If the integral Ι = ∫ e ^ x ^ 2 x ^ 3 dx = e ^ x ^ 2 f(x) + c , where c is the constant of integration and f(1) = 0, then the value of f(2) is equal to

Correct answer: C. (3)/(2) Let, x ^ 2 = t ⇒ 2xdx = dt ∴ Ι = ∫ e ^ t ⋅ (t)/(2)dt Using integration by parts, we get, Ι = (1)/(2)(t e ^ t - ∫ 1 ⋅ e ^ t dt) = (1)/(2)(t e ^ t - e ^ t ) + c = e ^ x ^ 2 ( x ^ 2 - 1 2 ) + c ∴ f(x) = x ^ 2 - 1 2 ⇒ f(2) = (3)/(2)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If the integral Ι = ∫ e ^ x ^ 2 x ^ 3 dx = e ^ x ^ 2 f(x) + c ,...

If the integral

Ι = ∫ {e}^{{x}^{2}}{x}^{3}dx

= {e}^{{x}^{2}}f\left(x\right) + c

, where

c

is the constant of integration and

f\left(1\right) = 0,

then the value of

f\left(2\right)

is equal to

Options

A $4$
B $\dfrac{5}{2}$
C $\dfrac{3}{2}$
D $3$

Correct answer

C. $\dfrac{3}{2}$

Step-by-step solution

Let,

{x}^{2} = t ⇒ 2xdx = dt

∴ Ι = ∫ {e}^{t} ⋅ \dfrac{t}{2}dt

Using integration by parts, we get,

Ι = \dfrac{1}{2}\left(t{e}^{t} - ∫ 1 ⋅ {e}^{t}dt\right)

= \dfrac{1}{2}\left(t{e}^{t} - {e}^{t}\right) + c

= {e}^{{x}^{2}}\left(\dfrac{{x}^{2} - 1}{2}\right) + c

∴ f\left(x\right) = \dfrac{{x}^{2} - 1}{2} ⇒ f\left(2\right) = \dfrac{3}{2}

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