Practicing S Chand Class 12 Maths Solutions Chapter 25 Application of Integrals Ex 25(a) is the ultimate need for students who intend to score good marks in examinations.

## S Chand Class 12 ICSE Maths Solutions Chapter 25 Application of Integrals Ex 25(a)

Question 1.

Using integration, find the area of the region bounded by

(i) the line 3 y = 2 x + 4, the x-axis and the lines x = 1 and x = 3.

(ii) 2 y = -x + 6, the x-axis and the lines x = 2 and x = 4.

Answer:

(i) The eqn. of given line be

3 y = 2 x + 4 ……………..(1)

eqn. (1) meets x-axis at A(-2, 0) and y-axis at B (0, \(\frac{4}{3}\)).

∴ required area = \(\int_1^3\) y d x

= \(\int_1^3\) \(\frac{2 x+4}{3}\) d x = \(\frac{1}{3}\) \(\frac{(2 x+4)^2}{2 \times 2}]_1^3\)

= \(\frac{1}{12}\) [(10)^{2} – 6^{2}]

= \(\frac{1}{12}\)(100 – 36)

= \(\frac{64}{12}\) = \(\frac{16}{3}\) sq. units

(ii) The eqn. of given line be

2y = -x + 6

it meets coordinate axes at A(6, 0) and B(0, 3).

∴ Required area = \(\int_2^4\) y d x

= \(\int_2^4\) \(\frac{-x+6}{2}\) d x

= \(\frac{1}{2}\) \(\frac{(6-x)^2}{-2}]_2^4 \)

= \( -\frac{1}{4}\) [4 – 16]

= 3 sq. units

Question 2.

(i) Using integration, find the area of the region bounded between the line x = 2 and the parabola y^{2} = 8 x.

(ii) Find the area of the region bounded by the parabola y^{2} = 12 x and its latus rectum.

(iii) Find the area enclosed between the co-ordinate axes and the curve y^{2} = 4 a(x + λ) in the second quadrant.

Answer:

(i) Clearly the figure is symmetrical about x-axis.

∴ area enclosed by the parabola

= 2 × area enclosed by parabola

and x = 2 in Ist quadrant

Divide the region in first quadrant into vertical strips each vertical strip has its lower end on x-axis and upper and on the parabola y = \(\sqrt{8 x}\). So the approximating rectangle has length |y| and width d x. This rectangle can move between x = 0 and x = 2.

[∵ y ≥ 0 ∴|y|=y]

∴ required area = 2 × \(\int_0^2\)|y| d y

= 2 \(\int_0^2\) y d y

= 2 \(\int_0^2\) √ 8 √ x d x

= 2 × 2 \(\sqrt{2}\) [\(\frac{2}{3}\) x^{3/2} ]^{2} _{0}

= \(\frac{8 \sqrt{2}}{3}\) [2^{3/2} – 0]

= \(\frac{32}{3}\) sq. units

(ii) Given eqn. of parabola be y^{2} = 12 x represents a right handed parabola with vertex at (0,0). Here 4 a = 12

⇒ a = 3

∴ required area = 2 × area of region OABO

= 2 × \(\int_0^3\) \(\sqrt{12 x}\) d x

= 2 × 2 \(\sqrt{3}\) \(\int_0^3\) x^{1/2} d x

= 4 \(\sqrt{3}\) \(\frac{x^{3 / 2}}{3 / 2}]_0^3\)

= \(\frac{8 \sqrt{3}}{3}\)[3 \(\sqrt{3}\) – 0]

= 24 sq. units

(iii) The given eqn. of parabola be

y^{2} = 4 a(x + λ)

represents a right handed parabola with vertex (-λ, 0) it meets y-axis at A(0, 2 \(\sqrt{a \lambda}\))

∴ required area = \(\int_{-\lambda}^0 2 \sqrt{a}\) \(\sqrt{x+\lambda}\) d x

[∵ y > 0 in 2nd quad ]

= 2 \(\sqrt{a}\) \(\frac{(x+\lambda)^{3 / 2}}{3 / 2}]_{-\lambda}^0\)

= \(\frac{4 \sqrt{a}}{3}\) λ^{3/2}

Question 3.

Draw the rough sketch of the curve y = \(\sqrt{3 x + 4 x}\) and find the area under the curve, above the x-axis and between x = 0 and x = 4.

Answer:

The eqn. of given curve y = \(\sqrt{3 x+4}\)

The table of values is given as under :

x | 0 | 4 | 2 |

y | 2 | 4 | √10 |

Plot the points (0, 2) ;(4, 4) and (2, \(\sqrt{10}\)) on graph paper and join then by free hand curve.

∴ Required area of region OABCO

= \(\int_0^4\) \(\sqrt{3 x+4}\) d x

= \(\frac{(3 x+4)^{3 / 2}}{\frac{3}{2} \cdot 3}]_0^4\)

= \(\frac{2}{9}[(16)^{3 / 2}-(4)^{3 / 2}]\)

= \(\frac{2}{9}\)[64 – 8]

= \(\frac{112}{9}\) sq. units

Question 4.

Sketch the graph of y = |x + 1|. Evaluate \(\int_{-4}^2\)|x + 1| d x. What does the value of the integral represent on the graph?

Answer:

For Graph ; For x < -1; y = f(x) = -(x + 1) For x = -1 ; y = 0 For x > -1 ; y = x + 1

∴ \(\int_{-4}^2\)|x + 1| d x

= \(\int_{-4}^{-1}\) – (x + 1) d x + \(\int_{-1}^2\) (x + 1) d x

∴ I = \(-\frac{(x+1)^2}{2}]_{-4}^1\) + \(\frac{(x+1)^2}{2}]_{-1}^2\)

= \(-\frac{1}{2}\)(0 – 9) + \(\frac{1}{2}\)(9 – 0)

= \(\frac{9}{2}\) + \(\frac{9}{2}\) = 9

When x = -4 < -1 ∴ y = -(-4 + 1) = 3 i.e. point becomes (-4, 3). When x = 2 > -1

∴ y = 2 + 1 = 3 i.e. point becomes (2, 3).

Further, area of ∆ ABC = \(|\frac{1}{2} \times(-3) \times(3)|\) = \(\frac{9}{2}\) sq. units

and area of ∆CDE = \(|\frac{1}{2} \times 3 \times 3|\) = \(\frac{9}{2}\) sq. units

∴ \(\int_{-4}^2\)|x + 1| d x

= sum of areas of two ∆ from -4 to -1 and -1 to 2 .

Question 5.

Find the area bounded by the curve y = x^{2} and the line y = 16. [Fig. 25.41]

Answer:

(i) The given curve y = x^{2}

and the line y = 16 intersects when 16 = x^{2}

⇒ x = ± 4

∴ Region R = {(x, y) ; x^{2} ≤ y ≤ 10 ; 0 ≤ x ≤ 4}

∴ Required area = \(\int_0^4\)[16 – x^{2} ] d x

= 16 x – \(\frac{x^3}{3}]_0^4\)

= 64 – \(\frac{64}{3}\)

= \(\frac{128}{3}\) sq. units

(ii) required area = \(\int_1^4\) x d y

= \(\int_1^4\) \(\frac{\sqrt{y}}{2}\) d y

= \(\frac{1}{2}\) \(\frac{y^{3 / 2}}{3 / 2}]_1^4\)

= \(\frac{1}{3}\)[4^{3/2} – 1] = \(\frac{7}{3}\) sq. units

Question 6.

Sketch the region lying in the first quadrant and bounded by y = 4 x^{2}, x = 0, y = 1 and y = 4. Find the area of the region, using integration.

Answer:

The given curve is y = 4 x^{2} be an upward parabola with vertex at (0, 0) and is symmetrical about y-axis.

∴ required area = area of region OABCO

= \(\int_1^4\) x d y [Taking Horizontal strips]

= \(\int_1^4 \) \(\frac{\sqrt{y}}{2}\) d y

[∵ y = 4 x^{2} ⇒ x = ± \(\frac{\sqrt{y}}{2}\) as region lies in 1st quadrant ∴ x > 0 i.e. x = \(\frac{\sqrt{y}}{2}\)]

= \(=\frac{1}{2}\) \(\frac{y^{3 / 2}}{3 / 2}]_1^4\)

= \(\frac{1}{3}\)[8 – 1]

= \(\frac{7}{3}\) square units

Question 7.

Find the gradients of the curve y = x^{2}(2 – x) at the points (0, 0) and (2, 0) and sketch the part of the curve of which both x and y are positive. Find the area between this part of the curve and the axis.

Answer:

Given eqn. of curve y = x^{2}(2 – x)

∴ \(\frac{d y}{d x}\) = 4 x – 3 x^{2}

Thus gradient of curve at (0, 0) = \(\frac{d y}{d x}\)_{(0,0)} = 0

and gradient of curve at (2, 0) = \(\frac{d y}{d x}\)_{(2,0)}

= 8 – 12 = -4

The table of values is given as under:

x | 0 | 1 | 2 | \(\frac{3}{2}\) |

y | 0 | 1 | 0 | \(\frac{9}{8}\) |

by plotting all these points on graph paper and join them by free hand curve.

∴ required area of region OBAO = \(\int_0^2\) x^{2}(2 – x) d x

= \(\frac{2 x^3}{3}\) – \(\frac{x^4}{4}]_0^2\)

= \(\frac{16}{3}\) – 4

= \(\frac{4}{3}\) sq. units

Question 8.

Find the area included between the curve y = 9 – x^{2} , the X-axis and the lines x = -2 and x = ± 2.

Answer:

The given curve

y = 9 – x^{2}

x^{2} = 9 – y = -(y – 9)

⇒ represents a downward parabola with vertex (0, 9).

∴ required area of region P A Q P

= \(\int_{-2}^2\) y d x

= \(\int_{-2}^2\)(9 – x^{2}) d x

= 9 x – \(\frac{x^3}{3}]_{-2}^2\)

= (18 – \(\frac{8}{3}\)) – (-18 + \(\frac{8}{3}\))

= 36 – \(\frac{16}{3}\)

= \(\frac{108-16}{3}\)

= \(\frac{92}{3}\) sq. units

Question 9.

Find the area bounded by the curve y^{2} = 4 x, the line y = 3, and the y-axis.

Answer:

The given curve y^{2} = 4 x represents a right handed parabola with vertex at origin O(0, 0). Now the line y = 3 meets the given curve y^{2} = 4 x when 9 = 4 x

i.e. x = \(\frac{9}{4}\) i.e. at point (\(\frac{9}{4}\), 3).

∴ R = {(x, y) ; 0 ≤ x ≤ \(\frac{y^2}{4}\) ; 0 ≤ y ≤3}

Thus required area = \(\int_0^3\) \(\frac{y^2}{4}\) d y

= \(\frac{1}{4}\) \(\frac{y^3}{3}]_0^3\)

= \(\frac{1}{12}\) × 27

= \(\frac{9}{4}\) sq. units

Aliter: Divide the region into vertical strips with upper end on line y = 3

and. low erend on given curve y^{2} = 4 × and 0 ≤ x ≤ \(\frac{9}{4}\)

∴ required area = \(\int_0^{9 / 4}\)[3 – 2 \(\sqrt{x}\)] dx = 3x – \(\frac{2 x^{3 / 2}}{3 / 2}]_0^{9 / 4}\)

= 3(\(\frac{9}{4}\) – 0) – \(\frac{4}{3}\)[\(\frac{9}{4})^{3 / 2}\) – 0]

= \(\frac{27}{4}\) – \(\frac{4}{3}\) × \(\frac{27}{8}\)

= \(\frac{27}{4}\) – \(\frac{9}{2}\)

= \(\frac{27-18}{4}\) = \(\frac{9}{4}\) sq. units

Question 10.

Find the area of the region bounded by y = -1, y = 2, x = y and x = 0.

Answer:

Given lines are

y = -1 ……………….(1)

y = 2 ……………….(2)

x = y ……………….(3)

x = 0 …………………(4)

and

Since y = -1 and y = 2 are the lines are parallel to x-axis and x = 0 represents y-axis.

The line y = x passes through origin and making a slope of 45° with positive direction of x-axis.

Divide the region into horizontal strips as shown in shaded portion.

R_{1} = {(x, y); 0 ≤ x ≤ y ; 0 ≤ y ≤ 2}

R_{2} = {(x, y) ; y ≤ x ≤ 0 ;-1 ≤ y ≤ 0}

∴ Required area = \(\int_{-1}^0\)|x| d y + \(\int_0^2\) x d y

= \(\int_{-1}^0\) – y d y + \(\int_0^2\) y d y

= \(-\frac{y^2}{2}]_{-1}^0\) + \(\frac{y^2}{2}]_0^2\)

= \(-\frac{1}{2}\)(0 – 1) + \(\frac{1}{2}\) (4 – 0)

= \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\) sq. units

Question 11.

Find the area bounded by the parabola y^{2} = 2 x and the ordinates x = 1 and x = 4.

Answer:

The given parabola y^{2} = 2 x be a right handed parabola with vertex (0, 0).

Divide the region into vertical strips with lower end on x-axis and upper end on given curve

y^{2} = 2 x .

∴ required area = 2 \(\int_1^4\) y d x

= 2 \(\int_1^4\) \(\sqrt{2}\) \(\sqrt{x}\) d x

= 2 \(\sqrt{2}\) \(\frac{x^{3 / 2}}{3 / 2}]_1^4\)

= \(\frac{4 \sqrt{2}}{3}\)[4^{3/2} – 1]

= \(\frac{4 \sqrt{2}}{3}\)[8 – 1]

= \(\frac{28 \sqrt{2}}{3}\) sq.units

Question 12.

Find the area bounded by the curve y^{2} = 4 a^{2} (x – 1) and the lines x = 1 and y = 4 a.

Answer:

Given eqn. of curve be y^{2} = 4 a^{2} (x – 1) which represents a right handed parabola with vertex (1, 0).

The line x = 1 meets given parabola at A(1, 0).

The line y = 4 a meets given parabola when 16 a^{2} = 4 a^{2} (x – 1)

⇒ x = 5 i.e. at B(5, 4 a).

∴ R = {(x, y) ; 4 a ≤ y ≤ 2 a \(\sqrt{x-1}\) ; 1 ≤ x ≤ 5}

Thus required area = \(\int_1^5\) [4 a – 2 a \(\sqrt{x-1}\)] d x

= 4 a x – \(\frac{2 a(x-1)^{3 / 2}}{3 / 2}]_1^5\)

= 4 a(5 – 1) – \(\frac{4 a}{3}\) [4^{3/2} – 0]

= 16 a – \(\frac{32 a}{3}\)

= \(\frac{16 a}{3}\) sq.units

Question 13.

Evaluate the area of the region bounded by the curve y = 2 \(\sqrt{1-x^2}\), and the X-axis, after drawings rough sketch of the same.

Answer:

Given equation of curve be y = 2 \(\sqrt{1-x^2}\)

⇒ y^{2} = 4 (1 – x^{2})

⇒ \(\frac{y^2}{4}\) + \(\frac{x^2}{1}\) = 1

So it represents an ellipse with major axis on y-axis and minor axis as x-axis. Clearly it contains even power of x and y so the given curve is symmetrical about both coordinate axis.

∴ area of whole ellipse = 4 × area of ellipse in first quadrant and coordinate axes.

Now, we divide the region in first quadrant into vertical strips each vertical strip has lower and on x-axis and upper end on ellipse. So the approximating rectangle has length |y| and width d x and clearly it move from x = 0 and x = 1.

∴ area of whole ellipse = 4 × \(\int_0^1\) |y| d x

= 4 \(\int_0^1\) y d x [∵ y ≥ 0]

= 4 \(\int_0^1\) 2 \(\sqrt{1-x^2}\) d x

= 8 \(\frac{x \sqrt{1-x^2}}{2}\) + \(\frac{1}{2}\) sin^{-1} x]^{1} _{0}

= 8[0 + \(\frac{1}{2}\) sin^{-1} (1) – 0 – 0]

= 8 × \(\frac{1}{2}\) × \(\frac{\pi}{2}\)

= 2 π square units

Here required area = area of region in first quadrant = \(\frac{1}{4}\) × 2π = \(\frac{\pi}{2}\) square units

Question 14.

Make a rough sketch of the curve 3 y = (2 x + 1)(2 – x), and calculate its gradient at the point where it meets the axis of y. Find the area bounded by the curve and the axis of x.

Answer:

Given eqn. of curve be

3y = (2 x + 1)(2 – x) ……………….(1)

The table of values is given as under :

The given eqn. (1) can be written as,

3y = -2 x^{2} + 3 x + 2

⇒ 3y = -2[x^{2} – \(\frac{3}{2}\) x + \(\frac{9}{16}\) – \(\frac{9}{16}\) – 1]

⇒ 3y = -2[(x – \(\frac{3}{4}\))^{2} – \(\frac{25}{16}\)]

⇒ (x – \(\frac{3}{4}\))^{2}

= \(\frac{25}{16}\) – \(\frac{3}{2}\)

y = \(-\frac{3}{2}\)(y – \(\frac{25}{16}\) × \(\frac{2}{3}\))

⇒ (x – \(\frac{3}{4}\))^{2}

= \(-\frac{3}{2}\) (y – \(\frac{25}{24}\))

which represents a downward parabola with vertex (\(\frac{3}{4}\), \(\frac{25}{24}\)).

Diff. (1) both sides w.r.t. x, we have

3 \(\frac{d y}{d x}\) = -4 x + 3

⇒ \(\frac{d y}{d x}\) = \(\frac{-4 x+3}{3}\)

curve given by eqn. (1) meets y-axis i.e. x = 0

∴ y = \(\frac{2}{3}\) i.e. at point (0, \(\frac{2}{3}\)).

∴ gradient of the curve at (0, \(\frac{2}{3}\))

= \(\frac{d y}{d x}\) (0, \(\frac{2}{3}\))

= \(-\frac{4 × 0 + 3}{3}\) = 1

Given curve meets x-axis at y = 0 i.e. (2 x + 1)(2 – x) = 0

⇒ x = -\frac{1}{2}, 2

∴ required area = \(\frac{1}{3}\) \(\int_{-1 / 2}^2\) (2 x + 1)(2 – x) d x

= \(\frac{1}{3}\) \(\int_{-1 / 2}^2\) (- 2 x^{2} + 3 x + 2) d x

= \(\frac{1}{3}\)[\(-\frac{2}{3}\) x^{3} + \(\frac{3 x^2}{2}\) + 2 x]^{-1/2} _{2}

= \(\frac{1}{3}\)[(\(-\frac{16}{3}\) + 6 + 4) – (\(\frac{1}{12}\) + \(\frac{3}{8}\) – 1)]

= \(\frac{1}{3}\) [\(\frac{14}{3}\) – \(\frac{1}{12}\) – \(\frac{3}{8}\) + 1]

= \(\frac{125}{72}\) sq. units

Question 15.

Given alongside figure shows a sketch of the curve y = (x – 1)(4 – x). If P is the point (2, 2), verify that the tangent at P passes through the origin O as shown. Calculate the area O A P enclosed between the tangent, the curve and the axis of x.

Answer:

The eqn. of given curve be

y = (x – 1)(4 – x)

∴ \(\frac{d y}{d x}\) = (x – 1)(-1) + (4 – x) 1

= -x + 1 + 4 – x = 5 – 2 x

∴ slope of tangent to given curve (1) at point (2, 2)

= \(\frac{d y}{d x}\) {(2,2)} = 5 – 4 = 1

Thus eqn. of tangent through the point (2, 2) be given by

y – 2 = 1(x – 2)

⇒ y = x

Clearly eqn. (2) passes through origin O(0, 0).

∴ area of shaded region OAPO

= area of ∆ OPQ – area of region APQA

= \(\frac{1}{2}\) × 2 × 2 – \(\int_1^2\) (x – 1)(4 – x) d x

[since curve (1) meets x-axis i.e. y = 0 ∴ x = 1, x = 4 ]

= 2 – \(\int_1^2\) (-x^{2} + 5 x – 4) d x

= 2 + \(\int_1^2\) (x^{2} – 5 x + 4) d x

= 2 + \(\frac{x^3}{3}\) – \(\frac{5 x^2}{2}\) + 4 x]^{2}^{1}

= 2 + \(\frac{8}{3}\) – 10 + 8 – \(\frac{1}{3}\) + \(\frac{5}{2}\) – 4]

= \(\frac{8}{3}\) – \(\frac{1}{3}\) + \(\frac{5}{2}\) – 4

= \(\frac{7}{3}\) – \(\frac{3}{2}\)

= \(\frac{5}{6}\) sq. units

Question 16.

The curve y = a x^{2} + b x + c passes through the points (1, 0),(2, 0) and its gradient at the point (2, 0) is 2 . Find the numerical value of the area included between the curve and the axis of x.

Answer:

The eqn. of given curve y = a x^{2} + b x + c eqn. (1) passes through the point (1, 0) and (2, 0).

∴ 0 = a + b + c

0 = 4 a + 2 b + c

Diff. eqn. (1) both sides w.r.t. x; ; we have

\(\frac{d y}{d x}\) = 2 a x + b

∴ gradient of curve at point (2, 0) = 2

⇒ ( \(\frac{d y}{d x}\)) (2,0) = 2

⇒ 4 a + b = 2

eqn. (3) -2 x eqn. (4) ; we have

4 a + 2 b + c – 8 a – 2 b + 4 = 0

⇒ -4 a + c = -4

⇒ 4 a – c = 4

eqn. (4) – eqn. (2) ; we have

3a – c = 2

eqn. (5) – eqn. (6) gives; a = 2

∴ from (6) ; c = 6 – 2 = 4

∴ from (4); b = 2 – 4 a = 2 – 8 = -6

Thus from (1); we have

y = 2 x^{2} – 6 x + 4

eqn. (7) meets x-axis when y = 0

∴ 2 x^{2} – 6 x + 4 = 0

⇒ (x – 1)(2 x – 4) = 0

⇒ x = 1,2

∴ required area = \(\int_1^2\) y d x

= \(\int_1^2\) (2 x^{2} – 6 x + 4) d x

= 2 \(\int_1^2\) (x^{2} – 3 x + 2) d x

= 2\(\frac{x^3}{3}\) – \(\frac{3 x^2}{2}\) + 2 x]^{2} _{1}

= |2\(\frac{8}{3}\) – 6 + 4 – \(\frac{1}{3}\) + \(\frac{3}{2}\) – 2]|

= 2[\(\frac{7}{3}\) + \(\frac{3}{2}\) – 4]

= |2\(\frac{14+9-24}{6}\)|

= \(\frac{1}{3}\) sq. units

Question 17.

The line y = 2 x meets the curve y^{2} = 4 x at the point O (the origin) and P, and P N is perpendicular to the y-axis. Prove that the area between the curve and O P is one-half the area enclosed by the lines O N, N P and the curve.

Answer:

eqn. of given line be

y = 2 x

y^{2} = 4 x

and eqn. of given curve be

eqn. (2) represents a right handed parabola with vertex at (0, 0) and line (1) meets curve (2) when 4 x^{2} = 4 x

⇒ 4 x(x – 1) = 0

⇒ x = 0,1

∴ from (1); y = 0, 2

Thus the point of intersection are O(0, 0) and P(1, 2).

∴ area betwen the curve and OP

A_{1} = \(\int_0^1\) [2 \(\sqrt{x}\) – 2 x] d x

= \(\frac{2 x^{3 / 2}}{3 / 2}-x^2\)]_0^1

= \(\frac{4}{3}\) – 1

= \(\frac{1}{3}\) sq. units

A_{2} = area of region ONPO

= area of ∆ ONP – area between curve and OP

= \(\frac{1}{2}\) × 2 × 1 – \(\frac{1}{3}\)

= \(\frac{2}{3}\) sq. units

∴ ⇒ \(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\)

= \(\frac{1}{2}\)

⇒ area between the curve and OP is one-half the area enclosed by the lines ON, NP and the curve.

Question 18.

Calculate the areas of the two parts into which the area enclosed by the x-axis and the curve y=18 x-3 x^2 is divided by the line x=4.

Answer:

Given eqn. of curve be

y = 18 x – 3 x^{2}

⇒ y = -3(x^{2} – 6 x + 9 – 9)

⇒ y = -3(x – 3)^{2} + 27

⇒ 3(x – 3)^{2} = 27 – y

⇒ (x – 3)^{2}

= \(-\frac{1}{3}\)(y – 27)

which represents a downward parabola with vertex (3, 27). Curve (1) meets x-axis at y = 0

∴18 x – 3 x^{2} =0

⇒ 3 x(6 – x) = 0

⇒ x = 0, 6

i.e. at points (0, 0) and Q(6, 0)

The line x = 4 meets curve (1)

When y = 72 – 48 = 24 i.e. at P(4, 24)

A_{1} = area of shaded region OAPRO

= \(\int_0^4\) y d x

= \(\int_0^4\) (18 x – 3 x^{2} ) d x

= 9 x^{2} – x^{3}]^{0} _{4}

= 144 – 64

= 80 sq. units

A_{2} = area of shaded region PQRP

= \(\int_4^6\) (18 x – 3 x^{2}) d x

= 9 x^{2} – x^{3}]^{4}_{6}

= (9 × 36 – 216) – (144 – 64) = 108 – 80

= 28 sq. units

Question 19.

Draw the rough sketch of y^{2} + 1 = x, x < 2 and find the area enclosed by the curve and the line x = 2.

Answer:

Given equation of curve be y^{2} + 1 = x

⇒ y^{2} = x – 1 Clearly it represents a parabola (right handed) with vertex (1, 0) and does not meeting y-axis at any point.

Further the given curve meets the line x = 2 at y^{2} + 1 = 2

⇒ y = ± 1 i.e. at points (2, + 1) and (2, – 1).

Clearly the given curve is symmetrical about x-axis.

∴ required area = 2 × area of region enclosed by parabola and line x = 2 in first quadrant.

Divide this region R into vertical strips with lower end on x axis and upper end on y = \(\sqrt{x-1}\) and corresponding rectangle move from x = 1 and x = 2

∴ required area = 2 \(\int_1^2\) |y| d y = 2 \(\int_1^2\) y d x

= 2 \(\int_1^2\) \(\sqrt{x-1}\) d x

= 2 \(\frac{(x-1)^{3 / 2}}{3 / 2}\)]^{2}_{1}

= \(\frac{4}{3}\)[1 – 0]

= \(\frac{4}{3}\) square units.

[∵ y ≥ 0 ∴ |y| = y]

Question 20.

Find the area bounded by the curve y = 4 – x^{2} and the line y = 0 and y = 3.

Answer:

We want to find the area of region bounded by curves given below:

and

y = 4 – x^{2}

y = 0, y = 3

eqn. (1) represents a parabola (downward) with vertex (0,4). y = 4 – x^{2} meets y = 3 when 3 = 4 – x^{2}

⇒ x^{2} = 1

⇒ x = ± 1

∴ points of intersection are ( ± 1,3).

∴ required area = 2 × area of region OABCO

Divide the region into small horizontal strips. Each strip has left end on y-axis and right end on curve y = 4 – x^{2}.

∴ required area = 2 \(\int_0^3\) x d y

= 2 \(\int_0^3\) \(\sqrt{4-y}\) d y

= 2 \(\frac{(4-y)^{3 / 2}}{-3 / 2}\)]^{0} _{3}

= \(-\frac{4}{3}\)[1 – 8]

= \(\frac{28}{3}\) sq. units

Question 21.

Make a rough sketch of the following curves. Also, find the area enclosed between the curves and the axes.

(i) y = cos x, 0 ≤ x ≤ \(\frac{\pi}{2}\)

(ii) y = cos 2 x, 0 ≤ x ≤ \(\frac{\pi}{4}\)

(iii) y = sin x, 0 ≤ x ≤ \(\frac{\pi}{2}\)

(iv) y = cos ^2 x, 0 ≤ x ≤ \(\frac{\pi}{2}\)

Answer:

(i) Given eqn. of curve be y = cos x ; 0 ≤ x ≤ \(\frac{\pi}{2}\)

Table of values is given as under:

x | 0 | \(\frac{\pi}{2}\) | \(\frac{\pi}{3}\) | \(\frac{\pi}{2}\) |

y | 1 | \(\frac{1}{2}\) | 0 | 0 |

∴ required area = \(\int_0^{\pi / 2}\) y d x

= \(\int_0^{\pi / 2}\) cos x d x

= sin x]^{π/2} _{0}

= sin \(\frac{\pi}{2}\) – sin 0

= 1 – 0 = 1 sq. units

(ii) Given eqn. of curve be

y = cos 2 x ; 0 ≤ x ≤ \(\frac{\pi}{4}\)

x | 0 | \(\frac{\pi}{6}\) | \(\frac{\pi}{4}\) |

y | 1 | \(\frac{1}{2}\) | 0 |

∴ required area = \(\int_0^{\pi / 4}\) y d x

= \(\int_0^{\pi / 4}\) cos 2 x d x

= \(\frac{\sin 2 x}{2}\)]^{π/4} _{0}

= \(\frac{1}{2}\) [sin \(\frac{\pi}{2}\) – 0]

= \(\frac{1}{2}\) sq. units

(iii) Equation of given curve be

y = sin x ; 0 ≤ x ≤ \(\frac{\pi}{2}\)

x | 0 | \(\frac{\pi}{6}\) | \(\frac{\pi}{4}\) | \(\frac{\pi}{3}\) | \(\frac{\pi}{2}\) |

y | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{\sqrt{3}}{2}\) | 1 |

∴ required area = \(\int_0^{\pi / 2}\) y dx = \(\int_0^{\pi / 2}\) sin x d x

= -cos x]^{π/2} _{0}

= -[0 – 1]

= 1 sq. units

(iv) Given curve is y = cos^{2} x

For rough sketch we construct a table of values as under :

x | 0 | \(\frac{\pi}{6}\) | \(\frac{\pi}{4}\) | \(\frac{\pi}{3}\) | \(\frac{\pi}{2}\) |

y | 0 | \(\frac{3}{4}\) | \(\frac{1}{2}\) | \(\frac{1}{4}\) | 1 |

∴ Required area = \(\int_0^{\pi / 2}\) y d x [using vertical strips]

= \(\int_0^{\pi / 2}\) cos^{2} x d x

= \(\frac{1}{2}\) \(\int_0^{\pi / 2}\) (1 + cos 2 x) d x

= \(\frac{1}{2}\) [x + \(\frac{\sin 2 x}{2}\)]^{π/2} _{0}

= \(\frac{\pi}{4}\) sq. units

Question 22.

Draw a rough sketch of the curve y = cos^{2} x in [0, π] and find the area enclosed by the the lines x = 0, x = π and the x-axis.

Answer:

Given eqn. of curve be y = cos^{2} x in [0, π]

x | 0 | \(\frac{\pi}{6}\) | \(\frac{\pi}{4}\) | \(\frac{\pi}{3}\) | \(\frac{\pi}{2}\) | \(\frac{2\pi}{3}\) | π |

y | 1 | \(\frac{3}{4}\) | \(\frac{1}{2}\) | \(\frac{1}{4}\) | 0 | \(\frac{1}{4}\) | 1 |

∴ Required area = \(\int_0^{\pi / 2}\) cos^{2} x + \(\int_{\pi / 2}^\pi\) cos^{2} x d x

= \(\int_0^\pi \) cos^{2} x d x

= \(\int_0^\pi\) \(\frac{1+\cos 2 x}{2}\) d x

= \(\frac{1}{2}\) [x + \(\frac{\sin 2 x}{2}\)]^{π} _{0}

= \(\frac{1}{2}\)[π + 0 – 0 – 0]

= \(\frac{\pi}{2}\) sq. units

Question 23.

Draw a rough sketch of the curve y = \(\frac{1}{2}\)π + 2 sin^{2} x and find the area between the x-axis, the curve and the ordinates x = 0 and x = π

Answer:

Given eqn. of curve be y = \(\frac{\pi}{2}\) + 2 sin^{2} x

x | 0 | \(\frac{\pi}{6}\) | \(\frac{\pi}{4}\) | \(\frac{\pi}{3}\) | \(\frac{\pi}{2}\) | \(\frac{2\pi}{3}\) | \(\frac{3\pi}{4}\) | \(\frac{5\pi}{6}\) | π |

y | \(\frac{\pi}{2}\) | \(\frac{\pi}{2}\)+\(\frac{1}{2}\) | \(\frac{\pi}{2}\)+1 | \(\frac{\pi}{2}\)+\(\frac{3}{2}\) | 3.57 | 3.07 | 2.57 | 2.07 | 1.57 |

The rough sketch for given curve is given as under.

Divide the region into vertical strips. Each vertical strip has lower end on x-axis and upper end on given curve.

Thus the corresponding rectangle has length |y| and width d x and this rectangle move from x = 0 and x = π.

∴ Required area = \(\int_0^\pi\) |y| d x = \(\int_0^\pi\) y d x

[∵ y ≥ 0 ∴ |y| = y]

= \(\int_0^\pi\) \(\frac{\pi}{2}\) + 2 sin^{2} x] d x

= \(\frac{\pi}{2}\) x]^{π} _{0} + \(\int_0^\pi \) 2[ \(\frac{1-\cos 2 x}{2}\)] d x

= \(\frac{\pi}{2}\)(π – 0) + [x – \(\frac{\sin 2 x}{2}\)]^{π} _{0}

= \(\frac{\pi^2}{2}\) + [π – 0 – 0 – 0]

= \(\frac{\pi^2}{2}\) + π

= \(\frac{\pi}{2}\) (π + 2) sq. units

Question 24.

Show that the area included between the x-axis and the curve a^{2} y = x^{2} (x + a) is \(\frac{a^2}{12}\)

Answer:

Clearly the curve a^{2} y = x^{2} (x + a)

meets x-axis i.e. y = 0, where x^{2} (x + a) = 0

⇒ x = 0,-a

∴ curve meets x-axis at (0,0) and (-a, 0).

∴ required area

= \(\int_{-a}^0\) y d x

= \(\int_{-a}^0\) \(\frac{1}{a^2}\) (x^{3} + a x^{2}) d x

= \(\frac{1}{a^2}\) \(\frac{x^4}{4}\) + \(\frac{a x^3}{3}\)]^{0} _{a}

= \(\frac{1}{a^2}\) [0 + 0 – \(\frac{a^4}{4}\) + \(\frac{a^4}{3}\)]

= \(\frac{a^2}{12}\) sq. units

Question 24.

Show that the area included between the x-axis and the curve a^{2} y = x^{2} (x + a) is \(\frac{a^2}{12}\)

Answer:

Clearly the curve a^{2} y = x^{2} (x + a)

meets x-axis i.e. y = 0, where x^{2} (x + a) = 0

⇒ x = 0,-a

∴ curve meets x-axis at (0,0) and (-a, 0).

∴ required area

= \(\int_{-a}^0\) y d x

= \(\int_{-a}^0\) \(\frac{1}{a^2}\) (x^{3} + a x^{2}) d x

= \(\frac{1}{a^2}\) \(\frac{x^4}{4}\) + \(\frac{a x^3}{3}\)]^{0} _{a}

= \(\frac{1}{a^2}\) [0 + 0 – \(\frac{a^4}{4}\) + \(\frac{a^4}{3}\)]

= \(\frac{a^2}{12}\) sq. units

Question 25.

(i) Calculate the area bounded by the curve y = x^{2} – 1, the x-axis and the line y = 8.

(ii). Calculate the approximate increase in this area, if the line y = 8 is changed to y = 8.01.

Answer:

(i) The given curve y = x^{2} – 1

⇒ x^{2} = y + 1

it represents an upward parabola with vertex (0, -1) and meets x-axis at y = 0.

∴ from (1), x^{2}

= 1 ⇒ x = ± 1

i.e. at points ( ± 1, 0).

The line y = 8 meets the curve (1) when x^{2} = 9

⇒ x = ± 3 i.e. at points ( ± 3, 8).

∴ Required area = 2 \(\int_0^8\) x d y

= 2 \(\int_0^8\) \(\sqrt{y+1}\) d y

= 2 \(\frac{(y+1)^{3 / 2}}{3 / 2}\)]^{0} _{8}

= \(\frac{4}{3}\) [9^{3/2} – 1]

= \(\frac{4}{3}\) × 26

= \(\frac{104}{3}\) sq. units

(ii) Thus required area = 2 \(\int_8^{8.01}\) x d y

= 2 \(\int_8^{8.01}\) \(\sqrt{y+1}\) d y

= 2 \(\frac{(y+1)^{3 / 2}}{3 / 2}\)]^{8} _{0.88}

= \(\frac{4}{3}\)[(9.01)^{3/2} – 9^{3/2} ]

= \(\frac{4}{3}\)[27.0450 – 27]

= 0.06 sq. units

Question 26.

Find the area of the region bounded by y = -1, y = 2, x = y^{3} and x = 0.

Answer:

The given curve be x = y^{3} and lines are y = -1 ; y = 2 and x = 0

The line y = 2 intersects the curve x = y^{3} at (8, 2)

and y = -1 meets curve at the point (-1, -1).

∴ Required area = \(\int_0^2\) x d y + \(\int_{-1}^0\) x d y

= \(\int_0^2 y^3\) d y + \(\int_{-1}^0\) (0 – y^{3}) d y

= \(\frac{y^4}{4}\)]_0^2 – \(\frac{y^4}{4}\)]^{-1} 2

= 4 – \(\frac{1}{4}\) [0 – 1]

= 4 + \(\frac{1}{4}\)

= \(\frac{19}{4}\) sq. units

Question 27.

Find the area bounded by x = a t^{2} , y = 2 a t between the ordinates corresponding to t = 1 and t = 2.

Answer:

Equation of given curve be x = a t^{2} ; y = 2 a t

On eliminating t, we have y^{2} = 4 a x which is a right handed parabola with vertex (0, 0).

When t = 1 ⇒ x = a, y = 2 a

When t = 2 ⇒ x = 4 a, y = 4 a

∴ required area = 2 × area of shaded region in Ist qudrant. Divide this region into vertical strips. Each vertical strip has lower end on x-axis and upper end on given curve. So the corresponding rectangle has length |y| and width d x and this rectangle move from x = a to x = 4 a.

∴ Required area = 2 \(\int_a^{4 a}\)|y| d x

= 2 \(\int_a^{4 a}\) 2 \(\sqrt{a x}\) d x [∵ y ≥ 0]

= 4 \(\sqrt{a}\) \(\frac{x^{3 / 2}}{3 / 2}\)]^{a} _{4a}

= \(\frac{8 \sqrt{a}}{3}\) [(4 a)^{3/2} – a^{3/2} ]

= \(\frac{8 \sqrt{a}}{3}\)[8 – 1] a^{3/2}

= \(\frac{56}{3}\) a^{2} sq. units