# 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

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).

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) $$f’(α) = \int_a^b \frac{d}{dα} f (x,α) dx$$

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) $$f’(α) = \int_a^b \frac{∂}{∂α} f (x,α) dx + f(b, α) \frac{db}{dα} – f(a, α) \frac{da}{dα}$$

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

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) $$\frac{3 sin a^3 – 3 sin^2 a}{6a}$$

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

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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