Integration by substitution with radicals

[adaschau2]

Homework Statement

2
h(x)=∫√(1+t^3) dt find h'(2)
x^2

Homework Equations

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).

 

[LCKurtz]

Homework Statement

2
h(x)=∫√(1+t^3) dt find h'(2)
x^2

Homework Equations

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).

Re-rendering your integral in tex:

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

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.

 

[adaschau2]

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?

 

[Bohrok]

Try reading the "First part" here, what LCKurtz wanted you to look up.
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part

 

[adaschau2]

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.

 

[Dick]

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?

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.

 

[adaschau2]

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.

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?

 

[Dick]

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?

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

 

[adaschau2]

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

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!

 

[Dick]

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!

Sign's wrong, dude. Other than that, fine!