Help with a tricky calculus question

Thomas_S_Hunterson

This question has got me well and truly stumped. I've tried several things but can't work out the integral. I'm probably missing something stupid.:o



Would anyone know how to do it?

Thanks
Sean

Fremen

Did you try splitting it up into two integrals?

Let F be your integrand. Then your question can be written as

d/dx(integral[0 to x^2](Fdt)) + d/dx(integral[x to 0](Fdt)

then the fundemental theorem of calculus should work after a change of variables

Thomas_S_Hunterson

I'm not quite sure I understand you, what should I be changing the variables to?

For the FTC to work, shouldn't it be differentiation and integration with respect to the same variable?

Fremen

d/dx(integral[0 to x^2](Fdt)):

set u = x^2
then d/dx(integral[0 to x^2](Fdt)) = (d/du(integral[0 to u](Fdt)))*du/dx

or 2x*F(u), which is equal to 2x*F(x^2)


d/dx(integral[x to 0](Fdt):

same as -1*d/dx(integral[0 to x](Fdt) = -F(x)

For the FTC to work, shouldn't it be differentiation and integration with respect to the same variable?

well, it's the same principle as this:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part

Michael Collins

Yeh that integral isn't solvable in terms of elementary functions like sin and cos. However, since you don't actually need to find an anti-derivative you don't need to actually solve the integral, just apply the chain rule.

e.g. let F(t) be the anti-derivative of your integrand f(t). So then the definate integral becomes:

F(u2) - F(u1) (where u2 and u1 are the upper and lower limits respectively)

but you want the derivative of this so

F'(x^2).2x - F'(x) ...by the chain rule

f(x^2).2x-f(x) ...where this is your integrand function, just with respect to t.

Thomas_S_Hunterson

Gotcha, wraps up neatly (I had gone and tried to solve the integral...things got messy), thanks a mil for the help.

Hopefully it'll come up in my exam on thurs now:p