# Changing limits of integration.

• Jul 20th 2010, 08:41 AM
momopeaches
Changing limits of integration.
This problem involves a change in variable of the limit of integration.

Consider the area bounded by $y=\frac{8x^2}{\sqrt(1-2x^2)}$, y=0, x=0, and x=.5. A definite integral of this area, using x as the variable of integration, is
$\int^x_0 \frac{8x^2}{\sqrt(1-2x^2)}dx$. The next part of the question says: make the substitution of the form $x = A sin\theta$ to transform the integral so that $\theta$ is the variable of integration.

I'm not sure how to start this one. I thought of simply solving for $\theta$, but I'm don't know how to transform the integrand.

*edit* forgot my dx. x.x
• Jul 20th 2010, 08:56 AM
yeKciM
hmmm... I'm very low on English but if i understood correctly, u need to change limits of integration... Isn't that for $\int\int$ ?
Just wondering if there's $dxdy$ at the end (Thinking)
• Jul 20th 2010, 09:02 AM
momopeaches
Oops! Forgot my dx at the end of the integral. Yes, I also need to change the limit of integration, in addition to the variable. I'm not sure how to do this, though.
• Jul 20th 2010, 09:16 AM
fobos3
Quote:

Originally Posted by momopeaches
This problem involves a change in variable of the limit of integration.

Consider the area bounded by $y=\frac{8x^2}{\sqrt(1-2x^2)}$, y=0, x=0, and x=.5. A definite integral of this area, using x as the variable of integration, is
$\int^x_0 \frac{8x^2}{\sqrt(1-2x^2)}$. The next part of the question says: make the substitution of the form $x = A sin\theta$ to transform the integral so that $\theta$ is the variable of integration.

I'm not sure how to start this one. I thought of simply solving for $\theta$, but I'm don't know how to transform the integrand.

First of all, you should always include $dx$ when you integrate. Secondly, here is a plot of the function.

Attachment 18256

I would assume you want to evaluate $\int_0^{0.5} \dfrac{8x^2}{\sqrt{1-2x^2}}\, dx$

The key is to think of it like a triangle(check triangle.pdf)Attachment 18257

Now $\dfrac{1}{\sqrt{1-2x^2}}=\dfrac{1}{\cos{\theta}}$

$8x^2=4\times 2x^2=4\sin^2 \theta$

And lastly you have $\dfrac{d}{dx}(\sin \theta)=\sqrt{2}\dfrac{dx}{dx}$

$\sqrt{2}\, dx=\cos(\theta)\, d\theta$

Now the last part is to change the limits of integration. You have $\sin \theta=\sqrt{2}x$ so $x=0,\theta=0$ and $x=0.5,\theta=\sin^{-1}\left(\sqrt{2}\times 0.5\right)$

Can you finish from here?

P.S.
You don't need to change the limits of integration if you don't want to. You can write it like this $\int_{x=0}^{x=0.5} f(\theta)\, d\theta$ and then change back to x when you've solved the integral before plugging in the numerical values and calculating the difference.
• Jul 20th 2010, 09:39 AM
momopeaches
I'm not sure that I follow. I understand about thinking of it as a triangle, but where did the $\frac{d}{dx}sin\theta=\sqrt(2)\frac{dx}{dx}$ come from?
• Jul 21st 2010, 07:22 AM
fobos3
Quote:

Originally Posted by momopeaches
I'm not sure that I follow. I understand about thinking of it as a triangle, but where did the $\frac{d}{dx}sin\theta=\sqrt(2)\frac{dx}{dx}$ come from?

You have $\sin \theta=\sqrt{2}x$ from the triangle. Then you differentiate both sides with respect to x

$\dfrac{d}{dx}(\sin \theta)=\sqrt{2}\dfrac{dx}{dx}$

Now you apply the chain rule

$\dfrac{d}{dx}(\sin \theta)=\dfrac{d\theta}{dx}\cos\theta=\sqrt{2}\dfr ac{dx}{dx}$

And finally you multiply both sides by dx

$d\theta\cos \teta=\sqrt{2}\,dx$
• Jul 21st 2010, 01:29 PM
momopeaches
Okay, I get that, and was able to write out my definite integral. The next part of the question asks to evaluate each of the integrals. Should I be getting the same or different answers for the two integrals? I got .40361 for the first answer, and .697067 for the second answer.
• Jul 22nd 2010, 04:02 AM
fobos3
Yes you should be getting the same answers. Here I'll show you a simpler example. Consider the integral

$\int_0^1 \dfrac{1}{\sqrt{1-x^2}}\,dx$

In your triangle put the hypotenuse equal to 1. The leg that is across the angle should be x so that $\sin \theta=x$

Now using Pythagoras theorem the other leg is $\sqrt{1-x^2}=\cos \theta$

Now you can write the integral like this $\int_0^1 \dfrac{1}{\sqrt{1-x^2}}=\int_{x=0}^{x=1}\dfrac{1}{\cos \theta}\,dx$

You have to express $dx$ in terms of $d\theta$

We have $x=\sin \theta$ and we differentiate both sides to get

$dx=\cos \theta \,d\theta$

Now we substitute dx in the integral to get $\int_{x=0}^{x=1}1\,d\theta=\left[\theta\right]_{x=0}^{x=1}$

What you do next is change the limits of integration. The limits are x=0 and x=1 but we are integrating with respect to $\theta$

You use $x=\sin\theta$ and you need to solve the two equations.

$\sin \theta=0 \therefore \theta=0$

and

$\sin \theta=1 \therefore \theta=\sin^{-1} 1=\dfrac{\pi}{2}$

You have to work in radians. The reason for that is because radians have a unit of length which is the same unit for x and y.

Now your integral becomes $\int_0^{\pi/2} 1\,d\theta=\left[\theta\right]_0^{\pi/2}=\dfrac{\pi}{2}$
• Jul 22nd 2010, 05:03 AM
General
Quote:

Originally Posted by fobos3
Yes you should be getting the same answers. Here I'll show you a simpler example. Consider the integral

$\int_0^1 \dfrac{1}{\sqrt{1-x^2}}\,dx$

In your triangle put the hypotenuse equal to 1. The leg that is across the angle should be x so that $\sin \theta=x$

Now using Pythagoras theorem the other leg is $\sqrt{1-x^2}=\cos \theta$

Now you can write the integral like this $\int_0^1 \dfrac{1}{\sqrt{1-x^2}}=\int_{x=0}^{x=1}\dfrac{1}{\cos \theta}\,dx$

You have to express $dx$ in terms of $d\theta$

We have $x=\sin \theta$ and we differentiate both sides to get

$dx=\cos \theta \,d\theta$

Now we substitute dx in the integral to get $\int_{x=0}^{x=1}1\,d\theta=\left[\theta\right]_{x=0}^{x=1}$

What you do next is change the limits of integration. The limits are x=0 and x=1 but we are integrating with respect to $\theta$

You use $x=\sin\theta$ and you need to solve the two equations.

$\sin \theta=0 \therefore \theta=0$

and

$\sin \theta=1 \therefore \theta=\sin^{-1} 1=\dfrac{\pi}{2}$

You have to work in radians. The reason for that is because radians have a unit of length which is the same unit for x and y.

Now your integral becomes $\int_0^{\pi/2} 1\,d\theta=\left[\theta\right]_0^{\pi/2}=\dfrac{\pi}{2}$

This is an improper integral. It should be calculated in terms of limits.
• Jul 22nd 2010, 06:39 AM
fobos3
Quote:

Originally Posted by General
This is an improper integral. It should be calculated in terms of limits.

It is true it is improper integral but after the trig substitution that is no longer the case. Besides the answer is correct check http://www.wolframalpha.com/input/?i=integrate[1%2Fsqrt[1-x^2]%2C{x%2C0%2C1}]
• Jul 22nd 2010, 11:51 AM
momopeaches
Okat, I understand how the problem works, but I still am getting the wrong answer. I don't know what I'm doing wrong. The integral that I'm trying to evaluate is $\int\frac{4sin^2(\theta)}{cos(\theta)}d\theta$ and the limits of integration are from 0 to $arcsin(\sqrt(2)*.5)$. What step did I mess up on?
• Jul 22nd 2010, 12:11 PM
fobos3
Your integral seems wrong. Did you replace $dx$ with $d\theta$ straight away. You can't do that. From my first post

$\sqrt{2}\, dx=\cos(\theta)\, d\theta$

$dx=\dfrac{1}{\sqrt{2}}\cos\theta\,d\theta$

Now you substitute that and you get $\dfrac{4}{\sqrt{2}}\int_0^{\sin^{-1}(0.5\times \sqrt{2})}\dfrac{\sin^2 \theta}{\cos\theta}\cos\theta \,d\theta=2\sqrt{2}\int_0^{\sin^{-1}(1/\sqrt{2})}\sin^2\theta\,d\theta\approx0.403614$

P.S.
$\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)=\dfrac{\pi}{4}$