Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Definite Integrals Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2B Definite Integrals Important Questions

Question 1.
$$\int_{2}^{3} \frac{2 x}{1+x^{2}} d x$$ [T.S. Mar. 16; May 06]
Solution:
I = $$\left[\ln \left|1+x^{2}\right|\right]_{2}^{3}$$
= ln 10 – ln 5
= ln (10/5)
= ln 2

Question 2.
$$\int_{0}^{\pi} \sqrt{2} \cdot \sqrt{2} \sqrt{\cos ^{2} \frac{\theta}{2} d \theta}$$ [A.P. Mar. 16; Mar. 05]
Solution:
I = $$\int_{0}^{\pi} \sqrt{2} \cdot \sqrt{2} \sqrt{\cos ^{2} \frac{\theta}{2} d \theta}$$
= $$\int_{0}^{\pi} 2 \cdot \cos \theta / 2 d \theta$$
= $$[4 \sin \theta / 2]_{0}^{\pi}$$
= 4 (sin $$\frac{\pi}{2}$$ – sin 0)
= 4

Question 3.
$$\int_{0}^{2}|1-x| d x$$ [A.P. Mar. 15; May 11]
Solution:

Question 4.
I = $$\int_{1}^{5} \frac{d x}{\sqrt{2 x-1}}$$ [T.S. Mar. 15]
Solution:
Let 2x – 1 = t2
2 dx = 2t dt
dx = t dt
UL : t = 3
LL : t = 1
I = $$\int_{1}^{3} \frac{t d t}{t}$$
= $$\int_{1}^{3} d t$$
= $$[\mathrm{t}]_{1}^{3}$$ = 3 – 1
= 2

Question 5.
I = $$\int_{0}^{1} \frac{x^{2}}{x^{2}+1} d x$$ [Mar. 11]
Solution:

Question 6.
$$\int_{0}^{2 \pi}$$ sin2x cos4; x dx [T.S. Mar. 15; Mar 14]
Solution:
sin2x cos4x is even function.

Question 7.
Evaluate $$\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x$$ [T.S. Mar. 16]
Solution:
Put x = a sin θ ⇒ dx = a cos θ . dθ
θ = 0 ⇒ x = 0, x = a ⇒ θ = $$\frac{\pi}{2}$$

Question 8.
Find $$\int_{-\pi / 2}^{\pi / 2}$$ sin2 x cos4 x dx [A.P. Mar. 16]
Solution:
f(x) is even
∴ $$\int_{-\pi / 2}^{\pi / 2}$$ sin2 x cos4 x dx = 2 $$\int_{0}^{\pi / 2} \sin ^{2} x \cdot \cos ^{4} x d x$$
= 2 . $$\frac{3}{6}$$ . $$\frac{1}{4}$$ . $$\frac{1}{4}$$ . $$\frac{\pi}{4}$$ = $$\frac{\pi}{16}$$

Question 9.
$$\int_{0}^{\pi / 2} \cdot \frac{\sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$$ [Mar. 14, 08]
Solution:

Question 10.
I = $$\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$$ [Mar. 08]
Solution:

= –$$\frac{1}{40} \ln \left[\frac{1 / 4}{9 / 4}\right]$$ = $$\frac{1}{40}$$ . 2ln . 3 = $$\frac{1}{20}$$ ln 3

Question 11.
y = x3 + 3, y = 0, x = -1, x = 2 [Mar. 05]
Solution:
Required area PABQ

Question 12.
x = 4 – y2, x = 0. [Mar. 11]
Solution:
The given parabola x = 4 – y2 meets, the x – axis at A (4, 0) and Y – axis at P(0, 2) and Q(6, -2).
The parabola is symmetrical about X – axis

Required area = 2 Area of OAP

Question 13.
Evaluate the following definite integrals.
(i) $$\int_{0}^{\pi / 2} \sin ^{4} x \cos ^{5} x d x$$
(ii) $$\int_{0}^{\pi / 2} \sin ^{5} x \cos ^{4} x d x$$
(iii) $$\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{4} x d x$$
Solution:
(i) $$\int_{0}^{\pi / 2} \sin ^{4} x \cos ^{5} x d x$$
= $$\frac{4}{9}$$ . $$\frac{2}{7}$$ . $$\frac{1}{5}$$ = $$\frac{8}{315}$$

(ii) $$\int_{0}^{\pi / 2} \sin ^{5} x \cos ^{4} x d x$$
= $$\frac{3}{9}$$ . $$\frac{1}{7}$$ . $$\frac{4}{5}$$ . $$\frac{2}{3}$$ = $$\frac{8}{315}$$

(iii) $$\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{4} x d x$$
= $$\frac{3}{10}$$ . $$\frac{1}{8}$$ . $$\frac{5}{6}$$ . $$\frac{3}{4}$$ . $$\frac{1}{2}$$ . $$\frac{\pi}{2}$$
= $$\frac{3}{512}$$ π

Question 14.
$$\int_{0}^{\pi / 2} \frac{d x}{4+5 \cos x}$$ [A.P. Mar. 16, 15]
Solution:

= $$\frac{1}{3}\left[\ln \frac{4}{2}\right]=\frac{1}{3} \ln 2$$

Question 15.
$$\int_{0}^{\pi} \frac{x}{1+\sin x} d x$$ [May 11]
Solution:

Question 16.
$$\int_{0}^{\pi} \frac{x \sin ^{3} x}{1+\cos ^{2} x} d x$$ [T.S. Mar. 15; Mar. 11]
Solution:

Question 17.
$$\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$$ [Mar. 07, 05]
Solution:
Put x = tan θ
dx = sec2 θ dθ
x = 0 ⇒ θ = 0
x = 1 ⇒ θ = $$\frac{\pi}{4}$$

Question 18.
$$\int_{0}^{\pi / 4} \log (1+\tan x) d x$$ [A.P. Mar. 16]
Solution:
I = $$\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$$

Question 19.
y = 4x – x2, y = 5 – 2x. [T.S. Mar. 16]
Solution:

y = 4x – x2 …………….. (i)
y = 5 – 2x ……………….. (ii)
y = -([x – 2]2) + 4
y – 4 = (x – 2)2
Solving equations (i) and (ii) we get
4x – x2 = 5 – 2x
x2 – 6x + 5 = 0
(x – 5) (x – 1) = 0
x = 1, 5
Required area = $$=\int_{1}^{5}\left(4 x-x^{2}-5+2 x\right) d x$$
= $$\int_{-1}^{5}\left(6 x-x^{2}-5\right) d x$$
= $$\left(3 x^{2}-\frac{x^{3}}{3}-5 x\right)_{1}^{5}$$
= (75 – $$\frac{125}{3}$$ – 25) – (3 – $$\frac{1}{3}$$ – 5)
= 50 – $$\frac{125}{3}$$ + 2 + $$\frac{1}{3}$$
= $$\frac{150-125+6+1}{3}$$ = $$\frac{32}{3}$$ sq. units.

Question 20.
y2 = 4x, y2 = 4(4 – x) [May 11]
Solution:
Equations of the curves are y2 = 4x ………………… (1)
y2 = 4(4 – x) …………………. (2)
Eliminating y, we get
4x = 4 (4 – x)
2x = 4 ⇒ x = 2
Substituting in equation (1), y2 = 8

= 2[$$\frac{4}{3}$$(2$$\sqrt{2}$$) – $$\frac{4}{3}$$(-2$$\sqrt{2}$$)]
= 2($$\frac{8 \sqrt{2}}{3}$$ + $$\frac{8 \sqrt{2}}{3}$$)
= 2($$\frac{16 \sqrt{2}}{3}$$) = $$\frac{32 \sqrt{2}}{3}$$ sq. units

Question 21.
Show that the area of the region bounded by $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 (ellipse) is it ab. also deduce the area of the circle x2 + y2 = a2. [Mar. 14, May 05]
Solution:

The ellipse is symmetrical about X and Y axis
Area of the ellipse = 4 Area of CAB
= 4 . $$\frac{\pi}{4}$$ ab

(from Prob. 8 in ex 10(a))
= πab
Substituting b = a, we get the circle
x2 + y2 = a2
Area of the circle = πa(a) = πa2 sq. units.

Question 22.
Evaluate $$\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$$ [Mar. 14]
Solution:
Let A = $$\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$$
Put x = $$\frac{\pi}{2}$$ – t, dx = – dt

= $$\int_{\pi / 6}^{\pi / 3} d x=(x)_{\pi / 6}^{\pi / 3}$$
= $$\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$$
A = $$\frac{\pi}{12}$$

Question 23.
Find $$\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\sin x} d x$$ [T.S. Mar. 16] [A.P. Mar. 15]
Solution:

= $$\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}$$
= 2 . tan $$\frac{\pi}{2}$$ – 2 . 0
= 2 – 1 = 2
2A = $$\pi(x)_{0}^{\pi}$$ – 2π = π(π) – 2 = π2 – 2π
A = $$\frac{\pi^{2}}{2}$$ – π

Question 24.
Find $$\int^{\pi} x \sin ^{7} x \cos ^{6} x d x .$$ [May 05] [T.S. Mar. 19]
Solution:

A = $$\pi \int_{0}^{\pi / 2} \sin ^{7} x \cos ^{6} x d x$$
= π . $$\frac{6}{17}$$ . $$\frac{3}{11}$$ . $$\frac{1}{9}$$ . $$\frac{6}{7}$$ . $$\frac{4}{5}$$ . $$\frac{2}{3}$$
= π $$\frac{16}{3003}$$

Question 25.
Evaluate $$\int_{1}^{2} x^{5} d x$$ dx
Solution:
$$\int_{1}^{2} x^{5} \cdot d x=\left[\frac{x^{6}}{6}\right]_{1}^{2}$$
= $$\frac{2^{6}}{6}-\frac{1}{6}=\frac{63}{6}=\frac{21}{2}$$

Question 26.
Evaluate $$\int_{0}^{\pi} \sin x d x$$
Solution:
$$\int_{0}^{\pi} \sin x d x=[-\cos x]_{0}^{\pi}$$
= – cos π – (- cos 0)
= + 1 + 1= 2

Question 27.
Evaluate $$\int_{0}^{a} \frac{d x}{x^{2}+a^{2}}$$
Solution:
$$\int_{0}^{a} \frac{d x}{x^{2}+a^{2}}=\left[\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{a}$$
= $$\frac{1}{a}$$ [tan-1 (1) – tan-1 (0)]
= $$\frac{1}{a}$$ ($$\frac{\pi}{4}$$ – 0) = $$\frac{\pi}{4 a}$$

Question 28.
Evaluate $$\int_{1}^{4} x \sqrt{x^{2}-1} d x$$
Solution:
g(x) = x2 – 1
f(t) = $$\sqrt{t}$$
g'(x) = 2x

Question 29.
Evaluate $$\int_{0}^{2} \sqrt{4-x^{2}} d x$$
Solution:
Let g(θ) = 2 sin θ ⇒ g'(θ) = 2 cos θ
f(x) = $$\sqrt{4-x^{2}}$$

Question 30.
Evaluate $$\int_{0}^{16} \frac{x^{1 / 4}}{1+x^{1 / 2}} d x$$
Solution:
Put t4 = x ⇒ dx = 4t3 . dt

Question 31.
Evaluate $$\int_{-\pi / 2}^{\pi / 2} \sin |x| d x$$
Solution:

Question 32.
Show that $$\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x$$
Solution:
f(x) = sinnx.

Question 33.
Evaluate $$\int_{0}^{\pi / 2} \frac{\cos ^{5 / 2} x}{\sin ^{5 / 2} x+\cos ^{5 / 2} x} d x$$
Solution:

= $$\int_{0}^{\pi / 2} d x=(x)_{0}^{\pi / 2}=\frac{\pi}{2}$$
A = $$\int_{0}^{\pi / 2} \frac{\cos ^{\frac{5}{2}} x}{\sin ^{\frac{5}{2}} x+\cos ^{\frac{5}{2}} x} d x=\frac{\pi}{4}$$

Question 34.
Show that $$\int_{0}^{\pi / 2} \frac{x}{\sin x+\cos x} d x=\frac{\pi}{2 \sqrt{2}} \log (\sqrt{2}+1)$$
Solution:

= $$\frac{\pi}{4 \sqrt{2}}$$ log ($$\sqrt{2}$$ + 1)2
= $$\frac{\pi}{4 \sqrt{2}}$$ 2 log ($$\sqrt{2}$$ + 1)
= $$\frac{\pi}{4 \sqrt{2}}$$ log ($$\sqrt{2}$$ + 1)

Question 35.
Evaluate $$\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$$ [Mar 14]
Solution:

Question 36.
Find $$\int_{-a}^{a}\left(x^{2}+\sqrt{a^{2}-x^{2}}\right) d x$$
Solution:

= 2($$\frac{a^{3}}{3}$$ – 0) + 2(0 + $$\frac{a^{2}}{3}$$ sin-1 (1) – 0 – 0)
= $$\frac{2a^{3}}{3}$$ + a2 . $$\frac{\pi}{2}$$

Question 37.
Find $$\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\sin x} d x$$ [T.S. Mar. 16] [A.P. Mar. 15]
Solution:

= $$\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}$$
= 2 . tan $$\frac{\pi}{2}$$ – 2 . 0
= 2 – 1 = 2
2A = $$\pi(x)_{0}^{\pi}$$ – 2π = π(π) – 2 = π2 – 2π
A = $$\frac{\pi^{2}}{2}$$ – π

Question 38.
Evaluate $$\int_{0}^{\pi / 2} x \sin x d x$$
Solution:
$$\int_{0}^{\pi / 2} x \cdot \sin x d x=(-x \cdot \cos x)_{0}^{\pi / 2}+\int_{0}^{\pi / 2} \cos x d x$$
= (0 – 0) + $$(\sin x)_{0}^{\pi / 2}$$
= sin $$\frac{\pi}{2}$$ – sin 0 = 1 – 0 = 1

Question 39.
Evaluate $$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{n}\left[\frac{n-i}{n+i}\right]$$ by using the method of finding definite integral as the limit of a sum.
Solution:

Question 40.
Evaluate $$\lim _{n \rightarrow \infty} \frac{2^{k}+4^{k}+6^{k}+\ldots .+(2 n)^{k}}{n^{k+1}}$$ by using the method of finding definite integral as the limit of a sum.
Solution:

Question 41.
Evaluate $$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{\frac{1}{n}}$$
Solution:

Question 42.
Let f: R → R be a continuous periodic function and T be the period of it. Then prove that for any positive integer n,
$$\int_{0}^{n T} f(x) d x=n \int_{0}^{T} f(x) d x$$ ………………. (1)
Solution:
Let k be an integer arid define
g : [kT, (k + 1)T] → [0, T] as g(t) = t – kT.
Then g'(t) = 1 for all t ∈ [kT, (k + 1)T].
Hence by $$\int_{g(c)}^{g(d)} f(t) d t=\int_{c}^{d} f(g(x)) g^{\prime}(x) d x,(f \circ g)$$
g’ is integrable on [kT, (k + 1 )T] and
$$\int_{k T}^{(k+1) T} f(g(t)) g^{\prime}(t) d t=\int_{0}^{T} f(x) d x$$ ………… (2)
We have f(g(t)) g'(t) = f(t – kT), 1 = f(t),
since f is periodic with T as the period.
Hence $$\int_{k T}^{(k+1) T} f(g(t)) g^{\prime}(t) d t=\int_{k T}^{(k+1) T} f(t) d t$$ ………….. (3)
Thus from (2) and (3),
$$\int_{k T}^{(k+1) T} f(t) d t=\int_{0}^{T} f(t) d t$$ ………………. (4)
Let us now prove eq. (1) by using the principle of mathematical induction.
For n = 1, clearly (1) is true.
Assume (1) is true for a positive integer m.

Hence eq. (1) is true for n = m + 1
Thus, eq. (1) ¡s true for any positive integer n, by the principle of mathematical induction.

Question 43.
Find
(i) $$\int_{0}^{\pi / 2} \sin ^{4} x d x$$
(ii) $$\int_{0}^{\pi / 2} \sin ^{7} x d x$$
(iii) $$\int_{0}^{\pi / 2} \cos ^{8} x d x$$
Solution:
(i) $$\int_{0}^{\pi / 2} \sin ^{4} x d x$$
= $$\frac{3}{4}$$ . $$\frac{1}{2}$$ . $$\frac{\pi}{2}$$ = $$\frac{3\pi}{16}$$

(ii) $$\int_{0}^{\pi / 2} \sin ^{7} x d x$$
= $$\frac{6}{7}$$ . $$\frac{4}{5}$$ . $$\frac{2}{3}$$ = $$\frac{16}{35}$$

(iii) $$\int_{0}^{\pi / 2} \cos ^{8} x d x$$
= $$\frac{7}{8}$$ . $$\frac{5}{6}$$ . $$\frac{3}{4}$$ . $$\frac{1}{2}$$ . $$\frac{\pi}{2}$$ = $$\frac{35\pi}{256}$$

Question 44.
Evaluate $$\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x$$ [T.S. Mar. 16]
Solution:
Put x = a sin θ ⇒ dx = a cos θ . dθ
θ = 0 ⇒ x = 0, x = a ⇒ θ = $$\frac{\pi}{2}$$
$$\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x=\int_{0}^{\pi / 2}(a \cos \theta)(a \cos \theta) d \theta$$
= a2 $$\int_{0}^{\pi / 2}$$ cos2 θ dθ
= a2 . $$\frac{1}{2}$$ . $$\frac{\pi}{2}$$ = $$\frac{\pi a^{2}}{4}$$

Question 45.
Evaluate the following definite integrals.
(i) $$\int_{0}^{\pi / 2}$$ sin4 x . cos5 x dx
(ii) $$\int_{0}^{\pi / 2}$$ sin5 x . cos4 x dx
(iii) $$\int_{0}^{\pi / 2}$$ sin6 x . cos4 x dx
Solution:
(i) $$\int_{0}^{\pi / 2}$$ sin4 x . cos5 x dx
= $$\frac{4}{9}$$ . $$\frac{2}{7}$$ . $$\frac{1}{5}$$ = $$\frac{8}{315}$$

(ii) $$\int_{0}^{\pi / 2}$$ sin5 x . cos4 x dx
= $$\frac{3}{9}$$ . $$\frac{1}{7}$$ . $$\frac{4}{5}$$ . $$\frac{2}{3}$$ = $$\frac{8}{315}$$

(iii) $$\int_{0}^{\pi / 2}$$ sin6 x . cos4 x dx
= $$\frac{3}{10}$$ . $$\frac{1}{8}$$ . $$\frac{5}{6}$$ . $$\frac{3}{4}$$ . $$\frac{1}{2}$$ . $$\frac{\pi}{2}$$
= $$\frac{3}{512}$$ π

Question 46.
Find $$\int_{0}^{2 \pi}$$ sin4 x . cos6 x dx [T.S. Mar. 19]
Solution:
f(x) = sin4 x . cos6 x dx
f(2π – x) = f(π – x) = f(x)
$$\int_{0}^{2 \pi}$$ sin4 x . cos6 x dx = 2 $$\int_{0}^{2 \pi}$$ sin4 x . cos6 x dx
= 4 $$\int_{0}^{2 \pi}$$ sin4 x . cos6 x dx
= 4 . $$\frac{5}{10}$$ . $$\frac{3}{8}$$ . $$\frac{1}{6}$$ . $$\frac{3}{4}$$ . $$\frac{1}{2}$$ . $$\frac{\pi}{2}$$
= $$\frac{3}{128}$$ π

Question 47.
Find $$\int_{-\pi / 2}^{\pi / 2}$$ sin2 x . cos4 x dx [A.P. Mar. 16, 19]
Solution:
f(x) is even
∴ $$\int_{-\pi / 2}^{\pi / 2}$$ sin2 x . cos4 x dx = 2 $$\int_{-\pi / 2}^{\pi / 2}$$ sin2 x . cos4 x dx
= 2 . $$\frac{3}{6}$$ . $$\frac{1}{4}$$ . $$\frac{1}{4}$$ . $$\frac{\pi}{4}$$
= $$\frac{\pi}{16}$$

Question 48.
Find $$\int_{0}^{\pi}$$ x sin7 x . cos6 x dx [May 05]
Solution:

$$\int_{0}^{\pi}$$ x sin7 x . cos6 x dx = 2 $$\int_{0}^{\pi}$$ sin7 x . cos6 x dx
A = π $$\int_{0}^{\pi}$$ x sin7 x . cos6 x dx
= π . $$\frac{6}{17}$$ . $$\frac{3}{11}$$ . $$\frac{1}{9}$$ . $$\frac{6}{7}$$ . $$\frac{4}{5}$$ . $$\frac{2}{3}$$
= π . $$\frac{16}{3003}$$

Question 49.
Find $$\int_{-a}^{a}$$ a2 (a2 – x2)3/2 dx
Solution:
f(x) = x2 (a2 – x2)
f(x) is even

Question 50.
Find $$\int_{0}^{1} x^{3 / 2} \sqrt{1-x} d x$$
Solution:
Put x = sin2 θ
dx = 2 sin θ . cos θ . dθ
x = 0 ⇒ θ = 0, x = 1 ⇒ θ = $$\frac{\pi}{2}$$
$$\int_{0}^{1} x^{3 / 2} \sqrt{1-x} d x$$
= $$\int_{0}^{\pi / 2} \sin ^{3} \theta \cdot \cos \theta .2 \sin \theta \cos \theta d \theta$$
= $$2 \int_{0}^{\pi / 2} \sin ^{4} \theta \cdot \cos ^{2} \theta d \theta$$
= 2 . $$\frac{3}{6}$$ . $$\frac{1}{4}$$ . $$\frac{1}{2}$$ . $$\frac{\pi}{2}$$ = $$\frac{\pi}{16}$$

Question 51.
Find the area under the curve f(x) = sin x in [0, 2π].
Solution:

f(x) = sin x,
We know that in [0, π], sin x ≥ 0 and [π, 2π], sin x ≤ 0
Required area = $$\int_{1}^{\pi}$$ sinx dx + $$\int_{\pi}^{2 \pi}$$ (-sinx) dx
= $$(-\cos x)_{0}^{\pi}[\cos x]_{\pi}^{2 \pi}$$
= – cos π + cos 0 + cos 2π – cos π
= -(-1) + 1 + 1-(-1) = 1 + 1 + 1 + 1
= 4.

Question 52.
Find the area under the curve f(x) = cos x in [0, 2π].
Solution:
We know that cos x ≥ 0 in (0, $$\frac{\pi}{2}$$) ∪ ($$\frac{3\pi}{2}$$, π) and ≤ 0 in $$\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$$

= sin $$\frac{\pi}{2}$$ – sin 0 – sin $$\frac{3\pi}{2}$$ + sin $$\frac{\pi}{2}$$ + sin 2π – sin $$\frac{3\pi}{2}$$
= 1 – 0 – (-1) + 1 + 0 – (-1)
= 1 + 1 + 1 + 1 = 4.

Question 53.
Find the area bounded by the parabola y = x2, the X-axis and the lines x = -1, x = 2.
Solution:

Question 54.
Find the area cut off between the line y = 0 and the parabola y = x2 – 4x + 3.
Solution:

Equation of the parabola is
y = x2 – 4x + 3
Equation of the line is y = 0
x2 – 4x + 3 = 0
(x – 1) (x – 3) = 0
x = 1, 3
The curve takes negative values for the values of x between 1 and 3.
Required area = $$\int_{1}^{3}$$ -(x2 – 4x + 3)dx
= $$\int_{1}^{3}$$ (-x2 + 4x – 3) dx
= $$\left(-\frac{x^{3}}{3}+2 x^{2}-3 x\right)_{1}^{3}$$
= (-9 + 18 – 9) – (-$$\frac{1}{3}$$ + 2 – 3)
= $$\frac{1}{3}$$ – 2 + 3 = $$\frac{4}{3}$$

Question 55.
Find the area bounded by y = sin x and y = cos x between any two consecutive points of intersection.
Solution:

= $$\frac{1}{\sqrt{2}}$$ + $$\frac{1}{\sqrt{2}}$$ + $$\frac{1}{\sqrt{2}}$$ + $$\frac{1}{\sqrt{2}}$$ = 4$$\frac{1}{\sqrt{2}}$$ = 2$$\sqrt{2}$$

Question 56.
Find the area of one of the curvilinear triangles bounded by y = sin x, y = cos x and X – axis.
Solution:

ln (0, $$\frac{\pi}{4}$$) cos x ≥ sin x and ($$\frac{\pi}{4}$$, $$\frac{\pi}{2}$$), cos x ≤ sin x.
Required area = $$\int_{0}^{\pi / 4}$$ sin x dx + $$\int_{\pi / 4}^{\pi / 2}$$ cos x dx
= $$(-\cos x)_{0}^{\pi / 4}+(\sin x)_{\pi / 4}^{\pi / 2}$$
= – cos $$\frac{\pi}{4}$$ + cos 0 + sin $$\frac{\pi}{2}$$ – sin $$\frac{\pi}{4}$$
= – $$\frac{1}{\sqrt{2}}$$ + 1 + 1 – $$\frac{1}{\sqrt{2}}$$
= 2(1 – $$\frac{1}{\sqrt{2}}$$) = 2 – $$\sqrt{2}$$

Question 57.
Find the area of the right angled triangle with base b and altitude h, using the fundamental theorem of integral calculus.
Solution:

OAB is a right angled triangle and ∠B = 90° take ‘O’ as the origin and OB as positive X-axis
If OB = band AB = h, then Co-ordinates of A are (b, h)
Equation of OA is y = $$\frac{h}{b}$$ x
Area of the triangle OAB = $$\int_{0}^{b} \frac{h}{b} x d x$$
= $$\frac{h}{b}\left(\frac{x^{2}}{2}\right)_{0}^{b}=\frac{h}{b} \cdot \frac{b^{2}}{2}=\frac{1}{2} b h .$$

Question 58.
Find the area bounded between the curves y2 – 1 = 2x and x = 0.
Solution:

= $$\left(-\frac{y^{3}}{3}+y\right)_{0}^{1}$$ = 1 – $$\frac{1}{3}$$ = $$\frac{2}{3}$$

Question 59.
Find the area enclosed by the curves y = 3x and y = 6x – x2.
Solution:

y = 6x – x2
The straight line y = 3x meets the parabola
y = 6x – x2
3x = 6x = x2
x2 – 3x = 0
x(x – 3) = 0
x = 0 or 3
Required area = $$\int_{0}^{3}\left(6 x-x^{2}-3 x\right) d x$$
= $$\int_{0}^{3}\left(3 x-x^{2}\right) d x=\left(\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{3}$$
= $$\frac{27}{2}$$ – $$\frac{27}{3}$$ = $$\frac{27}{6}$$ = $$\frac{9}{2}$$

Question 60.
Find the area enclosed between y = x2 – 5x and y = 4 – 2x.
Solution:
Equations of the curves are
y = x2 – 5x ………. (1)
y = 4 – 2x …………….. (2)
x2 – 5x = 4 – 2x
x2 – 3x – 4 = 0
(x + 1)(x – 4) = 0
x = -1, 4

= 44 – $$\frac{64}{3}$$ – $$\frac{3}{2}$$ – $$\frac{1}{3}$$
= $$\frac{264-128-9-2}{6}$$ = $$\frac{125}{6}$$

Question 61.
Find the area bounded between the curves y = x2, y = $$\sqrt{x}$$.
Solution:

= $$\left(\frac{2}{3} x \sqrt{x}-\frac{x^{3}}{3}\right)_{0}^{1}$$
= $$\frac{2}{3}$$ – $$\frac{1}{3}$$ = $$\frac{1}{3}$$

Question 62.
Find the area bounded between the curves y2 = 4ax, x2 = 4by (a > 0, b > 0).
Solution:
Equations of the given curves are
y2 = 4ax …………………… (1)
x2 = 4by ……………………. (2)
From equation (2), y = $$\frac{x^{2}}{4 b}$$
Substituting in (1) $$\left(\frac{x^{2}}{4 b}\right)^{2}$$ = 4ax
x4 = (16 b2) |4ax|