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

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

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

\int_{0}^{1} e^{\frac{2 x}{3}} d x

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

Decimal

1.42160 \dots

Solution steps

\int_{0}^{1} e^{\frac{2 x}{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

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

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

Simplify

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

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