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:


ANSWER

$$-4x^3 -1$$


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
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