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Question Number 147275 by ArielVyny last updated on 19/Jul/21

$$f\left(x\right)={\int }_0^x{e}^{t-\frac{t^2}{2}}dt$$$$showthat{\int }_0^{1}f\left(t\right)dt=\sqrt{e}-1$$

Answered by mindispower last updated on 19/Jul/21

$${\int }_0^{1}f\left(t\right)dt={\left[tf\left(t\right)\right]}_0^1-{\int }_0^1{tf}^{\prime }\left(t\right)dt$$$$=f\left(1\right)-{\int }_0^1{te}^{t-\frac{t^2}{2}}dt$$$$={\int }_0^1(1-t)e^{t-\frac{t^2}{2}}dt={\left[e^{t-\frac{t^2}{2}}\right]}_0^1=\sqrt{e}-1$$

Commented by ArielVyny last updated on 20/Jul/21

$$thanksir$$

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