Results 1 to 4 of 4

Math Help - Area of the Region Between Two Curves

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    17

    Area of the Region Between Two Curves

    Find the area of the region between y = x^2 and y = 4-3x from x = -1 to x = 4/3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by MathGeek06 View Post
    Find the area of the region between y = x^2 and y = 4-3x from x = -1 to x = 4/3
    Did you graph the bounded region? This will help you set up the integral.



    Recall that the area between two curves is A=\int_a^b \left(f(x)-g(x)\right) \,dx

    I hope this helps!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2008
    Posts
    17
    Yes, I graphed the bounded region but when I set up my integrals, I got a negative answer, which clearly was incorrect.

    I had an integral of ((4-3x) - x^3)) from x = -1 to x= 1 + the integral ((x^2) - (4-3x)) from x = 1 to x = 4/3.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by MathGeek06 View Post
    Yes, I graphed the bounded region but when I set up my integrals, I got a negative answer, which clearly was incorrect.

    I had an integral of ((4-3x) - x^3)) from x = -1 to x= 1 + the integral ((x^2) - (4-3x)) from x = 1 to x = 4/3.
    \int_{-1}^1 (4-3x-x^2)\,dx+\int_1^{\frac{4}{3}}(x^2+3x-4)\,dx

    \implies\left.\bigg[4x-\tfrac{3}{2}x^2-\tfrac{1}{3}x^3\bigg]\right|_{-1}^1+\left.\bigg[\tfrac{1}{3}x^3+\tfrac{3}{2}x^2-4x\bigg]\right|_1^{\frac{4}{3}}

    \implies\bigg[\left(4-\tfrac{3}{2}-\tfrac{1}{3}\right)-\left(-4-\tfrac{3}{2}+\tfrac{1}{3}\right)\bigg]+\bigg[\left(\tfrac{64}{81}+\tfrac{8}{3}-\tfrac{16}{3}\right)-\left(\tfrac{1}{3}+\tfrac{3}{2}-4\right)\bigg]=\color{red}\boxed{\frac{1235}{162}}

    Does this make sense?

    --Chris
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 10th 2009, 08:17 AM
  2. Replies: 8
    Last Post: February 23rd 2009, 04:40 PM
  3. Replies: 3
    Last Post: February 20th 2009, 04:46 AM
  4. Area of region bounded by these curves
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 19th 2009, 06:19 PM
  5. Replies: 2
    Last Post: September 6th 2008, 02:20 AM

Search Tags


/mathhelpforum @mathhelpforum