How Do You Solve These Integration Problems?

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In summary: We have:$\displaystyle L=\int_{\pi/3}^{\pi} \sqrt{1+(\frac{2}{x})^2},dx$In summary, In summary, the length of the curve $y=2\ln\left({\frac{x}{2}}\right)$, $\pi/3\le x\le\pi$, can be estimated using Simpson's rule with $n=6$, and the area of the surface obtained by rotating the curve about the $x-$axis can be found by using the arclength formula.
  • #1
ineedhelpnow
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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. $x=\frac{1}{3}(y^2+2)^{3/2}$, ${1}\le y\le{3}$


:eek: id really appreciate any help at all
 
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  • #2
Do you at least know the arclength formula?

$\displaystyle \begin{align*} \mathcal{l} = \int_a^b{ \sqrt{1 + \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 } \, \mathrm{d}x } \end{align*}$
 
  • #3
yes
 
  • #4
ineedhelpnow said:
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. $x=\frac{1}{3}(y^2+2)^{3/2}$, ${1}\le y\le{3}$


:eek: 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
 
  • #5
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
 
  • #6
having trouble integrating $\int_{\pi/3}^{\pi} \ \sqrt{1+{\frac{2}{x}}^{2}},dx$

btw that's $(2/x)^{2}$ not $(2^{2}/x)$
 
  • #7
I would consider a trigonometric substitution of the form:

\(\displaystyle \frac{2}{x}=\tan(\theta)\), and be sure to transform the limits and differential in accordance with the substitution...what do you get?
 
  • #8
ineedhelpnow said:
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 :)
 
  • #9
hold on. I am still workin on it.
 
  • #10
the way I am doing it, it seems like I am going to have to use substitution a couple times. is that right?
 
  • #11
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
 
  • #12
$y=2\ln\left({\frac{x}{2}}\right)$ 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?
 
  • #13
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:

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

Now, we want to make the substitution:

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

Now, from our substitution, we see that:

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

So, we use this to change our limits from $x$'s to $\theta$'s...what do we have after fully completing the substitution?
 
  • #14
ineedhelpnow said:
$y=2\ln\left({\frac{x}{2}}\right)$ 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 $\displaystyle \begin{align*} \sec{(\theta)} \end{align*}$.
 

FAQ: How Do You Solve These Integration Problems?

1. What is the purpose of integration in scientific applications?

Integration is used in scientific applications to determine the total value of a quantity that is continuously changing. It is a mathematical tool that helps to find the area under a curve, which can then be used to solve various real-world problems in fields such as physics, engineering, and economics.

2. What are some common applications of integration in physics?

Integration is used in physics to solve problems related to motion, such as finding the displacement, velocity, and acceleration of an object over time. It is also used to calculate the work done by a force, the energy stored in a system, and the momentum of an object.

3. How is integration used in engineering?

In engineering, integration is used to analyze and design systems that involve continuous changes, such as electrical circuits, fluid dynamics, and control systems. It is also used to determine the stress and strain in structures and to optimize designs for maximum efficiency.

4. What are some examples of applications of integration in economics?

Integration is used in economics to calculate the total revenue and total cost of a business, which can then be used to determine the profit or loss. It is also used to find the total utility and consumer surplus in microeconomics and to analyze economic growth and inflation in macroeconomics.

5. How is integration used in biology and medicine?

In biology and medicine, integration is used to analyze and model biological processes, such as the growth of a population, the spread of diseases, and the distribution of nutrients in the body. It is also used in medical imaging techniques, such as MRI and CT scans, to reconstruct 3D images from 2D data.

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