Page 2 of 2 FirstFirst 12
Results 16 to 26 of 26

Math Help - Area under a curve

  1. #16
    Member
    Joined
    Feb 2010
    From
    United States
    Posts
    155
    Thanks
    1
    Okay, well I still don't really know how to integrate functions like that, and not all of them are semicircles. For instance, integrating (2-x)^6....I'm not sure how you would do that other than the backwards chain rule, and I'm not sure how you do that either...
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Member
    Joined
    Feb 2010
    From
    United States
    Posts
    155
    Thanks
    1
    Quote Originally Posted by Ackbeet View Post
    Then 6.28 might not be the more accurate answer, but it is the better answer. In engineering, at least, you better give an answer that your boss is happy with. If he wants decimals, give him decimals. Otherwise, it won't be as useful to him.
    Haha yeah, I know what you mean. It's kind of irritating because technically it isn't as correct...but still. You're right.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Here is the integration you wanted

    \int^{2}_{-2}\sqrt{4-x^2}dx

    let x = 2\sin{\u} and dx = 2\cos{u}du

    and sub

    \int(\sqrt{4-4\sin^2{u}})(2\cos{u})du

    \int(4\cos^2{u})du

    The other method was a lot easier lol.

    \int(2\cos{2u}+2)du

     sin2u +2u

    2\sin{u}\cos{u} +2u

    u =arcsin(\frac{x}{2})

    \cos{u}=\sqrt{1-\frac{x^2}{4}}

    now plug all back in

    2(\frac{x}{2})(\sqrt{1-\frac{x^2}{4}})+2arcsin(\frac{x}{2})

    now just use your limits from -2 to 2
    Follow Math Help Forum on Facebook and Google+

  4. #19
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    So, was this a question to be answered solely with integration techniques,
    or any method?
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Member
    Joined
    Feb 2010
    From
    United States
    Posts
    155
    Thanks
    1
    It was supposed to be to practice integration, but it's all the same for getting the answer.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by bobsanchez View Post
    It was supposed to be to practice integration, but it's all the same for getting the answer.
    Have you moved on to trigonometric substitutions?

    If you draw a quarter-circle in the first quadrant, radius=2,
    pick any point on the circumference not on an axis,
    join the point to the origin (circle centre) and drop a perpendicular to the x-axis
    then you have a right-angled triangle.

    By Pythagoras' theorem

    \left(\sqrt{4-x^2}\right)^2+x^2=4=2^2

    Label the acute angle at the origin \theta

    \displaystyle\huge\ sin\theta=\frac{\sqrt{4-x^2}}{2}\ \Rightarrow\ \sqrt{4-x^2}=2sin\theta

    \displaystyle\huge\ cos\theta=\frac{x}{2}\ \Rightarrow\ x=2cos\theta\ \Rightarrow\ dx=-2sin\theta d\theta

    x=0,\ \theta=\frac{{\pi}}{2}

    x=2,\ \theta=0

    This allows you to integrate the trigonometric integral

    \displaystyle\huge\int_{\frac{{\pi}}{2}}^{0}2sin\t  heta(-2sin\theta)d\theta=4\int_{0}^{\frac{{\pi}}{2}}sin^  2\theta d\theta

    =\displaystyle\huge\ 2\int_{0}^{\frac{{\pi}}{2}}(1-cos2\theta)d\theta

    Double your final answer.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Member
    Joined
    Feb 2010
    From
    United States
    Posts
    155
    Thanks
    1
    Haven't done trigonometric subs yet...let me dissect your post and see what I get.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Newbie
    Joined
    Aug 2010
    Posts
    2

    Another idea...

    I did this using trig substitution.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,400
    Thanks
    1849
    Quote Originally Posted by bobsanchez View Post
    Haven't done trigonometric subs yet...let me dissect your post and see what I get.
    If you haven't then your teacher probably expected you to recognize that this was a semi-circle and use (1/2)\pi r^2.
    Follow Math Help Forum on Facebook and Google+

  10. #25
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by bobsanchez View Post
    Okay, well I still don't really know how to integrate functions like that, and not all of them are semicircles. For instance, integrating (2-x)^6....I'm not sure how you would do that other than the backwards chain rule, and I'm not sure how you do that either...
    In that case, you can integrate with respect to 2-x,
    or rename it "u".

    You would have a "power" integral if it was \int{x^6}dx

    u=2-x

    \frac{du}{dx}=-1

    du=-dx

    dx=-du

    Now, you have a "power" integral.

    \int{(2-x)^6}dx=-\int{u^6}du

    It's about getting it into a form you can integrate.

    The advantage of integrating the "semi-circle" function is..

    since you would know the area of a semicircle,
    then you will be "dead sure" your integration is incorrect if you do not get an answer of

    \frac{{\pi}r^2}{2}
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by bobsanchez View Post
    Oh, sorry. I had the question mark there because I didn't know where you were going with it, not whether or not it's a semicircle. Um...I don't really know. It's doesn't quite touch the x-axis according to my graphing calculator, and as a result it is sort of messing with all of the estimations I've attempted, and I don't know how to integrate a function like that.
    If you're studying calculus it is expected that:

    1. You can sketch graphs (like y = \sqrt{1 - x^2}) by hand, without having to use a graphing calculator. If you cannot do this you are strongly advised to review pre-calculus.

    2. You understand the limitations (due to resolution etc.) of graphs drawn using a graphics calculator.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. AREA under the curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 27th 2010, 11:39 PM
  2. Area under a curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 29th 2009, 11:52 AM
  3. Area under the curve
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 18th 2009, 11:34 AM
  4. Area below a curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 14th 2008, 09:43 AM
  5. area under the curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 31st 2007, 02:22 PM

Search Tags


/mathhelpforum @mathhelpforum