What is the integral from 3 to 7 of 5x^3e^{-x} ?

The integral from 3 to 7 of 5x^3e^{-x} is -5(-(78)/(e^3)+(538)/(e^7))

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integral from 3 to 7 of 5x^3e^{-x}

\int_{3}^{7} 5 x^{3} e^{- x} d x

- 5 (- \frac{78}{e ^{3}} + \frac{538}{e ^{7}})

Decimal

16.96399 \dots

Solution steps

\int_{3}^{7} 5 x^{3} e^{- x} d x

Take the constant out:

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

= 5 \cdot \int_{3}^{7} x^{3} e^{- x} d x

Apply u-substitution

: \int_{- 3}^{- 7} e^{u} u^{3} d u

= 5 \cdot \int_{- 3}^{- 7} e^{u} u^{3} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 5 (- \int_{- 7}^{- 3} e^{u} u^{3} d u)

Apply Integration By Parts

: [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- 7}^{- 3}

= 5 (- [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- 7}^{- 3})

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

= 5 (- [u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))]_{- 7}^{- 3})

Simplify

= - 5 [e^{u} u^{3} - 3 (e^{u} u^{2} - 2 (e^{u} u - e^{u}))]_{- 7}^{- 3}

Compute the boundaries:

- \frac{78}{e ^{3}} + \frac{538}{e ^{7}}

= - 5 (- \frac{78}{e ^{3}} + \frac{538}{e ^{7}})

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What is the integral from 3 to 7 of 5x^3e^{-x} ?
The integral from 3 to 7 of 5x^3e^{-x} is -5(-(78)/(e^3)+(538)/(e^7))