If int {{x^5}{e^{ - 4{x^3}}},dx = frac{1}{{48 | vedclass

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Question

(B) Let $I = \int x^5 e^{-4x^3} dx$. Substitute $t = -4x^3$,then $dt = -12x^2 dx$,which implies $x^2 dx = -\frac{1}{12} dt$. Also,$x^3 = -\frac{t}{4}$

Vedclass · Accepted Answer

$-4x^3 -1$

Vedclass · Answer

(B) Let $I = \int x^5 e^{-4x^3} dx$. Substitute $t = -4x^3$,then $dt = -12x^2 dx$,which implies $x^2 dx = -\frac{1}{12} dt$. Also,$x^3 = -\frac{t}{4}$. Substituting these into the integral: $I = \int x^3 \cdot e^{-4x^3} \cdot x^2 dx = \int (-\frac{t}{4}) e^t (-\frac{1}{12} dt) = \frac{1}{48} \int t e^t dt$. Using integration by parts: $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + C_1$. So,$I = \frac{1}{48} (t e^t - e^t) + C = \frac{1}{48} e^t (t - 1) + C$. Substituting $t = -4x^3$ back: $I = \frac{1}{48} e^{-4x^3} (-4x^3 - 1) + C$. Comparing this with the given form $\frac{1}{48} e^{-4x^3} f(x) + C$,we get $f(x) = -4x^3 - 1$.

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

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