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

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

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integral from 0 to c of 4e^{-x}

\int_{0}^{c} 4 e^{- x} d x

- 4 (e^{- c} - 1)

Solution steps

\int_{0}^{c} 4 e^{- x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{c} e^{- x} d x

Apply u-substitution

= 4 \cdot \int_{0}^{- c} - e^{u} d u

Take the constant out:

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

= 4 (- \int_{0}^{- c} e^{u} d u)

Use the common integral:

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

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

Simplify

= - 4 [e^{u}]_{0}^{- c}

Compute the boundaries:

e^{- c} - 1

= - 4 (e^{- c} - 1)

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