What is the integral from 0 to 1 of e^{3y} ?

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

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integral from 0 to 1 of e^{3y}

\int_{0}^{1} e^{3 y} d y

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

Decimal

6.36184 \dots

Solution steps

\int_{0}^{1} e^{3 y} d y

Apply u-substitution

: \int_{0}^{3} e^{u} \frac{1}{3} d u

= \int_{0}^{3} e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{3} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{3}

Compute the boundaries:

e^{3} - 1

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

Simplify

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

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