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

The integral from 0 to ln(2) of (x-1)e^{-x} is -1/2 ln(2)

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integral from 0 to ln(2) of (x-1)e^{-x}

\int_{0}^{l n (2)} (x - 1) e^{- x} d x

- \frac{1}{2} ln (2)

Decimal Notation

- 0.34657 \dots

Solution steps

\int_{0}^{l n (2)} (x - 1) e^{- x} d x

Apply Integration By Parts

: [- e^{- x} (x - 1) - \int - e^{- x} d x]_{0}^{l n (2)}

= [- e^{- x} (x - 1) - \int - e^{- x} d x]_{0}^{l n (2)}

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

= [- e^{- x} (x - 1) - e^{- x}]_{0}^{l n (2)}

Expand

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

= [- e^{- x} x]_{0}^{l n (2)}

Compute the boundaries:

- \frac{1}{2} ln (2)

= - \frac{1}{2} ln (2)

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\int_{- 1}^{1} (4 x - e^{- x})^{2} d x

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\int_{1}^{2} (i + 2 t j) d x

\int_{0}^{1} e^{x} - (e^{- x}) d x

What is the integral from 0 to ln(2) of (x-1)e^{-x} ?
The integral from 0 to ln(2) of (x-1)e^{-x} is -1/2 ln(2)