# How do you evaluate d /dx \int_[x]^{x^4} sqrt{t^2+t} dt?

Mar 6, 2017

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

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{4}} \sqrt{{t}^{2} + t} \mathrm{dt}$
$= \left[\setminus \sqrt{{\left({x}^{4}\right)}^{2} + {x}^{4}}\right] \cdot \left(4 {x}^{3}\right) - \left[\setminus \sqrt{{\left(x\right)}^{2} + x}\right] \cdot \left(1\right)$
**don't forget to chain!!

Mar 6, 2017

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{4}} \setminus \sqrt{{t}^{2} + t} \setminus \mathrm{dt} = 4 {x}^{3} \sqrt{{x}^{8} + {x}^{4}} - \sqrt{{x}^{2} + x}$

#### Explanation:

If asked to find the derivative of an integral using the fundamental theorem of Calculus, you should not evaluate the integral.

The Fundamental Theorem of Calculus tells us that:

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

(ie the derivative of an integral gives us the original function back).

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

(notice the upper and lower bounds are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral as follows:

${\int}_{x}^{{x}^{4}} \setminus \sqrt{{t}^{2} + t} \setminus \mathrm{dt} = {\int}_{0}^{{x}^{4}} \setminus \sqrt{{t}^{2} + t} - {\int}_{0}^{x} \setminus \sqrt{{t}^{2} + t} \setminus \mathrm{dt}$

We have arbitrary chosen the lower limit as $0$ wlog (any number will do!). The second integral is is now in the correct form, and we can directly apply the FTOC and write the derivative as:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{0}^{x} \setminus \sqrt{{t}^{2} + t} \setminus \mathrm{dt} = \sqrt{{x}^{2} + x}$

And using the chain rule we can write:

$\frac{d}{\mathrm{dx}} {\int}_{0}^{{x}^{4}} \setminus \sqrt{{t}^{2} + t} = \frac{d \left({x}^{4}\right)}{\mathrm{dx}} \frac{d}{d \left({x}^{4}\right)} {\int}_{0}^{{x}^{4}} \setminus \sqrt{{t}^{2} + t}$

Now, $\frac{d \left({x}^{4}\right)}{\mathrm{dx}} = 4 {x}^{3}$, And, using the FTOC, we have:

$\frac{d}{d \left({x}^{4}\right)} {\int}_{0}^{{x}^{4}} \setminus \sqrt{{t}^{2} + t} = \sqrt{{\left({x}^{4}\right)}^{2} + \left({x}^{4}\right)}$

Hence combining these trivial results we get:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{{x}^{4}} \setminus \sqrt{{t}^{2} + t} \setminus \mathrm{dt} = 4 {x}^{3} \sqrt{{x}^{8} + {x}^{4}} - \sqrt{{x}^{2} + x}$