Mathematics

# If $\displaystyle \int x^5e^{-4x^3}dx = \dfrac{1}{48}e^{-4x^3}f(x)+C$, where $C$ is a constant of integration, then $f(x)$ is equal to:

$-4x^3 -1$

##### SOLUTION
$\displaystyle \int x^5\cdot e^{-4x^3} dx = \dfrac{1}{48} e^{-4x^3} f(x) +c$
Put $x^3 = t$
$3x^2 dx = dt$
$\displaystyle \int x^3\cdot e^{-4x^3}\cdot x^2dx$
$\dfrac{1}{3}\displaystyle \int t\cdot e^{-4t} dt$
$\dfrac13\left[t\cdot \dfrac{e^{-4t}}{-4}-\int\dfrac{e^{-4t}}{-4}dt\right]$
$-\dfrac{e^{4t}}{48}[4t+1]+c$
$\dfrac{-e^{-4x^3}}{48} [4x^3 + 1]+c$
$\therefore f(x) = -1 -4x^3$
From the given options (1) is most suitable

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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