What is the integral from 0 to x of 0.1e^{-0.1x} ?

The integral from 0 to x of 0.1e^{-0.1x} is -e^{-0.1x}+1

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integral from 0 to x of 0.1e^{-0.1x}

\int_{0}^{x} 0.1 e^{- 0.1 x} d x

- e^{- 0.1 x} + 1

Solution steps

\int_{0}^{x} 0.1 e^{- 0.1 x} d x

Take the constant out:

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

= 0.1 \cdot \int_{0}^{x} e^{- 0.1 x} d x

Apply u-substitution

: \int_{0}^{- 0.1 x} - \frac{1}{0.1} e^{u} d u

= 0.1 \cdot \int_{0}^{- 0.1 x} - \frac{1}{0.1} e^{u} d u

Take the constant out:

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

= 0.1 (- \frac{1}{0.1} \cdot \int_{0}^{- 0.1 x} e^{u} d u)

Simplify

= 0.1 (- 10 \cdot \int_{0}^{- 0.1 x} e^{u} d u)

Use the common integral:

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

= 0.1 (- 10 [e^{u}]_{0}^{- 0.1 x})

Simplify

= - [e^{u}]_{0}^{- 0.1 x}

Compute the boundaries:

e^{- 0.1 x} - 1

= - (e^{- 0.1 x} - 1)

Simplify

= - e^{- 0.1 x} + 1

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What is the integral from 0 to x of 0.1e^{-0.1x} ?
The integral from 0 to x of 0.1e^{-0.1x} is -e^{-0.1x}+1