units

Area bounded by the circle and parabola

= 2[Area (OADO) + Area(ADBA)]

Thus, the correct answer is C.

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*= 2Ï€.**Solution**

The graph of y = sin x can be drawn as

∴ Required area = Area OABO + Area BCDB

= [ - cosÏ€ + cos 0] + |- cos 2Ï€ + cos Ï€|

= 1 + 1+ |(-1-1)|

= 2 + |-2|

= 2 + 2 = 4 units

**6. Find the area enclosed between the parabola**

*y*^{2}= 4*ax*and the line*y*=*mx.***Solution**

The area enclosed between the parabola, y

^{2}= 4*ax*and the line*y*=*mx*, is represented by the shaded area OABO asThe points of intersection of both the curves are (0, 0) and (4a/m

^{2}, 4a/m)

We draw AC perpendicular to x - axis.

Area OABO = Area OCABO - Area (Î”OCA)

**7. Find the area enclosed by the parabola 4**

*y*= 3*x*^{2}and the line 2*y*= 3*x*+ 12**Solution**

The area enclosed between the parabola, 4y = 3x

^{2}, and the line, 2y = 3x + 12, is represented by the shaded area OBAO asThe 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

= 45 - 18

= 27 units

**8. Find the area of the smaller region bounded by the ellipse x**

^{2}/9 + y^{2}/4 and the line x/3 + y/2 = 1.**Solution**

The area of the smaller region bounded by the ellipse, x

^{2}/9 + y^{2}/4 = 1, and the line x/3 + y/2 = 1 , is represented by the shaded region BCAB asArea BCAB = Area (OBCAO) - Area (OBAO)

**9. Find the area of the smaller region bounded by the ellipse x**

^{2}/a^{2}+ y^{2}/b^{2}= 1 and the line x/a + y/b = 1.**Solution**

The area of the smaller region bounded by the ellipse, x

^{2}/a^{2}+ y^{2}/b^{2}= 1 and the line, x/a + y/b = 1, is represented by the the shaded region BCAB asArea BCAB = Area (OBCAO) - Area (OBAO)

**10. Find the area of the region enclosed by the parabola**

*x*^{2}=*y*, the line*y*=*x*+ 2 and*x*-axis.**Solution**

The area of the region enclosed by the parabola,

*x*^{2}=*y*, the line, y = x + 2 and x - axis is represented by the shaded region OABCO asThe point of intersection of the parabola, x

^{2}= 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].

[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.

= 4(1/2)

= 2 units.

**12. Find the area bounded by curves {(x, y) : y ≥ x**

^{2}and y = |x|}.**Solution**

The area bounded by the curves, {(x, y) : y ≥ x

^{2}and y = |x|}, is represented by the shaded region asIt 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 Î”ABC 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 (Î”ABC) = 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: 2**

*x*+*y*= 4, 3*x*– 2*y*= 6 and*x*– 3*y*+ 5 = 0**Solution**

The given equations of lines are

2x + y= 4

3x -2 y = 6

And, x - 3y + 5 = 0

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 Î”ABC. AL and CM are the perpendiculars on x - axis.

Area (Î”ABC) = Area (ALMCA) - Area (ALB) - Area(CMB)

**15. Find the area of the region {(x, y) : y**

^{2}≤ 4x, 4x^{2}+ 4y^{2}≤ 9}**Solution**

The area bounded by the curves, {(x, y) : y

^{2}≤ 4x, 4x^{2}+ 4y^{2}≤ 9}, is represented asThe 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

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**

A. – 9

B. -15/4

C. 15/4

D. 17/4

*y*=*x*^{3}, the*x*-axis and the ordinates*x*= –2 and*x*= 1 isA. – 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| ,,**

[Hint:

A. 0

B. 1/3

C. 2/3

D. 4/3

*x*-axis and the ordinates*x*= –1 and*x*= 1 is given by[Hint:

*y*=*x*^{2}if*x*> 0 and*y*= –*x*^{2}if*x*< 0]A. 0

B. 1/3

C. 2/3

D. 4/3

**Solution**

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

= 1/3

∴ A(OAB) = 1/3

Thus, the total area = 1/3 + 1/3 = 2/3 units

Hence, the required area of the region is 2/3 units.

**18. Choose the correct answer The area of the circle**

*x*^{2}+*y*^{2}= 16 exterior to the parabola*y*^{2}= 6*x*is**Solution**

The given equations are

y

*x*^{2}+*y*^{2}= 16**...(1)**y

^{2}= 6x**...(2)**Area bounded by the circle and parabola

= 2[Area (OADO) + Area(ADBA)]

Area of circle = Ï€(r)

= Ï€(4)

^{2}= Ï€(4)

^{2}= 16Ï€ unitsThus, the correct answer is C.

**19. The area bounded by the**

(A) 2 ( 2 −1)

(B) √2-1

(C) √2+1

*y*-axis,*y*= cos*x*and*y*= sin*x*when 0 <= x <= Ï€2(A) 2 ( 2 −1)

(B) √2-1

(C) √2+1

**(D) √2**

**Solution**

The given equation are

y = cos x

And, y = sin x

Required area = Area (ABLA) + area (OBLO)

y = cos x

**...(1)**And, y = sin x

**...(2)**Required area = Area (ABLA) + area (OBLO)

Integrating by parts, we obtain

= √2 - 1 units

= √2 - 1 units

Thus, the correct answer is B.