NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.9

These NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.9 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.9

प्रश्न 1.
\(\int_{-1}^1\) (x + 1) dx
हल:
\(\int_{-1}^1\) (x + 1) dx
= \(\int_{-1}^1\) x dx + \(\int_{-1}^1\)dx
= \(\left[\frac{x^2}{2}\right]_{-1}^1\) + \([x]_{-1}^1\)
= [\(\frac{1}{2}\) – \(\frac{(-1)^2}{2}\)] + [1 – (-1)]
= \(\frac{1}{2}\) – \(\frac{1}{2}\) + 1 + 1 = 2

प्रश्न 2.
\(\int_{3}^2\) \(\frac{1}{x}\) dx
हल:
\(\int_{3}^2\) \(\frac{1}{x}\) dx = \([\log x]_2^3\)
= log 3 – log 2 = log \(\frac{3}{2}\)

प्रश्न 3.
\(\int_{2}^1\) (4x³ – 5x² + 6x + 9) dx
हल:
\(\int_{2}^1\) (4x³ – 5x² + 6x + 9) dx
= [4.\(\frac{x^4}{4}\) – 5.\(\frac{x^3}{3}\) + 6\(\frac{x^2}{2}\) + 9x]²₁
= [x⁴ – \(\frac{5}{3}\)x³ + 3x² + 9x]²₁
= (2⁴ – 1⁴) – \(\frac{5}{3}\)(2³ – 1³⁾ + 3(2² – 1²) + 9(2 – 1)
= 16 – 1 – \(\frac{5}{3}\)(8 – 1) + 3(4 – 1) + 9
= 15 – \(\frac{5}{3}\) × 7 + 9 + 9
= 15 – \(\frac{35}{3}\) + 18 = \(\frac{45-35+54}{3}\) = \(\frac{64}{3}\)

प्रश्न 4.
\(\int_0^{\pi / 4}\) sin 2x dx
हल:
\(\int_0^{\pi / 4}\) sin 2x dx
= [ – \(\frac{\cos 2 x}{2}\)]^π/4₀
= – \(\frac{1}{2}\)[cos2x]^π/4₀
= – \(\frac{1}{2}\)[cos2 × \(\frac{\pi}{4}\) – cos2 × 0]
= – \(\frac{1}{2}\)[cos\(\frac{\pi}{2}\) – cos0]
= – \(\frac{1}{2}\)[0 – 1] = \(\frac{1}{2}\)

प्रश्न 5.
\(\int_0^{\pi / 2}\)cos 2x dx
हल:
\(\int_0^{\pi / 2}\)cos 2x dx
= \(\frac{\sin 2 x}{2}\)^π/2 ₀
= \(\frac{1}{2}\)[sin2x]^π/2 ₀
= \(\frac{1}{2}\)[sin2 × \(\frac{\pi}{2}\) – sin2 × 0 ]
= \(\frac{1}{2}\)[sinπ – sin0]
= \(\frac{1}{2}\)[0 – 0] = 0

प्रश्न 6.
\(\int_4^5\) e^x dx
हल:
\(\int_4^5\) e^x dx = [e^x]⁵ ₄
= e⁵ – e⁴ = e⁴(e – 1)

प्रश्न 7.
\(\int_0^{\pi / 4}\) tan x dx
हल:
\(\int_0^{\pi / 4}\) tan x dx = [log secx]^π/4 ₀
= (log sec\(\frac{\pi}{2}\) – log sec 0)
= (log √2 – log 1)
= log 2^1/2 – 0 = \(\frac{1}{2}\)log 2

प्रश्न 8.
\(\int_{\pi / 6}^{\pi / 4}\) cosec x dx
हल:
\(\int_{\pi / 6}^{\pi / 4}\) cosec x dx
= [log (cosec x – cot x)]^π/4 _π/6
= log (cosec\(\frac{\pi}{4}\) – cot\(\frac{\pi}{4}\) – log (cosec\(\frac{\pi}{6}\) – cot\(\frac{\pi}{6}\))
= (log√2 – log 1) – log(2 – √3)
= log\(\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

प्रश्न 9.
\(\int_0^1\) \(\frac{d x}{\sqrt{1-x^2}}\)
हल:
\(\int_0^1\) \(\frac{d x}{\sqrt{1-x^2}}\)
[sin^-1 x]¹ ₀
(sin^-11 – sin^-10]
= \(\frac{\pi}{2}\)

प्रश्न 10.
\(\int_0^1\) \(\frac{d x}{1+x^2}\)
हल:
\(\int_0^1\) \(\frac{d x}{1+x^2}\)
= [tan^-1 x]¹ ₀
= tan^-1 1 – tan^-1 0
= \(\frac{\pi}{4}\) – 0 = \(\frac{\pi}{4}\)

प्रश्न 11.
\(\int_2^3\) \(\frac{d x}{x^2-1}\)
हल:
\(\int_2^3\) \(\frac{d x}{x^2-1}\) = \(\frac{1}{2}\)[log\(\left|\frac{x-1}{x+1}\right|\)]³
₂
= \(\frac{1}{2}\) [log \(\left|\frac{3-1}{3+1}\right|\) – log\(\left|\frac{2-1}{2+1}\right|\)]
= \(\frac{1}{2}\) [log\(\frac{2}{4}\) – log \(\frac{1}{3}\)]
= \(\frac{1}{2}\) [log\(\frac{1}{2}\) – log \(\frac{1}{3}\)]
= \(\frac{1}{2}\) log \(\frac{1 / 2}{1 / 3}\) = \(\frac{1}{2}\) log \(\frac{3}{2}\)

प्रश्न 12.
\(\int_0^{\pi / 2}\) cos² x dx
हल:
\(\int_0^{\pi / 2}\) cos² x dx
= \(\int_0^{\pi / 2}\) \(\left(\frac{1+\cos 2 x}{2}\right)\) dx
(∵ cos 2x = 2 cos² x – 1)
= \(\frac{1}{2}\)\(\int_0^{\pi / 2}\) dx – \(\frac{1}{2}\)\(\int_0^{\pi / 2}\) cos 2x dx
= \(\frac{1}{2}\) [x]^π/2 ₀ + \(\frac{1}{2}\)\(\frac{\sin 2 x}{2}\)^π/2 ₀
= \(\frac{1}{2}\)[\(\frac{\pi}{2}\) – 0] + \(\frac{1}{4}\)[sin2 × \(\frac{\pi}{2}\) – sin2 × 0]
= \(\frac{1}{2}\) × \(\frac{\pi}{2}\) + \(\frac{1}{4}\)(sinπ – sino)
= \(\frac{\pi}{4}\) + \(\frac{1}{4}\)(0 – 0) = \(\frac{\pi}{4}\) – 0 = \(\frac{\pi}{4}\)

प्रश्न 13.
\(\int_2^3\) \(\frac{x d x}{x^2+1}\)
हल:
\(\int_2^3\) \(\frac{x d x}{x^2+1}\)
माना x² + 1 = t
तब 2x dx = dt या xdx = \(\frac{1}{2}\) dt
जब x = 3 तब t = 3² + 1 = 10
जब x = 2 तब t = 2² + 1 = 5
∴ \(\int_2^3\) \(\frac{x d x}{x^2+1}\)
= \(\frac{1}{2}\) \(\int_10^5\) \(\frac{d t}{t}\)
= \(\frac{1}{2}\)[logt]¹⁰₅
= \(\frac{1}{2}\)[log10 – log 5]
= \(\frac{1}{2}\) log \(\frac{10}{5}\) = \(\frac{1}{2}\) log 2

प्रश्न 14.
\(\int_1^0\)\(\frac{2 x+3}{5 x^2+1}\) dx
हल:

प्रश्न 15.
\(\int_1^0\) xex² dx
हल:
\(\int_1^0\) xex² dx
माना x² = t तब 2x dx = dt या x dx = \(\frac{1}{2}\) dt
जब x = 1 t = 1
जब x = 0 t = 0
∴ \(\int_1^0\) xex² dx = \(\frac{1}{2}\)\(\int_1^0\)ex^t dt
= \(\frac{1}{2}\) [e^t]¹ ₀
= \(\frac{1}{2}\)[e^t – e⁰] = \(\frac{1}{2}\)(e – 1)

प्रश्न 16.
\(\int_2^1\)\(\frac{5 x^2}{x^2+4 x+3}\) dx
हल:

प्रश्न 17.
\(\int_0^{\pi / 4}\) (2sec² x + x³ + 2)dx
हल:
\(\int_0^{\pi / 4}\) (2sec² x + x³ + 2)dx
= 2\(\int_0^{\pi / 4}\)sec²x dx + \(\int_0^{\pi / 4}\)x³dx + 2\(\int_0^{\pi / 4}\) dx
= 2(tanx)^π/4 ₀ + \(\frac{x^4}{4}\)^π/4 ₀ + 2[x]^π/4 ₀
= 2[tan\(\frac{\pi}{4}\) – tan0] + \(\frac{1}{4}\)[\(\frac{\pi}{4}\)⁴ – 0] + 2[\(\frac{\pi}{4}\) – 0]
= 2[1 – 0] + \(\frac{1}{4}\)\(\frac{\pi}{4}\)⁴ + 2 × \(\frac{\pi}{4}\)
= 2 + \(\frac{1}{4}\)(\(\frac{\pi}{4}\)⁴) + \(\frac{\pi}{2}\) = 2 + \(\frac{\pi}{4}\) + \(\frac{\pi}{1024}\)

प्रश्न 18.
\(\int_0^\pi\)(sin²\(\frac{x}{2}\) – cos²\(\frac{x}{2}\))dx
हल:
\(\int_0^\pi\)(sin²\(\frac{x}{2}\) – cos²\(\frac{x}{2}\))dx
= – \(\int_0^\pi\)(cos²\(\frac{x}{2}\) – sin²\(\frac{x}{2}\))dx
= –\(\int_0^\pi\)cosx dx = – [sinx]^π ₀
= – [sinπ – sin 0] = 0 – 0 = 0

प्रश्न 19.
\(\int_0^2\)\(\frac{6 x+3}{x^2+4}\) dx
हल:
\(\int_0^2\)\(\frac{6 x+3}{x^2+4}\) dx
= \(\int_0^2\)\(\frac{6 x}{x^2+4}\)dx + \(\int_0^2\)\(\frac{3 d x}{x^2+4}\)
= 3\(\int_0^2\)\(\frac{2 x}{x^2+4}\)dx + \(\int_0^2\)\(\frac{d x}{x^2+4}\)
= 3[log|x² + 4|]² ₀ + 3 × \(\frac{1}{2}\)[tan^-1 \(\frac{x}{2}\)]² ₀
= x² + 4 = t
2x dx = dt
∴ ∫\(\frac{2 x}{x^2+4}\)dx = ∫\(\frac{d t}{t}\)
= log t = log|x² + 4|
= 3[log|2² + 4| – log|0² + 4|] + \(\frac{3}{2}\) × [tan^-1\(\frac{2}{2}\) – tan^-1\(\frac{0}{2}\)]
= 3 [log|8| – log|4|] + \(\frac{3}{2}\)|tan^-1| 1 – 0]
= 3 log\(\frac{8}{4}\) + \(\frac{3}{2}\) × \(\frac{\pi}{4}\)
= 3log2 + \(\frac{3 \pi}{8}\)

प्रश्न 20.
\(\int_0^1\)(xe^x + sin\(\frac{\pi x}{4}\))dx
हल:
\(\int_0^1\)(xe^x + sin\(\frac{\pi x}{4}\))dx
= \(\int_0^1\)xe^x dx + \(\int_0^1\)(sin \(\frac{\pi x}{4}\))dx ………….(1)
अब ∫xe^xdx = x∫e^xdx -∫(\(\frac{d}{d x}\) x∫e^xdx)dx
या ∫xe^xdx = xe^x – ∫1e^xdx
या ∫xe^xdx = xe^x – e^x …………(2)
तथा ∫sin\(\frac{\pi x}{4}\) dx = – \(\frac{\cos \left(\frac{\pi x}{4}\right)}{\pi / 4}\)
= – \(\frac{4}{\pi}\)cos (\(\frac{\pi x}{4}\)) …………..(3)
समीकरण (1), (2) तथा (3) से,
\(\int_0^1\)(xe^x + sin\(\frac{\pi x}{4}\))dx
= [xe^x – e^x]¹ ₀ – \(\frac{4}{\pi}\)[cos\(\frac{\pi x}{4}\)]¹ ₀
= (1e¹ – e¹) – (0.e⁰ – e⁰) – \(\frac{4}{\pi}\)[cos\(\frac{4}{\pi}\) – cos0]
= (e – e) – (0 – 1) – \(\frac{4}{\pi}\)[\(\frac{1}{\sqrt{2}}\) – 1]
= 1 – \(\frac{4}{\pi}\) × \(\frac{1}{\sqrt{2}}\) + \(\frac{4}{\pi}\) = 1 + \(\frac{4}{\pi}\) – \(\frac{2 \sqrt{2}}{\pi}\)

प्रश्न 21.
\(\int_1^{\sqrt{3}}\) \(\frac{d x}{1+x^2}\) dx बराबर है:
(A) \(\frac{\pi}{3}\)
(B) \(\frac{2 \pi}{3}\)
(C) \(\frac{\pi}{6}\)
(D) \(\frac{\pi}{12}\)
हल:

प्रश्न 22.
\(\int_0^{2 / 3}\) \(\frac{d x}{4+9x^2}\) dx बराबर है:
(A) \(\frac{\pi}{6}\)
(B) \(\frac{\pi}{12}\)
(C) \(\frac{\pi}{24}\)
(D) \(\frac{\pi}{4}\)
हल: