# How do I evaluate d/dx \int_5^(x^4) \sqrt{t^2 + t} dt?

Mar 6, 2017

To solve $\frac{d}{\mathrm{dx}} \setminus {\int}_{5}^{{x}^{4}} \sqrt{{t}^{2} + t} \mathrm{dt}$, we will use the Fundamental Theorem of Calculus.

$\frac{d}{\mathrm{dx}} \setminus {\int}_{5}^{{x}^{4}} \sqrt{{t}^{2} + t} \mathrm{dt}$
$= \left[\setminus \sqrt{{\left({x}^{4}\right)}^{2} + {x}^{4}}\right] \cdot \left(4 {x}^{3}\right)$ **don't forget to chain!!

Mar 6, 2017

$= 4 {x}^{5} \setminus \sqrt{{x}^{4} + 1}$

#### Explanation:

The FTC tells us that, for constant $a$ and parameter $u$:

$\frac{d}{\mathrm{du}} {\int}_{a}^{u} f \left(t\right) \mathrm{dt} = f \left(u\right)$

Now for the chaining bit.

If in fact $u = u \left(x\right)$, we can say that:

$\frac{d}{\mathrm{dc} o l \mathmr{and} \left(red\right) \left(x\right)} {\int}_{a}^{u \left(x\right)} f \left(t\right) \mathrm{dt}$

$= \frac{d}{\mathrm{du}} \left({\int}_{a}^{u} f \left(t\right) \mathrm{dt}\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

So for the specific question, where $u \left(x\right) = {x}^{4}$:

$\frac{d}{\mathrm{dx}} {\int}_{5}^{{x}^{4}} \sqrt{{t}^{2} + t} \setminus \mathrm{dt}$

$= \sqrt{{\left({x}^{4}\right)}^{2} + {x}^{4}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{4}\right)$

$= {x}^{2} \setminus \sqrt{{x}^{4} + 1} \cdot 4 {x}^{3}$

$= 4 {x}^{5} \setminus \sqrt{{x}^{4} + 1}$