What is the integral from 1 to x^2 of (t-4)e^{-t^2} ?

The integral from 1 to x^2 of (t-4)e^{-t^2} is -0.5(e^{-x^4}-1/e)-4(0.88622…\erf(x^2)-0.74682…)

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integral from 1 to x^2 of (t-4)e^{-t^2}

\int_{1}^{x^{2}} (t - 4) e^{- t^{2}} d t

- 0.5 (e^{- x^{4}} - \frac{1}{e}) - 4 (0.88622 \dots erf (x^{2}) - 0.74682 \dots)

Solution steps

\int_{1}^{x^{2}} (t - 4) e^{- t^{2}} d t

Expand

(t - 4) e^{- t^{2}} : e^{- t^{2}} t - 4 e^{- t^{2}}

= \int_{1}^{x^{2}} e^{- t^{2}} t - 4 e^{- t^{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{x^{2}} e^{- t^{2}} t d t - \int_{1}^{x^{2}} 4 e^{- t^{2}} d t

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

\int_{1}^{x^{2}} 4 e^{- t^{2}} d t = 4 (0.88622 \dots erf (x^{2}) - 0.74682 \dots)

= - \frac{1}{2} (e^{- x^{4}} - \frac{1}{e}) - 4 (0.88622 \dots erf (x^{2}) - 0.74682 \dots)

Simplify

= - 0.5 (e^{- x^{4}} - \frac{1}{e}) - 4 (0.88622 \dots erf (x^{2}) - 0.74682 \dots)

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What is the integral from 1 to x^2 of (t-4)e^{-t^2} ?
The integral from 1 to x^2 of (t-4)e^{-t^2} is -0.5(e^{-x^4}-1/e)-4(0.88622…\erf(x^2)-0.74682…)