What is the integral from 0 to 1 of e^{-x}(4-e^x) ?

The integral from 0 to 1 of e^{-x}(4-e^x) is -(-e+4)/e+2

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integral from 0 to 1 of e^{-x}(4-e^x)

\int_{0}^{1} e^{- x} (4 - e^{x}) d x

- \frac{- e + 4}{e} + 2

Solution steps

\int_{0}^{1} e^{- x} (4 - e^{x}) d x

Apply Integration By Parts

: [- e^{- x} (4 - e^{x}) - \int 1 d x]_{0}^{1}

= [- e^{- x} (4 - e^{x}) - \int 1 d x]_{0}^{1}

\int 1 d x = x

= [- e^{- x} (4 - e^{x}) - x]_{0}^{1}

Compute the boundaries:

- \frac{- e + 4}{e} + 2

= - \frac{- e + 4}{e} + 2

\int_{0}^{3} 0.3 e^{x} d x

\int_{0}^{2} e^{- r^{2}} r d r

\int_{0}^{2} r 4 - r^{2} d r

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

\int_{0}^{y} 2 e^{- (x + y)} d x

What is the integral from 0 to 1 of e^{-x}(4-e^x) ?
The integral from 0 to 1 of e^{-x}(4-e^x) is -(-e+4)/e+2