What is the integral from-2 to 2 of e^{x+x^2} ?

The integral from-2 to 2 of e^{x+x^2} is (\sqrt[4]{e}(sqrt(pi)\erfi(5/2)-sqrt(pi)\erfi(-3/2)))/2

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integral from-2 to 2 of e^{x+x^2}

\int_{- 2}^{2} e^{x + x^{2}} d x

\frac{4 e ( π erfi ( \frac{5}{2} ) - π erfi ( - \frac{3}{2} ) )}{2}

Solution steps

\int_{- 2}^{2} e^{x + x^{2}} d x

Complete the square

x + x^{2} : (x + \frac{1}{2})^{2} - \frac{1}{4}

= \int_{- 2}^{2} e^{(x + \frac{1}{2})^{2} - \frac{1}{4}} d x

Apply u-substitution

: \int_{- \frac{3}{2}}^{\frac{5}{2}} e^{u^{2} - \frac{1}{4}} d u

= \int_{- \frac{3}{2}}^{\frac{5}{2}} e^{u^{2} - \frac{1}{4}} d u

Simplify

e^{u^{2} - \frac{1}{4}} : e^{u^{2}} 4 e

= \int_{- \frac{3}{2}}^{\frac{5}{2}} e^{u^{2}} 4 e d u

Take the constant out:

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

= 4 e \cdot \int_{- \frac{3}{2}}^{\frac{5}{2}} e^{u^{2}} d u

Use the nonelementary integral:

\int e^{u^{2}} d u = \frac{π}{2} erfi (u)

= 4 e [\frac{π}{2} erfi (u)]_{- \frac{3}{2}}^{\frac{5}{2}}

Compute the boundaries:

\frac{π erfi ( \frac{5}{2} ) - π erfi ( - \frac{3}{2} )}{2}

= 4 e \frac{π erfi ( \frac{5}{2} ) - π erfi ( - \frac{3}{2} )}{2}

Simplify

= \frac{4 e ( π erfi ( \frac{5}{2} ) - π erfi ( - \frac{3}{2} ) )}{2}

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What is the integral from-2 to 2 of e^{x+x^2} ?
The integral from-2 to 2 of e^{x+x^2} is (\sqrt[4]{e}(sqrt(pi)\erfi(5/2)-sqrt(pi)\erfi(-3/2)))/2