What is the integral from 0 to x of 2/3 e^{-(2x)/3} ?

The integral from 0 to x of 2/3 e^{-(2x)/3} is -e^{-(2x)/3}+1

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integral from 0 to x of 2/3 e^{-(2x)/3}

\int_{0}^{x} \frac{2}{3} e^{- \frac{2 x}{3}} d x

- e^{- \frac{2 x}{3}} + 1

Solution steps

\int_{0}^{x} \frac{2}{3} e^{- \frac{2 x}{3}} d x

Take the constant out:

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

= \frac{2}{3} \cdot \int_{0}^{x} e^{- \frac{2 x}{3}} d x

Apply u-substitution

: \int_{0}^{- \frac{2 x}{3}} - \frac{3}{2} e^{u} d u

= \frac{2}{3} \cdot \int_{0}^{- \frac{2 x}{3}} - \frac{3}{2} e^{u} d u

Take the constant out:

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

= \frac{2}{3} (- \frac{3}{2} \cdot \int_{0}^{- \frac{2 x}{3}} e^{u} d u)

Use the common integral:

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

= \frac{2}{3} (- \frac{3}{2} [e^{u}]_{0}^{- \frac{2 x}{3}})

Simplify

\frac{2}{3} (- \frac{3}{2} [e^{u}]_{0}^{- \frac{2 x}{3}}) : - [e^{u}]_{0}^{- \frac{2 x}{3}}

= - [e^{u}]_{0}^{- \frac{2 x}{3}}

Compute the boundaries:

e^{- \frac{2 x}{3}} - 1

= - (e^{- \frac{2 x}{3}} - 1)

Simplify

= - e^{- \frac{2 x}{3}} + 1

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What is the integral from 0 to x of 2/3 e^{-(2x)/3} ?
The integral from 0 to x of 2/3 e^{-(2x)/3} is -e^{-(2x)/3}+1