Differential and Integral Calculus Questions and Answers – Differentiation Under Integral Sign

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This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Differentiation Under Integral Sign”.

1. When solved by the method of Differentiation for the given integral i.e \(\int_0^∞ \frac{x^{2}-1}{log⁡x} dx\) the result obtained is given by _______
a) log⁡4
b) log⁡3
c) 2log⁡3
d) log⁡8
View Answer

Answer: b
Explanation: To solve this problem let us assume the given function is dependent on α
Such that α=2 & thus \(f (α) = \int_0^∞ \frac{x^{α}-1}{log⁡x} dx\)
\(f’(α) = \int_0^1\frac{∂}{∂α} \left(\frac{x^{α}-1}{log⁡x}\right)dx\) …..Leibnitz rule
\( =\int_0^1\frac{x^α.log⁡x}{log⁡x} dx\)
\( = \int_0^1 x^α dx = [\frac{x^{α+1}}{α+1}]_0^1 = \frac{1}{α+1}\)
We have \(f’(α) = \frac{1}{α+1}\)
Thus \(f (α) = \int\frac{1}{α+1}dα+c\)
f (α) = log(α+1)+c
or f (α) = log(α+1) …… neglecting constant since the function is assumed
thus f (2) = log(2+1) = log(3).
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2. Which among the following correctly defines Leibnitz rule of a function given by \( f (α) = \int_a^b (x,α)dx\) where a & b are constants?
a) \(f’(α) = \frac{∂}{∂α}\int_a^b f (x,α) dx\)
b) \(f’(α) = \frac{d}{dα} \int_a^b f (α) dx\)
c) \(f’(α) = \int_a^b \frac{∂}{∂α} f (x,α) dx\)
d) \(f’(α) = \int_a^b \frac{d}{dα} f (x,α) dx\)
View Answer

Answer: c
Explanation: \(f’(α) = \int_a^b \frac{∂}{∂α} f (x,α) dx = \frac{d(f(α))}{dα} = \frac{d}{dα} \int_a^b f (x,α) dx.\)

3. Which among the following correctly defines Leibnitz rule of a function given by
\( f (α) = \int_a^b (x,α)dx\) where a & b are functions of α?
a) \(f’(α) = \int_a^b \frac{∂}{∂α} f(x,α) dx\)
b) \(f’(α) = \frac{d}{dα} \int_a^b f(x,α) dx\)
c) \(f’(α) = \int_a^b \frac{∂}{∂α} f (x,α) dx + f(b, α) \frac{da}{dα} – f(a, α) \frac{db}{dα}\)
d) \(f’(α) = \int_a^b \frac{∂}{∂α} f (x,α) dx + f(b, α) \frac{db}{dα} – f(a, α) \frac{da}{dα}\)
View Answer

Answer: d
Explanation: \(f’(α) = \int_a^b \frac{∂}{∂α} f (x,α) dx + f(b, α) \frac{db}{dα} – f(a, α) \frac{da}{dα}\) when a & b are constants
\(\frac{da}{dα} \& \frac{da}{dα} = 0\) which reduces the equation d into a
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4. Given \(f (a) = \int_a^{a^2} \frac{sin⁡ax}{x} dx \) what is the value of f’(a)?
a) \(\frac{sin⁡3a}{a}\)
b) \(\frac{3 sin⁡ a^3 – 2 sin a^2}{a}\)
c) \(\frac{3 sin a^2 – 4 sin a}{a}\)
d) \(\frac{3 sin a^3 – 3 sin^2 a}{6a}\)
View Answer

Answer: b
Explanation: Applying the Leibniz rule equation given by
\( f’ (α) = \int_p^q \frac{∂}{∂α} f (x,α) dx + f (q, α) \frac{dq}{dα} – f (p, α) \frac{dp}{dα} …..(1)\)
\(f (x,a) = \frac{sin⁡ax}{x}, p=a, q=a^2\) & further obtaining
\(f(q,a) = f(a^2,a) = \frac{sin⁡ a^3}{a^2}, \frac{dq}{da} = 2a\)
\(f(p,a) = f(a,a) = \frac{sin⁡ a^2}{a}, \frac{dp}{da} = 1\)
substituting all these values in (1) we get
\(f’(a) = \int_a^{a^2} \frac{∂}{∂α} (\frac{sin⁡ax}{x})dx + \frac{sin⁡ a^3}{a^2}.2a – \frac{sin⁡ a^2}{a}.1\)
\(\int_a^{a^2} \frac{1}{x} (cos(ax))(x)+ \frac{2 sin a^3 – sin^2 a}{a}\)
\([\frac{sin⁡ax}{x}]_a^{a^2} + \frac{2 sin a^3 – sin^2 a}{a} = \frac{sin⁡ a^3}{a} – \frac{sin⁡ a^2}{a} + \frac{2 sin a^3 – sin^2 a}{a}\)
thus \(f’(a) = \frac{3 sin⁡ a^3 – 2 sin a^2}{a}.\)

5. When solved by the method of Differentiation for the given integral i.e. \(\int_0^1 \frac{x^2-1}{log_2⁡x} dx \) the result obtained is given by _________
a) log⁡5
b) 3 log 3
c) log 4
d) 2 log 3
View Answer

Answer: a
Explanation: \(\int_0^1 \frac{x^2-1}{log_2⁡x} dx\) can also be written as \((log⁡2 \int_0^1 \frac{x^2-1}{log ⁡x} dx)…….(1).\)
Here during integration changing \(log_2⁡x = \frac{log⁡x}{log⁡2}\) and substituting we get (1) since logarithm to base ‘e’ can be easily integratable.
To solve this problem let us assume the given function is dependent on α
Such that α=2 & thus \(f (α) = log⁡2 \int_0^∞ \frac{x^{α}-1}{log⁡x} dx\)
\(f’(α) = log⁡2 \int_0^1\frac{∂}{∂α} \left(\frac{x^{α}-1}{log⁡x}\right)dx\) …..Leibniz rule
\( = log⁡2 \int_0^1\frac{x^α.log⁡x}{log⁡x} dx\)
\( = log⁡2 \int_0^1 x^α dx = [\frac{x^{α+1}}{α+1}]_0^1 = \frac{1}{α+1}\)
We have \(f’(α) = log⁡2.\frac{1}{α+1}\)
Thus \(f (α) = log⁡2\int\frac{1}{α+1}dα+c\)
f (α) = log⁡2 .log(α+1)+c
or f (α) = log(α+1+2) neglecting constant since the function is assumed
thus f (2) = log(2+3) = log(5).
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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