NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.11

These NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.11 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.11

प्रश्न 1.
\(\int_0^{\pi / 2}\) cos² xdx
हल:
माना I = \(\int_0^{\pi / 2}\) cos² x dx ……………….(1)
या
I = \(\int_0^{\pi / 2}\) cos²(\(\frac{\pi}{2}\) – x)dx
I = \(\int_0^{\pi / 2}\) sin² x dx
(1) तथा (2) को जोड़ने पर,
2I = \(\int_0^{\pi / 2}\) cos² xdx + \(\int_0^{\pi / 2}\) sin² x dx
= \(\int_0^{\pi / 2}\)(cos² x + sin² x) dx
= \(\int_0^{\pi / 2}\) 1 dx = [x]^π/2 ₀
2I = \(\frac{\pi}{2}\)
∴ I = \(\frac{\pi}{4}\)

प्रश्न 2.
\(\int_0^{\pi / 2}\) \(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx
हल:
माना
\(\int_0^{\pi / 2}\) \(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx …(1)
या
I = \(\int_0^{\pi / 2}\) \(\frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}\) dx (निश्चित समाकलन के गुणधर्म से)
या
I = \(\int_0^{\pi / 2}\)\(\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx …………..(2)
समीकरण (1) तथा (2) को जोड़ने पर,
2I = \(\int_0^{\pi / 2}\)\(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\)dx + \(\int_0^{\pi / 2}\)\(\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx
= \(\int_0^{\pi / 2}\)\(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx
= \(\int_0^{\pi / 2}\) dx = [x]^π/2 ₀
या 2I = \(\frac{\pi}{2}\)
∴ I = \(\frac{\pi}{4}\)

प्रश्न 3.
\(\int_0^{\pi / 2}\)\( \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x}\) dx
हल:

प्रश्न 4.
\(\int_0^{\pi / 2}\)\(\frac{\cos ^5 x d x}{\sin ^5 x+\cos ^5 x}\) dx
हल:
माना
I = \(\int_0^{\pi / 2}\)\(\frac{\cos ^5 x d x}{\sin ^5 x+\cos ^5 x}\) dx …………(1)
I = \(\int_0^{\pi / 2}\)\(\frac{\cos ^5\left(\frac{\pi}{2}-x\right) d x}{\sin ^5\left(\frac{\pi}{2}-x\right)+\cos ^5\left(\frac{\pi}{2}-x\right)}\) dx
या
I = \(\int_0^{\pi / 2}\)\(\frac{\sin ^5 x d x}{\cos ^5 x+\sin ^5 x}\) ………..(2)
(1) तथा (2) को जोड़ने पर,
2I = \(\int_0^{\pi / 2}\)\(\frac{\cos ^5 x d x}{\sin ^5 x+\cos ^5 x}\) + \(\int_0^{\pi / 2}\)\(\frac{\sin ^5 x}{\cos ^5 x+\sin ^5 x}\) dx
= \(\int_0^{\pi / 2}\)\(\frac{\sin ^5 x+\cos ^5 x}{\sin ^5 x+\cos ^5 x}\) dx
= \(\int_0^{\pi / 2}\)dx = [x]^π/2 ₀
या 2I = \(\frac{\pi}{2}\)
∴ I = \(\frac{\pi}{4}\)

प्रश्न 5.
\(\int_{-5}^5\) |x + 2| dx
हल:

प्रश्न 6.
\(\int_2^8\) |x – 5| dx
हल:
माना I = \(\int_2^8\) |x – 5| dx
= \(\int_2^5\)|x – 5|dx + \(\int_5^8\)|x – 5| dx
= \(\int_2^5\) – (x – 5) dx + \(\int_5^8\)(x – 5) dx
[अन्तराल [2, 5] में | x – 5 | = – (x – 5) क्योंकि x < 5 अर्थात् 2 < x < 5]
= –\(\int_2^5\) – (x – 5) dx + \(\int_5^8\)(x – 5) dx
= –\(\frac{x^2}{2}\) – 5x]⁵₂ + \(\frac{x^2}{2}\) – 5x]⁸₅
= –\(\frac{1}{2}\) [x² ]⁵₂ + 5[x]⁵ ₂ + \(\frac{1}{2}\)[x²]⁸₅ – 5[x]⁸₅
= –\(\frac{1}{2}\)(25 – 4) + 5(5 – 2) + \(\frac{1}{2}\)(64 – 25) – 5(8 – 5)
= –\(\frac{21}{2}\) + 15 + \(\frac{39}{2}\) – 15
= \(\frac{39-21}{2}\) = \(\frac{18}{2}\) = 9
∴ I = 9

प्रश्न 7.
\(\int_0^1\) x(1 – x)ⁿ dx
हल:
माना I = \(\int_0^1\) x(1 – x)ⁿ dx
तब
I = \(\int_0^1\)(1 – x) (1 – (1 – x)ⁿ dx
[ ∴\(\int_a^0\) f(x) dx = \(\int_a^0\)f(a – x)dx]
= \(\int_a^0\)(1 – x)(1 – 1 + x)ⁿ dx
= \(\int_a^0\)(1 – x)xⁿ dx
या I = \(\int_a^0\)(xⁿ – xⁿ+1)dx

प्रश्न 8.
\(\int_0^{\pi / 4}\) log (1+ tan x) dx
हल:
माना,

प्रश्न 9.
\(\int_0^2\) x√2 – x dx
हल:

प्रश्न 10.
\(\int_0^{\pi / 2}\)(2 log sin x – log sin 2x) dx
हल:
माना I = \(\int_0^{\pi / 2}\)(2 log sin x – log sin 2x) dx
= \(\int_0^{\pi / 2}\)[2log sin x – log (2 sin x cos x)] dx
= \(\int_0^{\pi / 2}\)[2logsinx – log2 – logsin.x – logcos.x] dx
= \(\int_0^{\pi / 2}\)[log sin x – log 2 – log cos.x] dx
= \(\int_0^{\pi / 2}\)log sin x dx – log 2\(\int_0^{\pi / 2}\) dx – \(\int_0^{\pi / 2}\)log cos x dx
= \(\int_0^{\pi / 2}\)log sin x dx – log2\(\int_0^{\pi / 2}\)dx – \(\int_0^{\pi / 2}\)log cos cos(\(\frac{\pi}{2}\) – x)dx
[∴ \(\int_0^a\) f(x) dx = [ \(\int_0^a\) f(a – x) dx]
= \(\int_0^{\pi / 2}\) log sin x dx – log 2 [x]^π/2₀ – \(\int_0^{\pi / 2}\) log sin x dx
= – log2 x \(\frac{\pi}{2}\) = –\(\frac{\pi}{2}\)log 2 = \(\frac{\pi}{2}\)log(2)^-1
= \(\frac{\pi}{2}\) log \(\frac{1}{2}\)
∴ I = \(\frac{\pi}{2}\) log \(\frac{1}{2}\)

प्रश्न 11.
\(\int_{-\pi / 2}^{\pi / 2}\) sin² x dx
हल:
माना I = \(\int_{-\pi / 2}^{\pi / 2}\) sin² x dx
∵ sin² x एक समफलन है।
∴ I = 2\(\int_0^{\pi / 2}\) sin² x dx ………..(1)
[∵ \(\int_{-a}^a\) f'(x) dx = 2 \(\int_0^a\) f(x) dx]
यदि f सम फलन है।
I = 2\(\int_0^{\pi / 2}\) [sin² (\(\frac{\pi}{2}\) – x)dx]
I = 2\(\int_0^{\pi / 2}\) cos² x dx ……(2)
[∵ \(\int_0^a\) ƒ (x) dx = \(\int_0^a\) ƒ (a − x) dx]
समीकरण (1) तथा (2) को जोड़ने पर,
2I = 2\(\int_0^{\pi / 2}\) sin² x dx + 2\(\int_0^{\pi / 2}\)cos² x dx
= 2[\(\int_0^{\pi / 2}\) (cos² x + sin²x)dx]
= 2\(\int_0^{\pi / 2}\) dx = 2[x]^π/2 ₀ = 2 × \(\frac{\pi}{2}\) = π
∴ 2I = π
∴ I = \(\frac{\pi}{2}\)

प्रश्न 12.
\(\int_0^\pi\) \(\frac{x}{1+\sin x}\) dx
हल:

प्रश्न 13.
\(\int_{-\pi / 2}^{\pi / 2}\) sin⁷ x dx
हल:
माना I = \(\int_{-\pi / 2}^{\pi / 2}\) sin⁷ x dx
तब f(-x) = sin⁷ (-x)
या
= [sin(-x)]⁷ = [- sin x]⁷
= f(-x) = – sin⁷ x = -f(x)
अर्थात् sin⁷ x विषम फलन है।
∵ \(\int_{-a}^a\) f(x) dx = 0
यदि f(x) = -f(x)
अर्थात् विषम फलन है।
∴ I = \(\int_{-\pi / 2}^{\pi / 2}\) sin⁷ x dx = 0

प्रश्न 14.
\(\int_0^{2 \pi}\) cos⁵ x dx
हल:
माना I = \(\int_0^{2 \pi}\) cos⁵ x dx
अब f(2π – x) = cos⁵ (2л – x)
= cos⁵ x = f(x)
∴ I = \(\int_0^{2 \pi}\) cos⁵ x dx
= 2 \(\int_0^\pi\)cos⁵ x dx ……….(1)
∴ I = 2\(\int_0^\pi\)cos⁵(π – x) dx
= 2\(\int_0^\pi\) – cos⁵ x dx …………(2)
[∴ \(\int_0^a (x)\) f(x) dx = [\(\int_0^a (x)\) ƒ (a – x) dx]
समीकरण (1) तथा (2) को जोड़ने पर,
2I = 2 \(\int_0^\pi\)cos⁵x dx – 2\(\int_0^\pi\) cos⁵ x dx = 0
∴ I = 0

प्रश्न 15.
\(\int_0^{\pi / 2}\) \(\frac{\sin x-\cos x}{1+\sin x \cos x}\) dx
हल:

प्रश्न 16.
\(\int_0^\pi\) log (1 + cos x) dx
हल:
माना I = \(\int_0^\pi\) log (1 + cos x) dx ……(i)
= \(\int_0^\pi\)log [1 + cos (π − x)] dx
I = \(\int_0^\pi\)log (1 – cos x) dx ……….(ii)
समीकरण (i) तथा (ii) को जोड़ने पर,
2I = \(\int_0^\pi\)log (1+ cos x) dx + \(\int_0^\pi\)log (1 – cos x) dx
= \(\int_0^\pi\)log (1 + cos x) (1 – cos x) dx
= \(\int_0^\pi\)log (1 – cos² x) dx
= \(\int_0^\pi\)log sin² x dx
= 2 \(\int_0^\pi\)long sin x dx
I = \(\int_0^\pi\) long sin x dx
= 2\(\int_0^{\pi / 2}\) log sin x dx = 2I₁ ……(iii)
[∴ \(\int_0^{2 a}\) f(x) dx = 2 \(\int_0^a\) f(x) dx]
जब f(2a – x) = f(x)
अब I₁ = \(\int_0^{\pi / 2}\) log sin x dx ………..(iv)
या
I₁ = \(\int_0^{\pi / 2}\) log sin(\(\frac{\pi}{2}\) – x)dx
या
I₁ = \(\int_0^{\pi / 2}\) log cos x dx ……….(v)
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
(iv) तथा (v) को जोड़ने पर,
2I₁ = \(\int_0^{\pi / 2}\) log sin x dx + \(\int_0^{\pi / 2}\) log cos x dx
= \(\int_0^{\pi / 2}\)[log sin x + log cos x] dx
= \(\int_0^{\pi / 2}\) log(sin x cos x) dx
= \(\int_0^{\pi / 2}\) log\(\left(\frac{2 \sin x \cos x}{2}\right)\) dx
= \(\int_0^{\pi / 2}\)[log (2 sin x cos x) – log2] dx
या
2I₁ = log sin 2x dx – log 2\(\int_0^{\pi / 2}\) dx
2I₁ = \(\int_0^{\pi / 2}\)log sin 2x – log 2[x]^π/2 ₀
= \(\int_0^{\pi / 2}\)log sin 2.x dx – log2[\(\frac{\pi}{2}\) – 0]
2I₁ = \(\int_0^{\pi / 2}\) log sin 2.x dx – \(\frac{\pi}{2}\)log 2 ………….(vi)
या
2I₁ = I₂ – \(\int_0^{\pi / 2}\)log2
अब I₂ = \(\int_0^{\pi / 2}\) log sin 2x dx
2x = t रखने पर,
2 dx = dt dx = \(\frac{1}{2}\) dt
जब x = 0 तब t = 0, जब t = π
I₂ = \(\int_0^\pi\) log sint \(\frac{d t}{2}\)
= \(\frac{1}{2}\)\(\int_0^\pi\)log sin t dt
= \(\frac{1}{2}\) × 2log (sin t) dt
= \(\int_0^{\pi / 2}\) log sin x dx = I₁
∴ I2 = I₁
समीकरण (vi) से,
2I₁ = I₁ – \(\frac{\pi}{2}\) log 2
∴ I₁ = –\(\frac{\pi}{2}\) log 2
I₁ का मान समीकरण (iii) में रखने पर,
I = 2I₁ = 2(-\(\frac{\pi}{2}\) log 2) = -π log 2
∴ I = -π log 2

प्रश्न 17.
\(\int_0^a\) \(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\) dx
हल:

प्रश्न 18.
\(\int_0^4\)|x – 1| dx
हल:

प्रश्न 19.
दर्शाइए कि
\(\int_0^a\) f(x) g(x) dx = 2\(\int_0^a\) f(x) dx
यदि f और g को f(x) = f(a – x) एवं g(x) + g(a – x) = 4 के रूप में परिभाषित किया गया है।
हल:
माना I = \(\int_0^a\) f(x) g(x) dx
I = \(\int_0^a\)f(a – x)g(a – x) dx ………..(1)
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
अब f(a – x) = f(x)
तथा g(x) + g(a – x) = 4
∴ g(a – x) = 4 – g(x)
f(a – x) तथा g(a – x) के मान समीकरण (1) में रखने पर,
I = \(\int_0^a\) f(x) [4 – g(x)] dx
= 4\(\int_0^a\) f(x) dx – \(\int_0^a\) f(x)g(x) dx
या I = 4\(\int_0^a\) f(x) dx – I
या 2I = 4\(\int_0^a\) f(x) dx
या I = 2\(\int_0^a\) f(x) dx

प्रश्न 20.
\(\int_{-\pi / 2}^{\pi / 2}\) (x³ + xcosx + tan⁵ x + 1)
(A) 0
(B) 2
(C) π
(D) 1
हल:
माना I = \(\int_{-\pi / 2}^{\pi / 2}\)(x³ + xcosx + tan⁵ x + 1) dx
= \(\int_{-\pi / 2}^{\pi / 2}\)(x³ + xcosx + tan⁵ x)dx + \(\int_{-\pi / 2}^{\pi / 2}\) dx
I = I₁ + \(\int_{-\pi / 2}^{\pi / 2}\) dx ………….(1)
पुन: f(x) = x³ + xcosx + tan⁵ x
तब f(-x) = (-x)³ + (-x) cos(-x) + tan⁵ x
= (-x)³ – xcosx – tan⁵ x
= -(x³ + xcosx + tan⁵ x)
= -f(x)
अर्थात् f विषम फलन है क्योंकि
f(-x) = -f(x)
∴ I₁ = \(\int_{-\pi / 2}^{\pi / 2}\)(x³ + xcosx + tan⁵ x)dx
= 0
[∵ \(\int_{-a}^a\) f(x) dx = 0 f विषम फलन है]
अब समीकरण (1) से,
I = I₁ + \(\int_{-\pi / 2}^{\pi / 2}\) dx = 0 + [x]^π/2 _π/2
= [\(\frac{\pi}{2}\) – (-\(\frac{\pi}{2}\)] = [\(\frac{\pi}{2}\) + \(\frac{\pi}{2}\)] = π
∴ I = π
अतः विकल्प (C) सही है।

प्रश्न 21.
\(\int_0^{\pi / 2}\) log\(\left(\frac{4+3 \sin x}{4+3 \cos x}\right)\) dx का मान है:
(A) 2
(B) \(\frac{3}{4}\)
(C) 0
(D) – 2
हल: