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

2 Answers
Jul 15, 2018


#= - 2xsqrt(x^6-2 x^2 +6 ) #


You are asking for:

#d/dx ( int_(x^2)^(-2) sqrt((x^3-2x+6) )dx )#

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:

  • #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:

#square = - sqrt(((x^2)^3-2(x^2)+6) )* 2x#

#= - 2xsqrt(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)#.


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 ) 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.