How Do You Solve These Integration Problems?

[ineedhelpnow]

hi. have a test comin up and i really need help with these questions.

1. find the length of the curve $y=2\ln\left({\frac{x}{2}}\right)$, $\pi/3\le x\le\pi$

2. use simpson's rule with $n=6$ to estimate the length of the curve $y=\sin\left({x}\right)$, $0\le x\le\pi$

3. find the area of the surface obtained by rotating the curve about the $x-$axis. $x=\frac{1}{3}(y^2+2)^{3/2}$, ${1}\le y\le{3}$

id really appreciate any help at all

 

[Prove It]

Do you at least know the arclength formula?

$\displaystyle \begin{align*} \mathcal{l} = \int_a^b{ \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 } \, \mathrm{d}x } \end{align*}$

 

[ineedhelpnow]

yes

 

[MarkFL]

hi. have a test comin up and i really need help with these questions.

1. find the length of the curve $y=2\ln\left({\frac{x}{2}}\right)$, $\pi/3\le x\le\pi$

2. use simpson's rule with $n=6$ to estimate the length of the curve $y=\sin\left({x}\right)$, $0\le x\le\pi$

3. find the area of the surface obtained by rotating the curve about the $x-$axis. $x=\frac{1}{3}(y^2+2)^{3/2}$, ${1}\le y\le{3}$

id really appreciate any help at all

I know you are relatively new here and as such probably are not aware of our rules designed to make MHB more efficient for both those asking questions and for those providing help, but we ask that you post no more than two questions in a thread (follow-up questions are okay). This way a thread is easier to follow and does not get potentially convoluted as more than one person may be trying to help you with different questions at the same time.

We also ask that you show what you have tried so we can offer better, more specific help. If you don't show your work, we really have no idea what you've tried and where you are stuck.

Can you show where you are stuck, or what your thoughts are on how to begin? This way we can guide you rather than simply work the problems which does you far less service. :D

 

[ineedhelpnow]

i was going to put 2 questions in one thread, another two in another thread and then one in another by instead i got lazy and just put 3 in one and 2 in the other.

i figured out the first two kinda so i just have to show the work but I am totally stuck on 3

 

[ineedhelpnow]

having trouble integrating $\int_{\pi/3}^{\pi} \ \sqrt{1+{\frac{2}{x}}^{2}},dx$

btw that's $(2/x)^{2}$ not $(2^{2}/x)$

 

[MarkFL]

I would consider a trigonometric substitution of the form:

$$\frac{2}{x}=\tan(\theta)$$, and be sure to transform the limits and differential in accordance with the substitution...what do you get?

 

[Prove It]

having trouble integrating $\int_{\pi/3}^{\pi} \ \sqrt{1+{\frac{2}{x}}^{2}},dx$

btw that's $(2/x)^{2}$ not $(2^{2}/x)$

OK so you have realized that this is the integral you need to evaluate. Good :)

$\displaystyle \begin{align*} \int_{\frac{\pi}{3}}^{\pi}{ \sqrt{1 + \left( \frac{2}{x} \right) ^2 }\,\mathrm{d}x } &= \int_{\frac{\pi}{3}}^{\pi}{ \sqrt{1 + \frac{4}{x^2}} \, \mathrm{d}x } \\ &= \int_{\frac{\pi}{3}}^{\pi}{ \sqrt{\frac{x^2 + 4}{x^2}}\,\mathrm{d}x } \\ &= \int_{\frac{\pi}{3}}^{\pi}{ \frac{\sqrt{x^2 + 4}}{x} \, \mathrm{d}x }\end{align*}$

Now a substitution of the form $\displaystyle \begin{align*} x = 2\sinh{(t)} \implies \mathrm{d}x = 2\cosh{(t)}\,\mathrm{d}t \end{align*}$ or $\displaystyle \begin{align*} x = 2\tan{(\theta)} \implies \mathrm{d}x = 2\sec^2{(\theta)}\,\mathrm{d}\theta \end{align*}$ might be appropriate :)

 

[ineedhelpnow]

hold on. I am still workin on it.

 

[ineedhelpnow]

the way I am doing it, it seems like I am going to have to use substitution a couple times. is that right?

 

[MarkFL]

Check out this thread to see a method for solving a similar integral

http://mathhelpboards.com/questions-other-sites-52/ktravenclaws-question-yahoo-answers-regarding-trigonometric-substitution-integral-8844.html

 

[ineedhelpnow]

$y=2\ln\left({\frac{x}{2}}\right)$ from ${\pi/3}$ to ${\pi}$ $\frac{dy}{dx}=\frac{2}{x}$

$L=\int_{\pi/3}^{\pi} \sqrt{1+(\frac{2}{x})^2},dx$
$=\int_{\pi/3}^{\pi} \sqrt{1+(\frac{4}{x^2})},dx$
$=\int_{\pi/3}^{\pi} \sqrt{\frac{x^2+4}{x^2}},dx$
$=\int_{\pi/3}^{\pi} \frac{\sqrt{x^2+4}}{x},dx$
$x=2\tan\left({\theta}\right)
dx=2(\sec\left({\theta}\right))^{2}$
$=\int_{\pi/3}^{\pi} \frac{\sqrt{4(\tan\left({\theta}\right))^2+4}}{2\tan\left({\theta}\right)}\times2(\sec\left({\theta}\right))^{2},d\theta$
let $u=\tan\left({\theta}\right)$ $du=(\sec\left({\theta}\right))^{2}$ $d\theta=(\cos\left({\theta}\right))^{2},du$

is the work up to there correct?

 

[MarkFL]

When you make a substitution, the limits of integration must be changed as well so that they are in terms of the new variable.

Let's pick up at this point:

$$I=\int_{\frac{\pi}{3}}^{\pi}\frac{\sqrt{x^2+4}}{x}\,dx$$

Now, we want to make the substitution:

$$x=2\tan(\theta)\,\therefore\,dx=2\sec^2(\theta)\,d\theta$$

Now, from our substitution, we see that:

$$\theta=\tan^{-1}\left(\frac{x}{2}\right)$$

So, we use this to change our limits from $x$'s to $\theta$'s...what do we have after fully completing the substitution?

 

[Prove It]

$y=2\ln\left({\frac{x}{2}}\right)$ from ${\pi/3}$ to ${\pi}$ $\frac{dy}{dx}=\frac{2}{x}$

$L=\int_{\pi/3}^{\pi} \sqrt{1+(\frac{2}{x})^2},dx$
$=\int_{\pi/3}^{\pi} \sqrt{1+(\frac{4}{x^2})},dx$
$=\int_{\pi/3}^{\pi} \sqrt{\frac{x^2+4}{x^2}},dx$
$=\int_{\pi/3}^{\pi} \frac{\sqrt{x^2+4}}{x},dx$
$x=2\tan\left({\theta}\right)
dx=2(\sec\left({\theta}\right))^{2}$
$=\int_{\pi/3}^{\pi} \frac{\sqrt{4(\tan\left({\theta}\right))^2+4}}{2\tan\left({\theta}\right)}\times2(\sec\left({\theta}\right))^{2},d\theta$
let $u=\tan\left({\theta}\right)$ $du=(\sec\left({\theta}\right))^{2}$ $d\theta=(\cos\left({\theta}\right))^{2},du$

is the work up to there correct?

The whole idea of using trigonometric substitution is to simplify your integrand! Why would you make a further substitution before simplifying?

You are correct up to $\displaystyle \begin{align*} \frac{\sqrt{4\tan^2{(\theta)} + 4}}{2\tan{(\theta)}}\cdot 2\sec^2{(\theta)} \end{align*}$, so now simplify...

$\displaystyle \begin{align*} \frac{\sqrt{4\tan^2{(\theta)} + 4}}{2\tan{(\theta)}}\cdot 2\sec^2{(\theta)} &= \frac{\sqrt{4 \left[ \tan^2{(\theta)} + 1 \right] }}{2\tan{(\theta)}} \cdot 2\sec^2{(\theta)} \\ &= \frac{\sqrt{4\sec^2{(\theta)}}}{2\tan{(\theta)}} \cdot 2\sec^2{(\theta)} \\ &= \frac{2\sec{(\theta)}}{2\tan{(\theta)}} \cdot 2\sec^2{(\theta)} \\ &= \frac{2\sec^3{(\theta)}}{\tan{(\theta)}} \\ &= \frac{2 \left[ 1 + \tan^2{(\theta)} \right] \sec{(\theta)} }{\tan{(\theta)}} \\ &= \frac{2\sec{(\theta)}}{\tan{(\theta)}} + \frac{2\tan^2{(\theta)}\sec{(\theta)}}{\tan{(\theta)}} \\ &= \frac{2}{\cos{(\theta)}} \cdot \frac{\cos{(\theta)}}{\sin{(\theta)}} + 2\tan{(\theta)}\sec{(\theta)} \\ &= \frac{2}{\sin{(\theta)}} + 2\tan{(\theta)}\sec{(\theta)} \\ &= \frac{2\sin{(\theta)}}{\sin^2{(\theta)}} + 2\tan{(\theta)}\sec{(\theta)} \\ &= - 2 \left[ \frac{-\sin{(\theta)}}{1 - \cos^2{(\theta)}} \right] + 2\tan{(\theta)}\sec{(\theta)} \end{align*}$

The first term can be integrated using the substitution $\displaystyle \begin{align*} u = \cos{(\theta)}\implies \mathrm{d}u = -\sin{(\theta)}\,\mathrm{d}\theta \end{align*}$ and the second you should recognise as the derivative of $\displaystyle \begin{align*} \sec{(\theta)} \end{align*}$.