What is the integral from 0 to x of 7e^{-7x} ?

The integral from 0 to x of 7e^{-7x} is -e^{-7x}+1

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integral from 0 to x of 7e^{-7x}

\int_{0}^{x} 7 e^{- 7 x} d x

- e^{- 7 x} + 1

Solution steps

\int_{0}^{x} 7 e^{- 7 x} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{x} e^{- 7 x} d x

Apply u-substitution

= 7 \cdot \int_{0}^{- 7 x} - \frac{1}{7} e^{u} d u

Take the constant out:

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

= 7 (- \frac{1}{7} \cdot \int_{0}^{- 7 x} e^{u} d u)

Use the common integral:

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

= 7 (- \frac{1}{7} [e^{u}]_{0}^{- 7 x})

Simplify

7 (- \frac{1}{7} [e^{u}]_{0}^{- 7 x}) : - [e^{u}]_{0}^{- 7 x}

= - [e^{u}]_{0}^{- 7 x}

Compute the boundaries:

e^{- 7 x} - 1

= - (e^{- 7 x} - 1)

Simplify

= - e^{- 7 x} + 1

What is the integral from 0 to x of 7e^{-7x} ?
The integral from 0 to x of 7e^{-7x} is -e^{-7x}+1