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Math Help - area under curve and line.

  1. #1
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    area under curve and line.

    The diagram shows a sketch of part of the curve with equation y=x2+1 and the line with equation y=7−x. The finite region R1 is bounded by the line and the curve. The finite region R2 is below the curve and the line and is bounded by the positive x- and y-axes as shown in the diagram.

    (a) Find the area of R1.
    (b) Find the area of R2.

     7-x = x^2 + 1

     0 = x^2 + x -6

     0 = (x+3) (x-2)

     x = 2 , x = -3


    Area of R1 =  \int_{-3}^2 [ 7-x-x^2 + 1] dx

    I can work this out, but How would I go about working the area of region 2? Becasue there's also the area of region 1 inside region 2.
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  2. #2
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    Hi

    Area of R1 =  \int_{-3}^2 [ 7-x-x^2 - 1] dx

    For R2, you need to split the integral into 2 parts : from 0 to 2 and from 2 to 7
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    Area of R1 =  \int_{-3}^2 [ 7-x-x^2 - 1] dx

    For R2, you need to split the integral into 2 parts : from 0 to 2 and from 2 to 7
    So would I minus the integrals after ? To get the area of R2?
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  4. #4
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    I got  \frac{310}{3} , which is not the right answer.

    Can you show me how to work it out ?

    thanks.
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  5. #5
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    Quote Originally Posted by running-gag View Post
    ...
    For R2, you need to split the integral into 2 parts : from 0 to 2 and from 2 to 7
    Quote Originally Posted by Tweety View Post
    I got  \frac{310}{3} , which is not the right answer.

    Can you show me how to work it out ?

    thanks.
    You have the functions:

    f(x)=x^2+1 and g(x)=7-x

    Split R_2 into two parts: One area is completely below the graph of the quadratic function, the second one is completely below the straight line. According to running-gag's post the area R_2 is:

    R_2=\int_0^2 f(x) dx +\int_2^7 g(x) dx

    R_2=\int_0^2 \left(x^2+1 \right)dx +\int_2^7 (7-x) dx

    I'll leave the rest for you. I've got R_2 = \dfrac{103}6
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  6. #6
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    Area of R2 =  \int_{0}^2 [ x^2 + 1] dx +  \int_{2}^7 [ 7 - x] dx
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  7. #7
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    Thanks I get it now.
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