# Integration by substitution with radicals

1. Jan 7, 2010

1. The problem statement, all variables and given/known data
2
h(x)=∫√(1+t^3) dt find h'(2)
x^2

2. Relevant equations

3. The attempt at a solution
I started out solving this equation by flipping x^2 and 2 and making the integral negative. From here on out, I'm lost. I've tried substituting u in for 1+t^3 and solving it that way, but I get caught up when I get du=3t^2 dt and h(x)=∫u^(1/2) du/(3t^2). How am I supposed to compensate for a t variable in the denominator? I can handle the rest of the problem if I can just find the antiderivative of √(1+t^3).

2. Jan 7, 2010

### LCKurtz

$$h(x) = \int_{x^2}^2 \sqrt{1+t^3}\ dt$$

It didn't ask you to find a simple formula for h(x). You only need h'. Look at what the fundamental theorem of calculus tells you about derivatives of integrals as functions of their limits.

3. Jan 7, 2010

If I understand this right, I'm setting f(x)= ∫√(1+t^3) dt, lower limit=2, upper limit=x
So f'(x)= √1+x^3 and g'(x)=2x. Going back to the original equation, h'(x)=-f'(g(x))g'(x)=√[1+(x^2)^3]2x=2x√(1+x^2)
h'(2)=2(2)√(1+2^2)=4√5

is that right?

4. Jan 7, 2010

### Bohrok

5. Jan 7, 2010

Okay, so if
$$h(x)= \int_{2}^x^2 \sqrt{1+t^3} \ dt$$
and $$\sqrt{1+t^3}$$ equals h'(t), then h(x)=h'(x^2)?
The solution is h'(x)=√(1+(x^2)^3) and h'(2)=√65?
This doesn't sound right, so I apologize for my ignorance.

Last edited: Jan 7, 2010
6. Jan 7, 2010

### Dick

About 50% of that is right. The rest is all garbled up. Lower limit x^2, upper limit 2. Start by fixing that. And how did (x^2)^3 turn into x^2? The -f'(g(x))*g'(x) is the center of the correct part.

7. Jan 7, 2010

I already did rearrange the limits, that's about the only part that I knew to do initially. I did a little research and found that the derivative with respect to x of the integral from 2 to x^2 of √(1+t^3) dt equals -f'(g(x))g'(x) (negative because I flipped the limits). Now if f(x)=√(1+x^3) and g(x)=x^2, then the resulting equation would be h'(x)=-[√(1+(x^2)^3)*2x]=-2x√(1+x^6). Is that correct reasoning? If not, what part did I mess up?

8. Jan 7, 2010

### Dick

Now that's correct. It's just that your previous post had some typos in it. So h'(2)=??

9. Jan 7, 2010

Yeah, sorry about that, I should really learn to use the LaTeX code better. And I believe the final answer would be $$h'(2)=4\sqrt{65}$$. There we go, that's a start. Thanks for all of your help!