<p>What is the antiderivative of f(x)=x^2 * sqrt(1-x)dx ?</p>
<p>Alright... so you're going to have to do a u substitution.
Set u=(1-x) so x=(1-u)
so du=-dx and dx=-du
So now you have:
∫(1-u)²√u (-du)
And you multiply that out:
∫(1-2u+u²)u^1/2 (-du)
And once again:
∫(-u^1/2+2u^3/2-u^5/2)(du)
Which is very easy to integrate as:
-(2/3)u^3/2+(4/5)u^5/2-(2/7)u^7/2+ C
Where C is some constant.
Remember that u=1-x, so:
∫(x)²√(1-x) (dx)= -(2/3)(1-x)^3/2+(4/5)(1-x)^5/2-(2/7)(1-x)^7/2+ C</p>
<p>Hope that helps!</p>
<p>wow, this is a long problem. I think I solved it, but there could be some mistakes. Use the integration by parts:</p>
<p>Lets call
antider ( x^2 * sqrt(1-x) dx ) = I
u = x^2 dv = sqrt(1-x)
u' = 2x v = -2/3 * (1-x) ^(3/2)
SO, I = uv - antider (vu') ```
I = (-2(x^2)/3) * (1-x) ^(3/2) - ``` antider (-4x/3 * (1-x) ^(3/2) dx) ``` = = (-2(x^2)/3) * (1-x) ^(3/2) -4/3 ``` antider (x * (1-x) ^(3/2) dx) ```
WHERE ``` antider (x * (1-x) ^(3/2) dx) ``` = u = x dv = (1-x)^(3/2) u' = 1 v= -2/5 * (1-x)^(5/2) ``` antider (x * (1-x) ^(3/2) dx) ``` = -2x/5 - ``` antider(-2/5 *(1-x)^5/2 dx) ```
``` antider(-2/5 *(1-x)^5/2 dx) ``` =
SO now we put everything together:
antider (x * (1-x) ^(3/2) dx)
= -2x/5 + (4/35 * (1-x)^(7/2)
I = (-2(x^2)/3) * (1-x) ^(3/2) +4/3* { 2x/5 + (4/35 * (1-x)^(7/2)) } + C</p>
<p>sorry, I din't know anyone else was typing the solution. We should see whose answer is right, they seem to be a little different, except mine approach is insanely harder (I didnt see any other way).</p>
<p>I just went back over my math, and it seems to be correct--- check it for me to make sure, but... u substitution is just so much easier. There's a lot less time to make mistakes. I thought about doing the whole integration by parts thing, but I saw that it was going to be a pain in the ass, so I didn't.</p>
<p>Try taking the derivative of your equation and see if it matches up to the original one. Mine does, I just checked.</p>
<p>Uh, no mine does not match. I guess the typing created a potential for many mistakes, I should have done it on paper. Sorry aw_rootbeer1, but galofdasouth's solution is about 10 times better than mine (which didn't even provide the right answer).</p>
<p>Just wait, ktoto, rootbeer will come back and check this and be like "Whoa" hehe.</p>
<p>"Woah!"
Thanks! That helped a lot.
Rootbeer</p>