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

The integral from 0 to 4 of y(e^{y^2}-1) is (e^{16}-17)/2

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

\int_{0}^{4} y (e^{y^{2}} - 1) d y

\frac{e ^{16} - 17}{2}

Decimal Notation

4443046.76025 \dots

Solution steps

\int_{0}^{4} y (e^{y^{2}} - 1) d y

Apply u-substitution

: \int_{0}^{16} \frac{e ^{u} - 1}{2} d u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{16} e^{u} d u - \int_{0}^{16} 1 d u)

\int_{0}^{16} e^{u} d u = e^{16} - 1

\int_{0}^{16} 1 d u = 16

= \frac{1}{2} (e^{16} - 1 - 16)

Simplify

\frac{1}{2} (e^{16} - 1 - 16) : \frac{e ^{16} - 17}{2}

= \frac{e ^{16} - 17}{2}

\int_{0}^{0.4} 30 x^{2} (1 - x^{2}) d x

\int_{1}^{4} \frac{1}{x} d x

\int_{2}^{4} 10 x d x

\int_{0}^{2 x} (y e^{x^{3}}) d y

\int_{y}^{y^{3}} e^{\frac{x}{y}} d x

What is the integral from 0 to 4 of y(e^{y^2}-1) ?
The integral from 0 to 4 of y(e^{y^2}-1) is (e^{16}-17)/2