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JEE Main 201910 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2019 · 10 January · Evening Shift · Mathematics · Indefinite Integration

If ∫ x ^ 5 e ^ - 4 x ^ 3 dx = (1)/(48) e ^ - 4 x ^ 3 f(x) + C ,...

If ∫x5e−4x3dx=148e−4x3f(x)+C∫ {x}^{5}{e}^{- 4{x}^{3}}dx = \dfrac{1}{48}{e}^{- 4{x}^{3}}f\left(x\right) + C, where CC is a constant of integration, then f(x)f\left(x\right) is equal to

Options

  1. A−4x3−1- 4{x}^{3} - 1
  2. B−2x3+1- 2{x}^{3} + 1
  3. C−2x3−1- 2{x}^{3} - 1
  4. D4x3+14{x}^{3} + 1

Correct answer

A. −4x3−1- 4{x}^{3} - 1

Step-by-step solution

We have, I=∫x5e−4x3dxI = ∫ {x}^{5}{e}^{- 4{x}^{3}}dx

Let, −4x3=t⇒−12x2dx=dt- 4{x}^{3} = t ⇒ - 12{x}^{2}dx = dt

⇒I=∫(−t4)et(−dt12)=148∫tetdt⇒ I = ∫ \left(- \dfrac{t}{4}\right){e}^{t}\left(- \dfrac{dt}{12}\right) = \dfrac{1}{48} ∫ t{e}^{t}dt

=148[tet−∫etdt]=(t−1)et48+C=(−4x3−1)e−4x348+C,= \dfrac{1}{48}\left[t{e}^{t} - ∫ {e}^{t}dt\right] = \dfrac{\left(t - 1\right){e}^{t}}{48} + C = \dfrac{\left(- 4{x}^{3} - 1\right){e}^{- 4{x}^{3}}}{48} + C,
where CC is the constant of integration.

Hence, f(x)=−4x3−1f\left(x\right) = - 4{x}^{3} - 1

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