What is the integral from 1 to 4 of e^{-x}(x-1) ?

The integral from 1 to 4 of e^{-x}(x-1) is -4/(e^4)+1/e

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integral from 1 to 4 of e^{-x}(x-1)

\int_{1}^{4} e^{- x} (x - 1) d x

- \frac{4}{e ^{4}} + \frac{1}{e}

Decimal

0.29461 \dots

Solution steps

\int_{1}^{4} e^{- x} (x - 1) d x

Apply Integration By Parts

: [- e^{- x} (x - 1) - \int - e^{- x} d x]_{1}^{4}

= [- e^{- x} (x - 1) - \int - e^{- x} d x]_{1}^{4}

\int - e^{- x} d x = e^{- x}

= [- e^{- x} (x - 1) - e^{- x}]_{1}^{4}

Expand

- e^{- x} (x - 1) - e^{- x} : - e^{- x} x

= [- e^{- x} x]_{1}^{4}

Compute the boundaries:

- \frac{4}{e ^{4}} + \frac{1}{e}

= - \frac{4}{e ^{4}} + \frac{1}{e}

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What is the integral from 1 to 4 of e^{-x}(x-1) ?
The integral from 1 to 4 of e^{-x}(x-1) is -4/(e^4)+1/e