# Thread: Arc Length Problem with e

1. ## Arc Length Problem with e

Explain why the two integrals are equal:
$\int_1^e\sqrt{1 + \frac{1}{x^2}}dx = \int_0^1\sqrt{1 + e^{2x}}dx$

All I was able to work out is the two original functions:
$f(x) = \frac{-1}{x}$

$g(x) = e^{2x}$

Looking at a graphing calculator, I can see the similarities, but I don't know how to articulate them mathematically.

2. Did you consider a change of variable (substitution)? For example, if, in the integral on the left, you let $x= e^{-t}$, so that $t= -ln(x)$, what does it become?

3. Originally Posted by HallsofIvy
Did you consider a change of variable (substitution)? For example, if, in the integral on the left, you let $x= e^{-t}$, so that $t= -ln(x)$, what does it become?
Dear HallsofIvy,

I think $x= e^{t}$ substitution on the left hand side integral will be more easy.

4. Well, if you do that you arrive at
$f(x) = e^{-2lnx}$
which is still not
$e^{2x}$

5. Sudharaka gave you the substitution, there's nothing more to add here.

6. Originally Posted by seuzy13
Well, if you do that you arrive at
$f(x) = e^{-2lnx}$
which is still not
$e^{2x}$

You make the substitution in the integral. Have you been taught how to integrate by making a substitution?

7. Originally Posted by mr fantastic
You make the substitution in the integral. Have you been taught how to integrate by making a substitution?
I think so, but for some reason I still don't see where this is going. I'm returning to calculus after being out of it since last May, so that might be the reason for my not understanding this right away.

The integration by substitution I remember looked something like this:
$\int_1^e\sqrt{1 + \frac{1}{u^2}}du$
Where u would be, I guess, e^t, and du would be e^t, but that doesn't help solve at all and even doesn't work because you can't divide by e^t outside the integral. (So obviously I don't understand what's being suggested here.)

8. Originally Posted by seuzy13
Explain why the two integrals are equal:
$\int_1^e\sqrt{1 + \frac{1}{x^2}}dx = \int_0^1\sqrt{1 + e^{2x}}dx$

All I was able to work out is the two original functions:
$f(x) = \frac{-1}{x}$

$g(x) = e^{2x}$

Looking at a graphing calculator, I can see the similarities, but I don't know how to articulate them mathematically.
$\displaystyle I = \int_{x=1}^{x=e} \sqrt{1 + \frac{1}{x^2}} \, dx$.

Substitute $x = e^u$.

Then:

1. $\displaystyle \frac{dx}{du} = e^u \Rightarrow dx = e^u du$.

2. $x = 1 \Rightarrow u = 0$ and $x = e \Rightarrow u = 1$.

3. $\displaystyle \sqrt{1 + \frac{1}{x^2}} = \sqrt{1 + e^{-2u}}$.

Then $\displaystyle I = \int_{u = 0}^{u = 1} \sqrt{1 + e^{-2u}} \, e^u \, du = \int_{u = 0}^{u = 1} \sqrt{e^{2u} + 1} \, du$.

But since it's a definite integral, u is just a dummy variable and you can replace the symbol u with the symbol x.

You need to thoroughly review and revise how to integrate because your instructor obviously assumes you know how to do things like this (otherwise you would not be asked to do questions like this). And most likely you need to review basic algebra (such as index laws etc.) too.

9. Originally Posted by mr fantastic
$\displaystyle I = \int_{x=1}^{x=e} \sqrt{1 + \frac{1}{x^2}} \, dx$.

Substitute $x = e^u$.

Then:

1. $\displaystyle \frac{dx}{du} = e^u \Rightarrow dx = e^u du$.

2. $x = 1 \Rightarrow u = 0$ and $x = e \Rightarrow u = 1$.

3. $\displaystyle \sqrt{1 + \frac{1}{x^2}} = \sqrt{1 + e^{-2u}}$.

Then $\displaystyle I = \int_{u = 0}^{u = 1} \sqrt{1 + e^{-2u}} \, e^u \, du = \int_{u = 0}^{u = 1} \sqrt{e^{2u} + 1} \, du$.

But since it's a definite integral, u is just a dummy variable and you can replace the symbol u with the symbol x.

You need to thoroughly review and revise how to integrate because your instructor obviously assumes you know how to do things like this (otherwise you would not be asked to do questions like this). And most likely you need to review basic algebra (such as index laws etc.) too.
Ah, I see. I didn't realize that the goal was to get the integral to look the same without actually doing any integrating. Thank you all for your help.