What is the integral from 0 to b of me^{-mx} ?

The integral from 0 to b of me^{-mx} is -e^{-bm}+1

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integral from 0 to b of me^{-mx}

\int_{0}^{b} m e^{- m x} d x

- e^{- bm} + 1

Solution steps

\int_{0}^{b} m e^{- m x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= m \cdot \int_{0}^{b} e^{- m x} d x

Apply u-substitution

: \int_{0}^{- mb} - \frac{e ^{u}}{m} d u

= m \cdot \int_{0}^{- mb} - \frac{e ^{u}}{m} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= m (- \frac{1}{m} \cdot \int_{0}^{- mb} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= m (- \frac{1}{m} [e^{u}]_{0}^{- mb})

Simplify

m (- \frac{1}{m} [e^{u}]_{0}^{- mb}) : - [e^{u}]_{0}^{- bm}

= - [e^{u}]_{0}^{- bm}

Compute the boundaries:

e^{- bm} - 1

= - (e^{- bm} - 1)

Simplify

= - e^{- bm} + 1

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What is the integral from 0 to b of me^{-mx} ?
The integral from 0 to b of me^{-mx} is -e^{-bm}+1