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

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

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

\int_{0}^{1} e^{x^{2}} (x^{3} + x) d x

\frac{1}{2} + \frac{e - 1}{2}

Decimal

1.35914 \dots

Solution steps

\int_{0}^{1} e^{x^{2}} (x^{3} + x) d x

Expand

e^{x^{2}} (x^{3} + x) : e^{x^{2}} x^{3} + e^{x^{2}} x

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

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

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

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

= \frac{1}{2} + \frac{e - 1}{2}

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