What is the integral of e^{-(x-8)^3}(x-8)^2 ?

The integral of e^{-(x-8)^3}(x-8)^2 is -1/3 e^{-(x-8)^3}+C

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integral of e^{-(x-8)^3}(x-8)^2

\int e^{- (x - 8)^{3}} (x - 8)^{2} d x

- \frac{1}{3} e^{- (x - 8)^{3}} + C

Solution steps

\int e^{- (x - 8)^{3}} (x - 8)^{2} d x

Apply u-substitution

: \int e^{- u^{3}} u^{2} d u

= \int e^{- u^{3}} u^{2} d u

Apply u-substitution

: \int - \frac{e ^{v}}{3} d v

= \int - \frac{e ^{v}}{3} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{3} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= - \frac{1}{3} e^{v}

Substitute back

= - \frac{1}{3} e^{- (x - 8)^{3}}

Add a constant to the solution

= - \frac{1}{3} e^{- (x - 8)^{3}} + C

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What is the integral of e^{-(x-8)^3}(x-8)^2 ?
The integral of e^{-(x-8)^3}(x-8)^2 is -1/3 e^{-(x-8)^3}+C