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

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 1

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

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 5.
I = \(\int_{0}^{1} \frac{x^{2}}{x^{2}+1} d x\) [Mar. 11]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 2

Question 6.
\(\int_{0}^{2 \pi}\) sin2x cos4; x dx [T.S. Mar. 15; Mar 14]
Solution:
sin2x cos4x is even function.
Inter 2nd Year Maths 2B Definite Integrals Important Questions 3

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 = \(\frac{\pi}{2}\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 4

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 . \(\frac{3}{6}\) . \(\frac{1}{4}\) . \(\frac{1}{4}\) . \(\frac{\pi}{4}\) = \(\frac{\pi}{16}\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 9.
\(\int_{0}^{\pi / 2} \cdot \frac{\sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x\) [Mar. 14, 08]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 5

Question 10.
I = \(\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\) [Mar. 08]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 6
= –\(\frac{1}{40} \ln \left[\frac{1 / 4}{9 / 4}\right]\) = \(\frac{1}{40}\) . 2ln . 3 = \(\frac{1}{20}\) ln 3

Question 11.
y = x3 + 3, y = 0, x = -1, x = 2 [Mar. 05]
Solution:
Required area PABQ
Inter 2nd Year Maths 2B Definite Integrals Important Questions 7

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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
Inter 2nd Year Maths 2B Definite Integrals Important Questions 8
Required area = 2 Area of OAP
Inter 2nd Year Maths 2B Definite Integrals Important Questions 9

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 10
= \(\frac{1}{3}\left[\ln \frac{4}{2}\right]=\frac{1}{3} \ln 2\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 15.
\(\int_{0}^{\pi} \frac{x}{1+\sin x} d x\) [May 11]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 11
Inter 2nd Year Maths 2B Definite Integrals Important Questions 12

Question 16.
\(\int_{0}^{\pi} \frac{x \sin ^{3} x}{1+\cos ^{2} x} d x\) [T.S. Mar. 15; Mar. 11]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 13
Inter 2nd Year Maths 2B Definite Integrals Important Questions 14

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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 = \(\frac{\pi}{4}\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 15
Inter 2nd Year Maths 2B Definite Integrals Important Questions 16

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 18.
\(\int_{0}^{\pi / 4} \log (1+\tan x) d x\) [A.P. Mar. 16]
Solution:
I = \(\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 17

Question 19.
y = 4x – x2, y = 5 – 2x. [T.S. Mar. 16]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 18
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
Inter 2nd Year Maths 2B Definite Integrals Important Questions 19
= 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

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 20
The ellipse is symmetrical about X and Y axis
Area of the ellipse = 4 Area of CAB
= 4 . \(\frac{\pi}{4}\) ab
Inter 2nd Year Maths 2B Definite Integrals Important Questions 21
(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
Inter 2nd Year Maths 2B Definite Integrals Important Questions 22
= \(\int_{\pi / 6}^{\pi / 3} d x=(x)_{\pi / 6}^{\pi / 3}\)
= \(\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)
A = \(\frac{\pi}{12}\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 23
Inter 2nd Year Maths 2B Definite Integrals Important Questions 24
= \(\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}\)
= 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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 25
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:
\(\int_{1}^{2} x^{5} \cdot d x=\left[\frac{x^{6}}{6}\right]_{1}^{2}\)
= \(\frac{2^{6}}{6}-\frac{1}{6}=\frac{63}{6}=\frac{21}{2}\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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
Inter 2nd Year Maths 2B Definite Integrals Important Questions 26

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) = \(\sqrt{4-x^{2}}\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 27

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 30.
Evaluate \(\int_{0}^{16} \frac{x^{1 / 4}}{1+x^{1 / 2}} d x\)
Solution:
Put t4 = x ⇒ dx = 4t3 . dt
Inter 2nd Year Maths 2B Definite Integrals Important Questions 28

Question 31.
Evaluate \(\int_{-\pi / 2}^{\pi / 2} \sin |x| d x\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 29
Inter 2nd Year Maths 2B Definite Integrals Important Questions 30

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.
Inter 2nd Year Maths 2B Definite Integrals Important Questions 31

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 33.
Evaluate \(\int_{0}^{\pi / 2} \frac{\cos ^{5 / 2} x}{\sin ^{5 / 2} x+\cos ^{5 / 2} x} d x\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 32
= \(\int_{0}^{\pi / 2} d x=(x)_{0}^{\pi / 2}=\frac{\pi}{2}\)
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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 33
Inter 2nd Year Maths 2B Definite Integrals Important Questions 34
= \(\frac{\pi}{4 \sqrt{2}}\) log (\(\sqrt{2}\) + 1)2
= \(\frac{\pi}{4 \sqrt{2}}\) 2 log (\(\sqrt{2}\) + 1)
= \(\frac{\pi}{4 \sqrt{2}}\) log (\(\sqrt{2}\) + 1)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 35.
Evaluate \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) [Mar 14]
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 35
Inter 2nd Year Maths 2B Definite Integrals Important Questions 36

Question 36.
Find \(\int_{-a}^{a}\left(x^{2}+\sqrt{a^{2}-x^{2}}\right) d x\)
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 37
= 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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 38
Inter 2nd Year Maths 2B Definite Integrals Important Questions 39
= \(\left(2 \tan \frac{x}{2}\right)_{0}^{\pi / 2}\)
= 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

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 40

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 41

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 42

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.
Inter 2nd Year Maths 2B Definite Integrals Important Questions 43
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) \(\int_{0}^{\pi / 2} \cos ^{8} x d x\)
= \(\frac{7}{8}\) . \(\frac{5}{6}\) . \(\frac{3}{4}\) . \(\frac{1}{2}\) . \(\frac{\pi}{2}\) = \(\frac{35\pi}{256}\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 44
\(\int_{0}^{\pi}\) 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
Inter 2nd Year Maths 2B Definite Integrals Important Questions 45

Inter 2nd Year Maths 2B Definite Integrals Important Questions

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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 46
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 \(\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\)
Inter 2nd Year Maths 2B Definite Integrals Important Questions 47
= 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.

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 53.
Find the area bounded by the parabola y = x2, the X-axis and the lines x = -1, x = 2.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 48

Question 54.
Find the area cut off between the line y = 0 and the parabola y = x2 – 4x + 3.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 49
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 = \(\frac{4}{3}\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 55.
Find the area bounded by y = sin x and y = cos x between any two consecutive points of intersection.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 50
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = 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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 51
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:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 52
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 = \(\int_{0}^{b} \frac{h}{b} x d x\)
= \(\frac{h}{b}\left(\frac{x^{2}}{2}\right)_{0}^{b}=\frac{h}{b} \cdot \frac{b^{2}}{2}=\frac{1}{2} b h .\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 58.
Find the area bounded between the curves y2 – 1 = 2x and x = 0.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 53
= \(\left(-\frac{y^{3}}{3}+y\right)_{0}^{1}\) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 59.
Find the area enclosed by the curves y = 3x and y = 6x – x2.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 54
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
Inter 2nd Year Maths 2B Definite Integrals Important Questions 55
= 44 – \(\frac{64}{3}\) – \(\frac{3}{2}\) – \(\frac{1}{3}\)
= \(\frac{264-128-9-2}{6}\) = \(\frac{125}{6}\)

Inter 2nd Year Maths 2B Definite Integrals Important Questions

Question 61.
Find the area bounded between the curves y = x2, y = \(\sqrt{x}\).
Solution:
Inter 2nd Year Maths 2B Definite Integrals Important Questions 56
= \(\left(\frac{2}{3} x \sqrt{x}-\frac{x^{3}}{3}\right)_{0}^{1}\)
= \(\frac{2}{3}\) – \(\frac{1}{3}\) = \(\frac{1}{3}\)

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|
Inter 2nd Year Maths 2B Definite Integrals Important Questions 57
Inter 2nd Year Maths 2B Definite Integrals Important Questions 58