What is the integral from 1 to 3 of x^3e^{-5/3 x} ?

The integral from 1 to 3 of x^3e^{-5/3 x} is (-19116e^{5/3}+2346e^5)/(625e^{20/3)}

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integral from 1 to 3 of x^3e^{-5/3 x}

\int_{1}^{3} x^{3} e^{- \frac{5}{3} x} d x

\frac{- 19116 e ^{\frac{5}{3}} + 2346 e ^{5}}{625 e ^{\frac{20}{3}}}

Decimal Notation

0.50287 \dots

Solution steps

\int_{1}^{3} x^{3} e^{- \frac{5}{3} x} d x

Apply Integration By Parts

: [- \frac{3}{5} e^{- \frac{5 x}{3}} x^{3} - \int - \frac{9}{5} e^{- \frac{5 x}{3}} x^{2} d x]_{1}^{3}

= [- \frac{3}{5} e^{- \frac{5 x}{3}} x^{3} - \int - \frac{9}{5} e^{- \frac{5 x}{3}} x^{2} d x]_{1}^{3}

\int - \frac{9}{5} e^{- \frac{5 x}{3}} x^{2} d x = - \frac{9}{5} (- \frac{3}{5} e^{- \frac{5 x}{3}} x^{2} + \frac{6}{5} (- \frac{3}{5} e^{- \frac{5 x}{3}} x - \frac{9}{25} e^{- \frac{5 x}{3}}))

= [- \frac{3}{5} e^{- \frac{5 x}{3}} x^{3} - (- \frac{9}{5} (- \frac{3}{5} e^{- \frac{5 x}{3}} x^{2} + \frac{6}{5} (- \frac{3}{5} e^{- \frac{5 x}{3}} x - \frac{9}{25} e^{- \frac{5 x}{3}})))]_{1}^{3}

Simplify

= [- \frac{3}{5} e^{- \frac{5 x}{3}} x^{3} + \frac{9}{5} (- \frac{3}{5} e^{- \frac{5 x}{3}} x^{2} + \frac{6}{5} (- \frac{3}{5} e^{- \frac{5 x}{3}} x - \frac{9}{25} e^{- \frac{5 x}{3}}))]_{1}^{3}

Compute the boundaries:

\frac{- 19116 e ^{\frac{5}{3}} + 2346 e ^{5}}{625 e ^{\frac{20}{3}}}

= \frac{- 19116 e ^{\frac{5}{3}} + 2346 e ^{5}}{625 e ^{\frac{20}{3}}}

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What is the integral from 1 to 3 of x^3e^{-5/3 x} ?
The integral from 1 to 3 of x^3e^{-5/3 x} is (-19116e^{5/3}+2346e^5)/(625e^{20/3)}