# How do you use the second fundamental theorem of Calculus to find the derivative of given int sqrt(6 + r^3) dr from [7, x^2]?

Oct 19, 2016

$\frac{d}{\mathrm{dx}} {\int}_{7}^{{x}^{2}} \sqrt{6 + {r}^{3}} \mathrm{dr} = 2 x \sqrt{6 + {x}^{6}}$

#### Explanation:

Let $F \left(r\right) + C = \int \sqrt{6 + {r}^{3}} \mathrm{dr}$, that is, the function such that $\frac{d}{\mathrm{dr}} F \left(r\right) = \sqrt{6 + {r}^{3}}$. The second fundamental theorem of calculus states that

${\int}_{a}^{b} \sqrt{6 + {r}^{3}} \mathrm{dr} = F \left(b\right) - F \left(a\right)$

With that, we have

$\frac{d}{\mathrm{dx}} {\int}_{7}^{{x}^{2}} \sqrt{6 + {r}^{3}} \mathrm{dr} = \frac{d}{\mathrm{dx}} \left(F \left({x}^{2}\right) - F \left(7\right)\right)$

$= \frac{d}{\mathrm{dx}} F \left({x}^{2}\right) - \frac{d}{\mathrm{dx}} F \left(7\right)$

$= \sqrt{6 + {\left({x}^{2}\right)}^{3}} \left(\frac{d}{\mathrm{dx}} {x}^{2}\right) - 0$

$= 2 x \sqrt{6 + {x}^{6}}$

where in the second to last step we used the chain rule along with the fact that $F \left(7\right)$ is a constant, and thus has a derivative of $0$.