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

##### 2 Answers

#### Explanation:

You are asking for:

That looks odd, right(?!?!), because one letter,

It can be written with a new **dummy** variable for the integration, which we label as

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

The FTC says that:

#d/dx ( int_(a)^(x) f(Omega) \ dOmega ) = f(x)#

Add in a variable limit and the Chain Rule:

#d/dx ( int_(a)^(g(x)) f(Omega) \ dOmega ) = f(g(x)) * g'(x) #

Reverse the limits:

#d/dx ( int_(g(x))^(a) f(Omega) dOmega ) = - f(g(x)) * g'(x) #

So our answer is:

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:

#### 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

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.