Maths MCQs for Class 12 with Answers Chapter 7 Integrals
Integrals Class 12 Maths MCQs Pdf
1. Given ∫ 2x dx = f(x) + C, then f(x) is
Answer/Explanation
Answer: c
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2.
(a) sin² x – cos² x + C
(b) -1
(c) tan x + cot x + C
(d) tan x – cot x + C
Answer/Explanation
Answer: d
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3.
(a) 2(sin x + x cos θ) + C
(b) 2(sin x – x cos θ) + C
(c) 2(sin x + 2x cos θ) + C
(d) 2(sin x – 2x cos θ) + C
Answer/Explanation
Answer: a
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4. ∫cot²x dx equals to
(a) cot x – x + C
(b) cot x + x + C
(c) -cot x + x + C
(d) -cot x – x + C
Answer/Explanation
Answer: d
Explaination: (d), ∫ (cosec²x -1)dx = -cot x – x + C
5.
(a) log |sin x + cos x|
(b) x
(c) log |x|
(d) -x
Answer/Explanation
Answer: d
Explaination:
6. If ∫ sec²(7 – 4x)dx = a tan (7 – 4x) + C, then value of a is
(a) 7
(b) -4
(c) 3
(d) \(-\frac{1}{4}\)
Answer/Explanation
Answer: d
Explaination:
(d), ∫sec²(7 – 4x)dx = \(\frac{\tan (7-4 x)}{-4}\) + C = –\(\frac{1}{4}\) tan (7 – 4x) + C.
7. The value of X for which
(a) 1
(b) loge4
(c) loe4 e
(d) 4
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Answer: b
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8.
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Answer: c
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9.
then value of a is equal to
(a) 3
(b) 6
(c) 9
(d) 1
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Answer: c
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10.
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11.
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Answer: c
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12.
(a) I1 > I2
(b) I2 > I1
(c) I1 = I2
(d) I1 > 2I2
Answer/Explanation
Answer: b
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13. If a is such that \(\int_{0}^{a} x d x\) ≤ a + 4, then
(a) 0 ≤ a ≤ 4
(b) -2 ≤ a ≤ 0
(c) a ≤ -2 or a ≤ 4
(d) -2 ≤ a ≤ 4
Answer/Explanation
Answer: d
Explaination:
(d), as \(\int_{0}^{a}\) x dx ≤ a + 4
⇒ \(\frac{a²}{2}\) ≤ a + 4
⇒ a² – 2a — 8 ≤ 0
⇒ (a – 1)² ≤ (3)²
⇒ -3 ≤ a – 1 ≤ 3
⇒ -2 ≤ a ≤ 4
14. If \(\frac{d}{dx}\) f(x) = g(x), then antiderivative of g(x) is ________ .
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f(x), as \(\frac{d}{dx}\) f(x) = g(x)
⇒ ∫ g(x)dx = f(x).
15. Derivative of a function is unique but a function can have infinite antiderivatives. State true or false.
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Explaination: True, as ∫ f(x)dx = g(x) + C, C is constant can take different values but \(\frac{d}{dx}\) [g(x) + C] =f(x) only
16.
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Explaination: \(\frac{2}{3}\) ∫ cosec x . cot x dx = –\(\frac{2}{3}\) ∫ cosec x + C
17. Find ∫(ax + b)3dx [AI 2011] Answer:Answer/Explanation
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18. If ∫(ax + b)² dx = f(x) + C, find f(x)
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19. We have \(\frac{d}{dx}\)(3x² + sin x – ex) = 6x + cos x -ex. Represent the expression in the form of anti derivative.
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\(\frac{d}{dx}\) (3x² + sin x – ex) = 6x + cos x – ex
⇒ ∫ (6x + cos x – ex) = 3x² + sin x – ex
20.
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21.
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22. Evaluate ∫ (sin x + cos x)² dx
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∫ (sin x + cos x)² dx = ∫ (sin²x + cos²x + 2sin x cos x)dx
= ∫(1 + sin 2x)dx = x – \(\frac{\cos 2 x}{2}\) + C
23.
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24.
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25. Find ∫(ex log a + ea log x + ea log a)dx
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26. Evaluate \(\int e^{\frac{1}{2} \log x} d x\).
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27.
(a) 3x + x3 + C
(b) log |3x + x3| + C
(c) 3x²+ 3x loge 3 +C
(d) log |3x² + 3x loge 3| + C
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Answer: d
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28.
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29.
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30.
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31. Find ∫ sec² (7 – x)dx
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32. Find \(\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x\)
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33. Find ∫2x sin(x² + 1) dx
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34.
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37.
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40.
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41.
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42.
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43.
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44. Evaluate ∫ sec4 x tan x dx
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45.
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46.
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47. Find ∫ cot x . log(sin x) dx [NCERT] Answer:Answer/Explanation
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48.
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49. Find ∫(ex + 3x)² (ex + 3)dx
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50.
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51.
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52. Find ∫ (cosx – sinx)² dx
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53. Evaluate \(\int \sqrt{1+\sin \frac{x}{4}} d x\)
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54.
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55.
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56.
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57.
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58.
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59.
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60. ∫ ex sec x(1 + tan x)dx = ________ + C.
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ex sec x, as ∫ex (sec x + sec x tan x) dx,
i.e. f(x) = sec x
f'(x) = sec x tan x,
using formula ∫ ex {f(x) + f'(x)}dx
= ex f(x) + C
61. If \(\int_{-1}^{4}\) f(x) dx =4 and \(\int_{2}^{4}\) (3 – f(x))dx = 7, then the value of \(\int_{-2}^{-1}\) f(x) dx is ________ .
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62.
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63.
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64.
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65.
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66. If \(\int_{0}^{a}\) 3x² dx = 8 write the value of a. [Foreign 2017] Answer:Answer/Explanation
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67. Evaluate. \(\int_{2}^{3}\) 3x dx [Delhi 2017] Answer:Answer/Explanation
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68.
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69. \(\int_{0}^{2a}\)f(x)dx = 2 \(\int_{0}^{a}\) f(x)dx if f(2a -x)= f(x). State true or false.
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Explaination: True; result
70.
then value of a is ________ .
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71.
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72. \(\int_{-1}^{1}\) |(1 – x)| dx is equal to ________ .
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73.
is equal to 0.State true or false.
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74. The value of \(\int_{0}^{\pi}\) | cos x|dx is 2. State true or false.
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75. The value of \(\int_{-\pi}^{\pi}\) sin3x cos²x dx is ________ .
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Explaination: 0, as f(x) = sin3 x. cos² x dx is an odd function
76.
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77.
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78.
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79.
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80. Evaluate \(\int_{-1}^{1}\) x|x|dx
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Consider \(\int_{-1}^{1}\) x|x| dx
f(x) = x|x|, f(-x) = (-x)|-x| = -x|x| = -f(x)
Odd function.
∴ \(\int_{-1}^{1}\) x|x|dx = 0
[using \(\int_{-a}^{a}\) f(x) = 0, if f(x) is odd function]
81. Evaluate \(\int_{0}^{2\pi}\) cos5x dx [Foreign 2017] Answer:Answer/Explanation
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82.
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83. Evaluate \(\int_{0}^{1}\) [2x]dx
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84. Evaluate \(\int_{1}^{4}\) f(x) dx, where
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85.
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86. Evaluate \(\int_{-\pi}^{\pi}\) (sin-93 x + x295) dx
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\(\int_{-\pi}^{\pi}\) (sin-93 x + x295)dx,f(x) is odd function as f(-x) = -f(x)
∴ \(\int_{-\pi}^{\pi}\) (sin-93 x + x295) dx=0
87.
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88.
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89.
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90. \(\int_{1}^{e}\) log x. dx
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91. Evaluate \(\int_{0}^{1}\) x(1 – x)89dx
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92.
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93. Evaluate \(\int_{0}^{1}\) x²(1 -x)ndx
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94.
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95. Evaluate \(\int_{0}^{\pi}\) |cos x|dx [DoE] Answer:Answer/Explanation
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