What is the integral from 1 to 0 of e^{x^2-x} ?

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

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integral from 1 to 0 of e^{x^2-x}

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

- \frac{4 e ( π erfi ( \frac{1}{2} ) - π erfi ( - \frac{1}{2} ) )}{2}

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Complete the square

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

= - \int_{0}^{1} e^{(x - \frac{1}{2})^{2} - \frac{1}{4}} d x

Apply u-substitution

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

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

Simplify

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

= - \int_{- \frac{1}{2}}^{\frac{1}{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{1}{2}}^{\frac{1}{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{1}{2}}^{\frac{1}{2}}

Compute the boundaries:

\frac{π erfi ( \frac{1}{2} ) - π erfi ( - \frac{1}{2} )}{2}

= - 4 e \frac{π erfi ( \frac{1}{2} ) - π erfi ( - \frac{1}{2} )}{2}

Simplify

= - \frac{4 e ( π erfi ( \frac{1}{2} ) - π erfi ( - \frac{1}{2} ) )}{2}

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