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Math Help - Area between curves

  1. #1
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    Area between curves

    I have some wrong answers .. help?

    Calculate the area between y-sin(x) and y=cos(x) , -pi<or equal to x <or equal to 2pi.
    I have -1.9984.

    Calculate the area between y=x^2 and y=3x+4.
    I have -18.7.

    Find the area of the region enclosed between y=2sin(x) and y=2cos(x) from x=0 and x=0.8pi
    I have .6334465.

    Can someone tell me what i did wrong?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by keemariee View Post
    I have some wrong answers .. help?

    Calculate the area between y-sin(x) and y=cos(x) , -pi<or equal to x <or equal to 2pi.
    I have -1.9984.

    Calculate the area between y=x^2 and y=3x+4.
    I have -18.7.

    Find the area of the region enclosed between y=2sin(x) and y=2cos(x) from x=0 and x=0.8pi
    I have .6334465.

    Can someone tell me what i did wrong?
    We want to compute the area of

    \int_{-\pi}^{2\pi}f(x)-g(x) were f(x) and g(x) are to be determined

    First off did you realize taht you must split this into four integrals?
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by keemariee View Post
    Calculate the area between y=x^2 and y=3x+4.
    I have -18.7.
    First, we need to know our boundaries, so we want to know where the lines cross. At these points, they will have the same (x,y) value.

    So x^2 = y = 3x+4
    this gives us x^2 = 3x+4
    x^2-3x-4 = 0
    (x-4)(x+1)=0

    So at x=4, and x=-1, these two graphs will cross. Then these are our boundaries, so we will integrate from -1 to 4.

    Now we need to know which function is above the other, substituting in x=0 (because it is a point inside of our boundaries) we get that y=3x+4 is above.

    So we know that we will take the integral of 3x+4 and then subtract the integral of x^2 (if you don't understand why to do this, look at the attached image, it should be rather clear.

    so A = \int_{-1}^4 (3x+4-x^2)~dx

    = \left[\frac 32x^2+4x-\frac 13x^3 \right]_{-1}^4

    = \frac 32*16 + 16 - \frac 13*64 - \frac 32 +4 +\frac 13

    = \frac {125}6
    Attached Thumbnails Attached Thumbnails Area between curves-areabetween.jpg  
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