Class 12 Maths MCQ – Definite Integral

This set of Class 12 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Definite Integral”.

1. In \(\int_a^b\)f(y) dy, what is ‘a’ called as?
a) Integration
b) Upper limit
c) Lower limit
d) Limit of an integral
View Answer

Answer: b
Explanation: In \(\int_a^b\)f(y) dy ‘a’ is the called as lower limit and ‘b’ is called the upper limit of the integral. The fuction f in \(\int_a^b\)f(y) dy is called the integrand. The letter ‘y’ is a dummy symbol and can be replaced by any other symbol.

2. The value of \(\int_0^{\pi }\)sin ⁡y dy is 2.
a) True
b) False
View Answer

Answer: a
Explanation: \(\int_0^{\pi }\)sin⁡ y dy = [-cos y]y0
= – cos π – (- cos 0)
= -(-1)-(-1)
= 2

3. Compute ∫cos(x)-\(\frac {3}{x4}\)dx.
a) sin(x)+\(\frac {3}{4}\)x-7+c
b) sec(x)+\(\frac {3}{4}\)x-3+c
c) sin(x)+\(\frac {3}{4}\)x-3
d) sin(x)+\(\frac {3}{4}\)x-3+c
View Answer

Answer: d
Explanation: ∫cos(x)-\(\frac {3}{x4}\)dx = ∫cos(x) dx−∫3(x-4) dx
= sin(x)+\(\frac {3}{4}\)x-3+c
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4. What is the value of \(\int_2^3\)cos⁡(x)-\(\frac {3}{x4}\)dx .
a) sin (3) – sin (2)
b) sin (3) – sin (9) – \(\frac {19}{288}\)
c) sin (8) – sin (2) – \(\frac {19}{288}\)
d) sin (3) – sin (2) – \(\frac {19}{288}\)
View Answer

Answer: d
Explanation: \(\int_2^3\)cos⁡(x)-\(\frac {3}{x4}\)dx = \(\int_2^3\)sin(x) dx + \(\int_2^3 \frac {3}{4}\)x-3 dx
= (sin (3) + \(\frac {3}{4}\)3-3) – (sin (2) + \(\frac {3}{4}\)2-3)
= sin (3) – sin (2) – \(\frac {19}{288}\)

5. Evaluate \(\int_7^9\)cos⁡(x)dx.
a) 8 (-sin 9 – sin 7)
b) 8 (sin 9 + sin 7)
c) 8 (sin 9 – sin 7)
d) 7 (sin 9 – sin 7)
View Answer

Answer: c
Explanation: \(\int_7^9\)8cos⁡(x)dx = 8 \(\int_7^9\)cos⁡(x)dx
= 8 (cos x)97
= 8 (sin 9 – sin 7)
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6. Compute \(\int_2^3 \frac {cos⁡x-sin⁡x}{4}\)dx.
a) \(\frac {1}{4}\) (sin 2 + cos 3 – sin 3 – cos 2)
b) \(\frac {1}{4}\) (sin 3 – cos 3 – sin 2 – cos 2)
c) \(\frac {1}{4}\) (sin 3 + cos 3 – sin 2 – cos 2)
d) \(\frac {1}{4}\) (sin 3 + cos 3 + sin 2 – cos 2)
View Answer

Answer: c
Explanation: \(\int_2^3 \frac {cos⁡x-sin⁡x}{4}\)dx = \(\frac {1}{4}\) [sin x – (- cos x)]32
= \(\frac {1}{4}\) (sin x + cos x)32
= \(\frac {1}{4}\) (sin 3 + cos 3) – \(\frac {1}{4}\) (sin 2 + cos 2)
= \(\frac {1}{4}\) (sin 3 + cos 3 – sin 2 – cos 2)

7. What is y in \(\int_a^b\)f(y) dy called as?
a) Random variable
b) Dummy symbol
c) Integral
d) Integrand
View Answer

Answer: b
Explanation: In \(\int_a^b\)f(y) dy ‘a’ is the called as lower limit and ‘b’ is called the upper limit of the integral. The fuction ‘f’ in \(\int_a^b\)f(y) dy is called the integrand. The letter ‘y’ is a dummy symbol and can be replaced by any other symbol.
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8. The value of \(\int_1^2\)1y5 dy is_____
a) 10.5
b) 56
c) 9
d) 23
View Answer

Answer: a
Explanation: \(\int_1^2\)1y5 dy = (y6/6)21
= \(\frac {64}{6} – \frac{1}{6}\)
= 10.5

9. The value of \(\int_1^2\)1y5/5dy is _____
a) 12
b) 2.1
c) 21
d) 11.1
View Answer

Answer: b
Explanation: \(\int_1^2\)1y5/5dy = \(\frac {1}{5}\)(y6/6)21
= \(\frac {1}{5}(\frac {64}{6} – \frac{1}{6})\)
= 2.1
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10. Evaluate \(\int_0^{\pi }\)sin⁡x dx.
a) 2
b) 6
c) 17
d) 3
View Answer

Answer: a
Explanation: \(\int_0^{\pi }\)sin⁡x dx = [- cos x]x0
= – cos π – (-cos 0)
= -(-1)-(-1)
= 2

11. Evaluate \(\int_2^3\)cosx dx.
a) 38.2
b) sin (9) – sin (4)
c) 89.21
d) sin (3) – sin (2)
View Answer

Answer: d
Explanation: \(\int_2^3\)cosx dx = (sin x)32
= sin (3) – sin (2)

12. Compute \(\int_2^3\)2ex dx.
a) 2(e9 – e4)
b) 84.32
c) 2(e3 – e2)
d) 83.25
View Answer

Answer: c
Explanation: \(\int_2^3\)2ex dx = 2(ex)32 dx
= 2(e3 – e2)

13. In \(\int_b^a\)f(x) dx, b called as lower limit and a is called as upper limit.
a) False
b) True
View Answer

Answer: b
Explanation: In \(\int_a^b\)f(y) dy ‘a’ is the called as lower limit and ‘b’ is called the upper limit of the integral. The fuction ‘f’ in \(\int_a^b\)f(y) dy is called the integrand. The letter ‘y’ is a dummy symbol and can be replaced by any other symbol.

14. Compute \(\int_3^6\)9 ex dx.
a) 30.82
b) 9(e6 – e3)
c) 11.23
d) 81(e6 – e3)
View Answer

Answer: b
Explanation: \(\int_3^6\)9 ex dx = 9(ex)63 dx
= 9(e6 – e3)

15. Evaluate \(\int_3^7\)sin(t)-2cos(t)dt.
a) cos(7) – 2sin(7) + (cos(3) + 2sin(3)
b) -17
c) 12
d) cos(7) – 2sin(7) – (cos(3) + 2sin(3)
View Answer

Answer: d
Explanation: \(\int_3^7\)sin(t)-2cos(t)dt = (cos(t)−2sin(t))73
= (cos(7) – 2sin(7)) – (cos(3) – 2sin(3))
= cos(7) – 2sin(7) – (cos(3) + 2sin(3)

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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