# Question #674da

Nov 25, 2016

$\frac{d}{\mathrm{dx}} {\int}_{0}^{2 x} \left({e}^{t} + 2 t\right) \mathrm{dt} = 2 {e}^{2 x} + 8 x$

#### Explanation:

$\frac{d}{\mathrm{dx}} {\int}_{0}^{2 x} \left({e}^{t} + 2 t\right) \mathrm{dt} = \frac{d}{\mathrm{dx}} \left[{\int}_{0}^{2 x} {e}^{t} \mathrm{dt} + {\int}_{0}^{2 x} 2 t \mathrm{dt}\right]$

$= \frac{d}{\mathrm{dx}} \left[{\left({e}^{t}\right)}_{0}^{2 x} + {\left({t}^{2}\right)}_{0}^{2 x}\right]$

$= \frac{d}{\mathrm{dx}} \left({e}^{2 x} - {e}^{0} + {\left(2 x\right)}^{2} - {0}^{2}\right)$

$= \frac{d}{\mathrm{dx}} \left({e}^{2 x} + 4 {x}^{2} - 1\right)$

$= 2 {e}^{2 x} + 8 x$

Note that we could have avoided calculating the integral and derivatives by letting $F \left(t\right) = \int \left({e}^{t} + 2 t\right) \mathrm{dt}$, meaning $F ' \left(t\right) = {e}^{t} + 2 t$. Then we have

$\frac{d}{\mathrm{dx}} {\int}_{0}^{2 x} \left({e}^{t} + 2 t\right) \mathrm{dt} = \frac{d}{\mathrm{dx}} \left[F \left(2 x\right) - F \left(0\right)\right]$

$= 2 F ' \left(2 x\right) - 0$

$= 2 \left[{e}^{2 x} + 2 \left(2 x\right)\right]$

$= 2 {e}^{2 x} + 8 x$