Results 1 to 4 of 4

Math Help - Integral Question

  1. #1
    Junior Member
    Joined
    Sep 2005
    Posts
    62

    Integral Question

    Hey guys I need help with theses two question. I don't know how to do them and I'm just stuck, Just losted

    Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.

    and



    Compute the indefinite integral
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by killasnake View Post
    Hey guys I need help with theses two question. I don't know how to do them and I'm just stuck, Just losted


    and
    1. Evaluate \int_{-4}^4 \sqrt{16-x^2} dx in term of the area
    of a region.

    Well this is the area of the upper half of the circle x^2+y^2=4^2, so its area is

    \frac{\pi r^2}{2}=\frac{\pi 4^2}{2}=8\pi

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by killasnake View Post
    Hey guys I need help with theses two question. I don't know how to do them and I'm just stuck, Just losted


    and
    Find:

    \int 5 x^4 + 3 \sec(x) tan(x)\ dx

    Look at the integrand:

    <br />
5 x^4 + 3 \sec(x) \tan(x)=5x^4 +3 \frac{\sin(x)}{(\cos(x))^2}=<br />
\frac{d}{dx} \left[ x^5 + 3\frac{1}{\cos(x)}\right]<br />

    So by the fundamental theorem of calculus:

    <br />
\int 5 x^4 + 3 \sec(x) tan(x)\ dx = x^5 + 3\frac{1}{\cos(x)} + C<br />

    RonL
    Last edited by CaptainBlack; November 22nd 2006 at 08:07 AM. Reason: correct error
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, killasnake!

    Exactly where is your difficulty?
    . . There are formulas for that second integral . . .

    \int\left[5x^4 + 3\sec(x)\tan(x)\right]\,dx
    We have: . \underbrace{\int5x^4\,dx}_\downarrow \;+ \;3\!\underbrace{\int\sec(x)\tan(x)\,dx}_\downarro  w
    . . . . . . . . = \;\;x^5 \quad\;\;+ \qquad\;\;3\sec(x) + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integral question
    Posted in the Calculus Forum
    Replies: 9
    Last Post: September 17th 2011, 10:02 PM
  2. Integral Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 18th 2011, 06:26 PM
  3. Double Integral and Triple Integral question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 3rd 2010, 12:47 PM
  4. Integral Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 5th 2010, 09:33 PM
  5. Integral Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 14th 2009, 08:26 PM

Search Tags


/mathhelpforum @mathhelpforum