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