Results 1 to 12 of 12

Math Help - Changing limits of integration.

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    11

    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
    Last edited by momopeaches; July 20th 2010 at 08:59 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    11
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    118
    Thanks
    5
    Quote Originally Posted by momopeaches View Post
    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.

    Changing limits of integration.-plot.jpg

    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)Changing limits of integration.-triangle.pdf

    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    Posts
    11
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    118
    Thanks
    5
    Quote Originally Posted by momopeaches View Post
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2010
    Posts
    11
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    118
    Thanks
    5
    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}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by fobos3 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    118
    Thanks
    5
    Quote Originally Posted by General View Post
    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}]
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    May 2010
    Posts
    11
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jun 2008
    From
    Plymouth
    Posts
    118
    Thanks
    5
    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}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Changing the limits of integration for this u-sub...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 27th 2011, 04:33 AM
  2. Changing the Limits of a Summation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 3rd 2010, 09:53 PM
  3. Changing integral limits?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 14th 2009, 06:37 PM
  4. Changing the Order of Integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2009, 03:06 PM
  5. Changing integration limits
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 10th 2008, 12:42 PM

Search Tags


/mathhelpforum @mathhelpforum