# Stumped on Definite Integral Question

1. Jun 19, 2012

### limelightdevo

1. The problem statement, all variables and given/known data
Definite integral, from 0 to 3, of Square root of 1+t^3.

2. Relevant equations
Tried substitution

3. The attempt at a solution
There should be a simple way to do this but I can't seem to figure it out. Tried the substitution and whatnot but couldn't reach an answer.

2. Jun 19, 2012

### algebrat

u-sub won't work because the derivative of the inside is not outside.

trig-sub won't work because it should be t^2.

You might try u^2=1+t^3, though I haven't tried it so no idea.

i would say maybe some extra special clever application of integration by parts, but i wouldn't bet on it.

How did the problem arise, are you sure this is the problem you are supposed to do?

Is it possible that using a computer is expected?

3. Jun 19, 2012

### NewtonianAlch

MAPLE output for both the definite and indefinite integral looks ridiculous with all kinds of special functions.

4. Jun 19, 2012

### Millennial

Same with Wolfram. This integral is not expressable in elementary terms (verified by Risch's algorithm.) It involves the elliptic integral function. If that is acceptable, you can do the problem; but it will still be ridiculously long.

5. Jun 19, 2012

### algebrat

As much as I have a warm place in my heart for MAPLE (created at UW?), I would just plug it in online at wolfram. It mentions something about the hypergeometric series, and about 7.3.

6. Jun 19, 2012

### NewtonianAlch

Haha yea, I use both to be honest, certain things are easier for me to find on MAPLE and in this instance I had it open already.

7. Jun 19, 2012

### limelightdevo

Well, the actual question is different from the question I asked. I didn't want to post the actual one because I partially solved it. Here it is though if there is in fact a different approach. I did find the derivative and was trying to get f^-1(0) by saying f^-1(0) = x so f(x) = 0, but couldn't solved the integral to be able to equal it to 0 to get x.

If f(x) = integral, from 3 to x , of square root of (1+t^3) dt, find (f^-1)'(0).

By the way, I would like to have the simplest and cleanest approach to this to be efficient.

Thanks

8. Jun 19, 2012

### Curious3141

Let $y = f(x) = \int_3^x \sqrt{1+t^3}dt$

Use the Fundamental Theorem of Calculus to figure out $\frac{dy}{dx} = f'(x)$

Now, if $y = f(x)$, then $f^{-1}(y) = x$ and ${(f^{-1})}'(y) = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{f'(x)}$.

You're asked to determine ${(f^{-1})}'(0)$, so first figure out what (obvious) value of $x$ would make $y$ zero. The rest is trivial.