250+ TOP MCQs on Differentiation Under Integral Sign and Answers

Differential and Integral Calculus Multiple Choice Questions 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
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_0^1frac{∂}{∂α} left(frac{x^{α}-1}{log⁡x}right)dx) …..Leibnitz rule
( =int_0^1frac{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) = intfrac{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)
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})
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

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})
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
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_0^1frac{∂}{∂α} left(frac{x^{α}-1}{log⁡x}right)dx) …..Leibniz rule
( = log⁡2 int_0^1frac{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⁡2intfrac{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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