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

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

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

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

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

Solution steps

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

Complete the square

x^{2} - 2 x : (x - 1)^{2} - 1

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

Apply u-substitution

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

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

Simplify

e^{u^{2} - 1} : e^{u^{2}} e^{- 1}

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

Take the constant out:

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

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Use the nonelementary integral:

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

= \frac{1}{e} [\frac{π}{2} erfi (u)]_{- 1}^{1}

Compute the boundaries:

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

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

Simplify

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

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