What is the integral from 2x to x^5 of e^{-t^2} ?

The integral from 2x to x^5 of e^{-t^2} is (sqrt(pi))/2 \erf(x^5)-(sqrt(pi))/2 \erf(2x)

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integral from 2x to x^5 of e^{-t^2}

\int_{2 x}^{x^{5}} e^{- t^{2}} d t

\frac{π}{2} erf (x^{5}) - \frac{π}{2} erf (2 x)

Solution steps

\int_{2 x}^{x^{5}} e^{- t^{2}} d t

Use the nonelementary integral:

\int e^{- t^{2}} d t = \frac{π}{2} erf (t)

= [\frac{π}{2} erf (t)]_{2 x}^{x^{5}}

Compute the boundaries:

\frac{π}{2} erf (x^{5}) - \frac{π}{2} erf (2 x)

= \frac{π}{2} erf (x^{5}) - \frac{π}{2} erf (2 x)

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\int_{0}^{4} \frac{1}{64 x ^{2} + 25} d x

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What is the integral from 2x to x^5 of e^{-t^2} ?
The integral from 2x to x^5 of e^{-t^2} is (sqrt(pi))/2 \erf(x^5)-(sqrt(pi))/2 \erf(2x)