What is the integral from 0 to x of 0.25e^{-0.25t} ?

The integral from 0 to x of 0.25e^{-0.25t} is -e^{-0.25x}+1

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integral from 0 to x of 0.25e^{-0.25t}

\int_{0}^{x} 0.25 e^{- 0.25 t} d t

- e^{- 0.25 x} + 1

Solution steps

\int_{0}^{x} 0.25 e^{- 0.25 t} d t

Take the constant out:

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

= 0.25 \cdot \int_{0}^{x} e^{- 0.25 t} d t

Apply u-substitution

: \int_{0}^{- 0.25 x} - \frac{1}{0.25} e^{u} d u

= 0.25 \cdot \int_{0}^{- 0.25 x} - \frac{1}{0.25} e^{u} d u

Take the constant out:

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

= 0.25 (- \frac{1}{0.25} \cdot \int_{0}^{- 0.25 x} e^{u} d u)

Simplify

= 0.25 (- 4 \cdot \int_{0}^{- 0.25 x} e^{u} d u)

Use the common integral:

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

= 0.25 (- 4 [e^{u}]_{0}^{- 0.25 x})

Simplify

= - [e^{u}]_{0}^{- 0.25 x}

Compute the boundaries:

e^{- 0.25 x} - 1

= - (e^{- 0.25 x} - 1)

Simplify

= - e^{- 0.25 x} + 1

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What is the integral from 0 to x of 0.25e^{-0.25t} ?
The integral from 0 to x of 0.25e^{-0.25t} is -e^{-0.25x}+1