f(x)=∫₀^x3 t³dt

Integral of 0 to x^3 t^3 dt

What I know tells me to do this:

Let u=x^3
du=3x^2dx

Divide by 3x^2 on both sides

(1/3x^2)du=dx

Going back to f(x), I replace 't' with 'u' and multiply by the derivative of 'u' due to the Chain Rule.

(u^3)1/3x^2

Now I substitute 'u' with x^3 and get:

(x^9)/(3x^2)


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I submit my homework electronically, and it tells me if the answer is correct or not. Obviously it's telling me that's not the right answer. Looking at Wolfram Alpha, I find (x^12)/4 but that isn't the answer either.

Help?

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UPDATE: Well, this thread was useless. Just realized my mistake. Although I guess the question is now, why is it that you don't go all the way with the chain rule this time, because I don't get that.

If you tell me how you typed that I'll give you a hand.

Quote from: SexyPiranha on December 08, 2014, 08:58:40 PM

If you tell me how you typed that I'll give you a hand.

∫^{you're a faggot}_{you can't even into formatting}

no stop I'll cri

Not sure why you think you need the chain rule. Just integrate the function, then evaluate it on the interval. The integral is t^4/4, from 0 to x^2.

Should be x^8/4.