What is the integral from 0 to 4 of (e^{y^2}y)/2 ?

The integral from 0 to 4 of (e^{y^2}y)/2 is (e^{16}-1)/4

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integral from 0 to 4 of (e^{y^2}y)/2

\int_{0}^{4} \frac{e ^{y^{2}} y}{2} d y

\frac{e ^{16} - 1}{4}

Decimal Notation

2221527.38012 \dots

Solution steps

\int_{0}^{4} \frac{e ^{y^{2}} y}{2} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{4} e^{y^{2}} y d y

Apply u-substitution

: \int_{0}^{16} \frac{e ^{u}}{2} d u

= \frac{1}{2} \cdot \int_{0}^{16} \frac{e ^{u}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{16} e^{u} d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} [e^{u}]_{0}^{16}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [e^{u}]_{0}^{16} : \frac{1}{4} [e^{u}]_{0}^{16}

= \frac{1}{4} [e^{u}]_{0}^{16}

Compute the boundaries:

e^{16} - 1

= \frac{1}{4} (e^{16} - 1)

Simplify

= \frac{e ^{16} - 1}{4}

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What is the integral from 0 to 4 of (e^{y^2}y)/2 ?
The integral from 0 to 4 of (e^{y^2}y)/2 is (e^{16}-1)/4