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}{logx} dx) the result obtained is given by _______
a) log4
b) log3
c) 2log3
d) log8
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}{logx} dx)
(f’(α) = int_0^1frac{∂}{∂α} left(frac{x^{α}-1}{logx}right)dx) …..Leibnitz rule
( =int_0^1frac{x^α.logx}{logx} 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{sinax}{x} dx ) what is the value of f’(a)?
a) (frac{sin3a}{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{sinax}{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{sinax}{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{sinax}{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_2x} dx ) the result obtained is given by _________
a) log5
b) 3 log 3
c) log 4
d) 2 log 3
Answer: a
Explanation: (int_0^1 frac{x^2-1}{log_2x} dx) can also be written as ((log2 int_0^1 frac{x^2-1}{log x} dx)…….(1).)
Here during integration changing (log_2x = frac{logx}{log2}) 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) = log2 int_0^∞ frac{x^{α}-1}{logx} dx)
(f’(α) = log2 int_0^1frac{∂}{∂α} left(frac{x^{α}-1}{logx}right)dx) …..Leibniz rule
( = log2 int_0^1frac{x^α.logx}{logx} dx)
( = log2 int_0^1 x^α dx = [frac{x^{α+1}}{α+1}]_0^1 = frac{1}{α+1})
We have (f’(α) = log2.frac{1}{α+1})
Thus (f (α) = log2intfrac{1}{α+1}dα+c)
f (α) = log2 .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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