# Problem Application of Integrals(Arc Legth) - I'm stuck

• SclayP
In summary, you are integrating the function f(x) = e^xbetween 0 and 1, but you seem to be having trouble with substitution.
SclayP
So I'm stuck resolving this integral...

$\int^{1}_{0} \sqrt{e^{2x} + 1} \, dx$
$$u=2x ; du=2dx$$
$\frac{1}{2}\int^{2}_{0} \sqrt{e^u + 1} \, du$
$$t=e^u+1 ; dt=e^u du$$
$\frac{1}{2}\int^{e^2 + 1}_{2} \frac{\sqrt{t}}{t-1} \, dt$
$$v=\sqrt{t} ; dv=\frac{1}{2\sqrt{t}}$$
$\int \frac{v^3}{v^2 -1} \, dv$

And here i don't know how to proceed, i red i have to do long division but i don't know why... please help me.

Thanks.

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SclayP said:
So I'm stuck resolving this integral...

$\int^{1}_{0} \sqrt{e^{2x} + 1} \, dx$
$$u=2x ; du=2dx$$
$\frac{1}{2}\int^{2}_{0} \sqrt{e^u + 1} \, du$
$$t=e^u+1 ; dt=e^u du$$
$\frac{1}{2}\int^{e^2 + 1}_{2} \frac{\sqrt{t}}{t-1} \, dt$
$$v=\sqrt{t} ; dv=\frac{1}{2\sqrt{t}}$$
$\int \frac{v^3}{v^2 -1} \, dv$

And here i don't know how to proceed, i red i have to do long division but i don't know why... please help me.

Thanks.

Please show us the function whose arc length you need to find. If you have made a mistake in setting up the integral, then there's no point in trying to figure out how to integrate it. In other words, please provide the original problem statement.

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Mark44 said:
Please show us the function whose arc length you need to find. If you have made a mistake in setting up the integral, then there's no point in trying to figure out how to integrate it. In other words, please provide the original problem statement.

Sorry, you're right the funcion is
$$f(x)=e^x$$
between 0 and 1

Wouldn't you need to parametrize the function first? I.e, let ##\vec\gamma(t) = (t, e^t)##?

Karnage1993 said:
Wouldn't you need to parametrize the function first? I.e, let ##\vec\gamma(t) = (t, e^t)##?

In the first place i wouldn't know how to do it, and in the other hand i don't think it's necesary i mean, i think the way i did it it's correct but I'm stuck in that step. I used the next formula

$L=\int \sqrt{(\frac{dy}{dx})^{2} + 1}$

SclayP,
Your work looks OK, but I didn't check it. Assuming that it actually is OK, here's a link to a wiki topic on long division - http://en.wikipedia.org/wiki/Polynomial_long_division

You want to write v3/(v2 - 1) as a polynomial plus a proper rational expression. It will turn out to v + (proper rational expression)

Well essentially, your formula is the magnitude of the derivative of ##\vec\gamma(t)##, which is what we'll use. Nevertheless, you seem to be having trouble using substitution. For example, on your second substitution, you don't have a ##e^u## to borrow to create a ##dt##.

Karnage1993 said:
Well essentially, your formula is the magnitude of the derivative of ##\vec\gamma(t)##, which is what we'll use. Nevertheless, you seem to be having trouble using substitution. For example, on your second substitution, you don't have a ##e^u## to borrow to create a ##dt##.

I don't need one i just say
$$t=e^{u}+1 ; dt=e^{u}du$$ and so $$du=\frac{dt}{e^{u}}=\frac{dt}{t-1}$$

Ahh, I see. I don't use this technique that often so it wasn't immediately obvious to me.

Okay, let's work on ##\displaystyle\int \frac{v^3}{v^2 - 1} dv##.

Using long division, you can rewrite ##\displaystyle\frac{v^3}{v^2 - 1}## as ##\displaystyle v + \frac{v}{v^2 - 1}##. This should make integrating much easier.

Karnage1993 said:
Ahh, I see. I don't use this technique that often so it wasn't immediately obvious to me.

Okay, let's work on ##\displaystyle\int \frac{v^3}{v^2 - 1} dv##.

Using long division, you can rewrite ##\displaystyle\frac{v^3}{v^2 - 1}## as ##\displaystyle v + \frac{v}{v^2 - 1}##. This should make integrating much easier.

Yes, i'v already done that and then i integrated by part the integral $\int\frac{v}{(v-1)(v+1)}$ and i come up with this $\frac{v^2}{2} |^{\sqrt{e^2 +1}}_{\sqrt{2}} + \frac{1}{2} \int^{\sqrt{e^2 +1}}_{\sqrt{2}}\frac{1}{v-1} + \frac{1}{v+1}\, dv$

So when i integrate 1/(v-1) and 1/(v+1) it's left ln|v-1| and ln|v+1| but then when i evaluate the function the resault it's wrong (i know because i check with the calculator the first integral)...I don't know what the problem is...

You don't need to split the fraction ##\displaystyle\frac{v}{v^2 -1}##. Just let ##w = v^2 - 1## and ##\displaystyle \frac{1}{2} dw = v dv##. In this case, you borrow the ##v## to form the ##dw##.

##\displaystyle\frac{1}{2} \int \frac{1}{w} \ dw = \frac{1}{2} ln(v^2 - 1)##.

From this point, simply substitute back for ##v, t, u## and solve the integral from 0 to 1. You may have made a mistake during the changing of the limits.

Karnage1993 said:
You don't need to split the fraction ##\displaystyle\frac{v}{v^2 -1}##. Just let ##w = v^2 - 1## and ##\displaystyle \frac{1}{2} dw = v dv##. In this case, you borrow the ##v## to form the ##dw##.

##\displaystyle\frac{1}{2} \int \frac{1}{w} \ dw = \frac{1}{2} ln(v^2 - 1)##.

well yes you are right be we ended up with the same resault because ln(x^2 -1) equals ln[(x-1)(x+1)] equals ln(x-1) + ln(x+1)...

but knowing that we came to the same and assuming it's all correct why is the length not the same that the one that give my calculator...

##\begin{align*}\displaystyle \frac{v^2}{2} + \frac{1}{2} ln(v^2 - 1) & = \frac{t}{2} + \frac{1}{2} ln(t - 1)\\ & = \frac{e^u + 1}{2} + \frac{1}{2} ln(e^u + 1 - 1)\\ & = \frac{e^{2x} + 1}{2} + \frac{1}{2} ln(e^{2x})\\ & = \frac{e^{2x} + 1}{2} + \frac{1}{2} |2x| \\ & = \frac{e^{2x} + 1}{2} + |x| \end{align*}##

Evaluating from 0 to 1, we have:

##\displaystyle \frac{e^{2} + 1 - e^{1} - 1}{2} + 2 = \frac{e^{2} - e^{1}}{2} + 2##

Is this what you get?

Karnage1993 said:
##\begin{align*}\displaystyle \frac{v^2}{2} + \frac{1}{2} ln(v^2 - 1) & = \frac{t}{2} + \frac{1}{2} ln(t - 1)\\ & = \frac{e^u + 1}{2} + \frac{1}{2} ln(e^u + 1 - 1)\\ & = \frac{e^{2x} + 1}{2} + \frac{1}{2} ln(e^{2x})\\ & = \frac{e^{2x} + 1}{2} + \frac{1}{2} |2x| \\ & = \frac{e^{2x} + 1}{2} + |x| \end{align*}##

Evaluating from 0 to 1, we have:

##\displaystyle \frac{e^{2} + 1 - e^{1} - 1}{2} + 2 = \frac{e^{2} - e^{1}}{2} + 2##

Is this what you get?

Yes, that is what I GET but calculator says that the resault it is 2...and what we get by doing it "by hand" it 4.19 so, there is a problem in the proceedure.

I haven't looked over your work, but this can also be done via the following substitution:
##e^x=\tan u\\
e^x \;dx=\sec^2u\;du\\
\tan u\;dx=\sec^2u\;du\\
dx=\dfrac{\sec^2u}{\tan u}du##

Okay, I've went through your substitutions and noticed the error.

##\displaystyle v = \sqrt{t}, dv = \frac{1}{2\sqrt{t}}\ dt##
##\displaystyle 2\sqrt{t}\ dv = dt##
##\displaystyle 2v \ dv = dt##

Which means, ##\displaystyle\frac{1}{2} \int \frac{\sqrt{t}}{t - 1} \ dt = \frac{1}{2} \int \frac{v}{v^2 - 1} 2v \ dv = \int \frac{v^2}{v^2 - 1}\ dv##.

So the ##v^3## should be a ##v^2##!

Karnage1993 said:
Okay, I've went through your substitutions and noticed the error.

##\displaystyle v = \sqrt{t}, dv = \frac{1}{2\sqrt{t}}\ dt##
##\displaystyle 2\sqrt{t}\ dv = dt##
##\displaystyle 2v \ dv = dt##

Which means, ##\displaystyle\frac{1}{2} \int \frac{\sqrt{t}}{t - 1} \ dt = \frac{1}{2} \int \frac{v}{v^2 - 1} 2v \ dv = \int \frac{v^2}{v^2 - 1}\ dv##.

So the ##v^3## should be a ##v^2##!

Oh! hahaha i can't believe it was that, and i can't believe i didnt notice while doing it...well thank you for you help...

PD: I know it's late but...sorry for my english

## 1. What is the concept of arc length in the context of problem application of integrals?

The arc length is the distance along a curve or arc. In problem application of integrals, it refers to finding the length of a curve using integral calculus.

## 2. How do I approach solving problems involving arc length?

To solve problems involving arc length, you first need to determine the function that defines the curve. Then, you can use an integral to find the arc length by integrating the square root of the sum of the squares of the derivative of the function with respect to the independent variable.

## 3. What is the significance of finding the arc length in real-world applications?

Finding the arc length is important in various fields such as physics, engineering, and architecture. It allows us to calculate the distance traveled by an object along a curved path, which is crucial in designing and analyzing real-world structures and systems.

## 4. Can you provide an example of a problem involving arc length?

Sure, for example, you are trying to find the length of a race track that has a curved section. You can use the concept of arc length to calculate the distance an athlete needs to run in order to complete one lap on the track.

## 5. What are some common mistakes to avoid when solving problems related to arc length?

One common mistake is forgetting to take the square root of the derivative when setting up the integral. It is also important to pay attention to the limits of integration and use the correct units when giving the final answer. Additionally, make sure to double-check your calculations to avoid any arithmetic errors.

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