Mastering Definite Integrals: Solving for the Square Root of 1+t^3 from 0 to 3

[limelightdevo]

Homework Statement

Definite integral, from 0 to 3, of Square root of 1+t^3.

Homework Equations

Tried substitution

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.

 

[algebrat]

Homework Statement

Definite integral, from 0 to 3, of Square root of 1+t^3.

Homework Equations

Tried substitution

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.

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?

 

[NewtonianAlch]

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

 

[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.

 

[algebrat]

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

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.

 

[NewtonianAlch]

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.

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.

 

[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

 

[Curious3141]

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

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.