What is the integral from 1 to 3 of x^3e^{-3/2 x} ?

The integral from 1 to 3 of x^3e^{-3/2 x} is -(2(-201e^3+1479))/(81e^{9/2)}

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integral from 1 to 3 of x^3e^{-3/2 x}

\int_{1}^{3} x^{3} e^{- \frac{3}{2} x} d x

- \frac{2 ( - 201 e ^{3} + 1479 )}{81 e ^{\frac{9}{2}}}

Decimal Notation

0.70170 \dots

Solution steps

\int_{1}^{3} x^{3} e^{- \frac{3}{2} x} d x

Apply u-substitution

: \int_{- \frac{3}{2}}^{- \frac{9}{2}} \frac{16 e ^{u} u ^{3}}{81} d u

= \int_{- \frac{3}{2}}^{- \frac{9}{2}} \frac{16 e ^{u} u ^{3}}{81} d u

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

= - \int_{- \frac{9}{2}}^{- \frac{3}{2}} \frac{16 e ^{u} u ^{3}}{81} d u

Take the constant out:

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

= - \frac{16}{81} \cdot \int_{- \frac{9}{2}}^{- \frac{3}{2}} e^{u} u^{3} d u

Apply Integration By Parts

: [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- \frac{9}{2}}^{- \frac{3}{2}}

= - \frac{16}{81} [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- \frac{9}{2}}^{- \frac{3}{2}}

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

= - \frac{16}{81} [u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))]_{- \frac{9}{2}}^{- \frac{3}{2}}

Compute the boundaries:

- \frac{201}{8 e ^{\frac{3}{2}}} + \frac{1479}{8 e ^{\frac{9}{2}}}

= - \frac{16}{81} (- \frac{201}{8 e ^{\frac{3}{2}}} + \frac{1479}{8 e ^{\frac{9}{2}}})

Simplify

= - \frac{2 ( - 201 e ^{3} + 1479 )}{81 e ^{\frac{9}{2}}}

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What is the integral from 1 to 3 of x^3e^{-3/2 x} ?
The integral from 1 to 3 of x^3e^{-3/2 x} is -(2(-201e^3+1479))/(81e^{9/2)}