What is the integral from 0 to 1 of e^{-kx} ?

The integral from 0 to 1 of e^{-kx} is -(e^{-k}-1)/k

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integral from 0 to 1 of e^{-kx}

\int_{0}^{1} e^{- k x} d x

- \frac{e ^{- k} - 1}{k}

Solution steps

\int_{0}^{1} e^{- k x} d x

Apply u-substitution

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

= \int_{0}^{- k} - \frac{e ^{u}}{k} d u

Take the constant out:

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

= - \frac{1}{k} \cdot \int_{0}^{- k} e^{u} d u

Use the common integral:

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

= - \frac{1}{k} [e^{u}]_{0}^{- k}

Simplify

- \frac{1}{k} [e^{u}]_{0}^{- k} : - \frac{[ e ^{u} ] _{0}^{- k}}{k}

= - \frac{[ e ^{u} ] _{0}^{- k}}{k}

Compute the boundaries:

e^{- k} - 1

= - \frac{e ^{- k} - 1}{k}

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\int_{0}^{1} - 2 x^{2} d x

What is the integral from 0 to 1 of e^{-kx} ?
The integral from 0 to 1 of e^{-kx} is -(e^{-k}-1)/k