# How do you calculate the derivative of intsqrt((x^3-2x+6) )dx from [x^2,-2]?

Jul 15, 2018

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

#### Explanation:

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

That looks odd, right(?!?!), because one letter, $x$, is being totally overloaded.

It can be written with a new dummy variable for the integration, which we label as $\Omega$:

d/dx ( int_(x^2)^(-2) sqrt((Omega^3-2Omega+6) ) qquad dOmega ) qquad square

You can, from here, evaluate the integral, and apply the interval. Or you can differentiate under the integral sign.

The FTC says that:

• $\frac{d}{\mathrm{dx}} \left({\int}_{a}^{x} f \left(\Omega\right) \setminus \mathrm{dO} m e g a\right) = f \left(x\right)$

Add in a variable limit and the Chain Rule:

• $\frac{d}{\mathrm{dx}} \left({\int}_{a}^{g \left(x\right)} f \left(\Omega\right) \setminus \mathrm{dO} m e g a\right) = f \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Reverse the limits:

• $\frac{d}{\mathrm{dx}} \left({\int}_{g \left(x\right)}^{a} f \left(\Omega\right) \mathrm{dO} m e g a\right) = - f \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\square = - \sqrt{\left({\left({x}^{2}\right)}^{3} - 2 \left({x}^{2}\right) + 6\right)} \cdot 2 x$

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

Jul 16, 2018

x is used as variable in integration and also in one of the limits. It is derivative with respect to x in the integrand. So, integrand is the derivative. Answer: $\sqrt{{x}^{3} - 2 x + 6}$.

#### Explanation:

It is a good convention not to use x in the limits, as dummy variable

in the integrand.

If the question is to calculate the derivative with respect to x of

$\int \sqrt{{X}^{3} - 2 X + 6} \mathrm{dX}$, then it is the derivative of the

integral that is a function of x (not X ). In this form, it is derivative

with respect to x of the integral evaluated between the limits.