What is the integral from 0 to 1 of (2x+3)e^{-5x} ?

The integral from 0 to 1 of (2x+3)e^{-5x} is -(27)/(25e^5)+17/25

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integral from 0 to 1 of (2x+3)e^{-5x}

\int_{0}^{1} (2 x + 3) e^{- 5 x} d x

- \frac{27}{25 e ^{5}} + \frac{17}{25}

Decimal Notation

0.67272 \dots

Solution steps

\int_{0}^{1} (2 x + 3) e^{- 5 x} d x

Apply Integration By Parts

: [- \frac{1}{5} e^{- 5 x} (2 x + 3) - \int - \frac{2}{5} e^{- 5 x} d x]_{0}^{1}

= [- \frac{1}{5} e^{- 5 x} (2 x + 3) - \int - \frac{2}{5} e^{- 5 x} d x]_{0}^{1}

\int - \frac{2}{5} e^{- 5 x} d x = \frac{2}{25} e^{- 5 x}

= [- \frac{1}{5} e^{- 5 x} (2 x + 3) - \frac{2}{25} e^{- 5 x}]_{0}^{1}

Simplify

[- \frac{1}{5} e^{- 5 x} (2 x + 3) - \frac{2}{25} e^{- 5 x}]_{0}^{1} : [\frac{1}{25} (- 10 e^{- 5 x} x - 17 e^{- 5 x})]_{0}^{1}

= [\frac{1}{25} (- 10 e^{- 5 x} x - 17 e^{- 5 x})]_{0}^{1}

Compute the boundaries:

- \frac{27}{25 e ^{5}} + \frac{17}{25}

= - \frac{27}{25 e ^{5}} + \frac{17}{25}

\frac{d}{d x} (\frac{e ^{x} ( x ^{5} + 3 )}{x ^{10} + 9})

\frac{s}{s ^{2} + a ^{2}}

\int \frac{6 x ^{3} + 4 x ^{2} - 47 x - 27}{x ^{2} - 9} d x

n = 0 \sum \infty 4^{n}

\int_{- \infty}^{- 4} \frac{3}{x ^{4}} d x

What is the integral from 0 to 1 of (2x+3)e^{-5x} ?
The integral from 0 to 1 of (2x+3)e^{-5x} is -(27)/(25e^5)+17/25