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

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

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

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

π erfi (4) - \frac{e ^{16} - 1}{4}

Solution steps

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

Expand

e^{y^{2}} (2 - \frac{y}{2}) : 2 e^{y^{2}} - \frac{e ^{y^{2}} y}{2}

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

Apply the Sum Rule:

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

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

\int_{0}^{4} 2 e^{y^{2}} d y = π erfi (4)

\int_{0}^{4} \frac{e ^{y^{2}} y}{2} d y = \frac{e ^{16} - 1}{4}

= π erfi (4) - \frac{e ^{16} - 1}{4}

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