# FInd numeric value of int_1^esqrt(lnx)dx-int_0^1e^(x^2)dx?

Jul 10, 2018

$\approx - 0.207$

#### Explanation:

$I = {\underbrace{{\int}_{1}^{e} \sqrt{\ln x} \mathrm{dx}}}_{{I}_{1}} \textcolor{red}{-} {\underbrace{{\int}_{0}^{1} {e}^{{x}^{2}} \mathrm{dx}}}_{{I}_{2}}$

For ${I}_{1}$:

Let: $y = \sqrt{\ln x}$

• $\implies \left\{\begin{matrix}x = {e}^{{y}^{2}} \\ \mathrm{dx} = 2 y {e}^{{y}^{2}} \mathrm{dy}\end{matrix}\right.$

$\therefore {I}_{1} = {\int}_{0}^{1} \setminus 2 {y}^{2} {e}^{{y}^{2}} \mathrm{dy}$

Switching the letter of the dummy variable $y$ back to $x$.

$I = {I}_{1} \textcolor{red}{-} {I}_{2}$

$= \left({I}_{1} + {I}_{2}\right) - 2 {I}_{2}$

$= {\int}_{0}^{1} \setminus {\underbrace{2 {x}^{2} {e}^{{x}^{2}} + {e}^{{x}^{2}}}}_{= \frac{d}{\mathrm{dx}} \left(x {e}^{{x}^{2}}\right)} \mathrm{dx} - 2 {\underbrace{{\int}_{0}^{1} {e}^{{x}^{2}} \mathrm{dx}}}_{= \frac{\sqrt{\pi}}{2} \text{erf} i \left(1\right) \approx 1.463}$

$\approx e - 2 \left(1.463\right) \approx - 0.207$