Calculus help!!

the answer is.

this integral cannot be expressed in terms of elementary functions.

the integral exists, but it's very complicated and can't be expressed simply as a function of t
 
^shitty. yeah the original problem was find F'(x) if f(x) = integral[x^2,0] of (1-t^3)^(-1/2) dt

OR

d/dx {integral[x^2,0] of (1-t^3)^(-1/2) dt}

which is fairly straight forward and simple. I started it out by trying to find the integral and was like whaat!? so I thought I'd throw it at you guys to see if that part was possible or not. Thanks for the help!
 
can't help you. but this is funny.

xzibitfunctionvm3.jpg


 
probably needs a u-sub or something. it actually looks like a trig sub, but i left my list of trig subs at school
 
second line is wrong i think.

its so hard to do calc on the computer. i need to write it down on paper for it to make any sense in my head.
 
yeah you need to do u-sub, its like 1/a square root of blah blah right? so what ever is inside the square root is u and then du is teh derivative of that, and then you integrate for you and plug u back in

sorry its kinda hard to do calculus on the internet, but hope it helps
 
do a u substitution of like t=u^(2/3).then its just like the derivative of inverse sin i believe.then take the integral so its inverse sin.Don't forget to change your limits of integration when you do your substitution.
At least that what I'd do.
 
Fuck I'm an idiot, I think if you do that you'll have a random u^(/1/3) hanging out n top from changing dt to du. So, I'm done.
 
What if you use the thing that d/dx(int(a,g(x)) of f(t)dt=g(g(x))g'(x)
then it equals int of 2x/(1-x^6)^(1/2) if you treat your original problem as the integral of the left side of the equation right?
I think thats right and that last part is an easy integral.
Probably wrong again though
 
im pretty sure its a trig substitution. so that whole function would go to cos(t) or something to that nature. not sure though my calc book is in newyork lol.
 
Okay so you are looking for F' and you have F...you just plug in the x^2 into the t value. that gives you the derivitive. depending on the book that is the Fundamental therom of calculus 1 or 2...
 
If that is the question i don't think you have to integrate at all. You use one of the fundamental theorems of calculus to find that the derivative is (1-t^3)^-(1/2) then you plug in x^2 and use the chain rule to find your solution
 
no. this is wrong. you are trying to use the second fundamental theorem of calculus which can not be applied to this integral. that is for finding derivatives of integrals where the upper bound is a function.

the problem is integrate (indefinately) (1-t^3)^-1/2....if your in calc 2 you can use binomial expansion where m = -1/2. you can take the binomial expansion for (1-t)^-1/2 and sub t^3 in for t as your last step
 
i didnt see this post. my first post is wrong. the second fundamental theorem is how you do it if your lookin for the derivative
 
shit, this just reminded me i haven't done calc in a year and i'm gonna need it for a class next semester. fuck
 
so to let all of you knowww:

substitute x^2 for t

take the derivative

multiply by the derivative of the upper limit

That's what I learned in school today mommy
 
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