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Math Help - Area under two curves by integration

  1. #1
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    Area under two curves by integration

    I want to find the area between two curves (see graph attached). But I want to find the area in function of "r", can anybody help?. The first curve is a line (x+y=1), and the second has the equation xy=r
    Attached Thumbnails Attached Thumbnails Area under two curves by integration-area-int.gif  
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  2. #2
    Senior Member DivideBy0's Avatar
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    r is just a constant, isn't it?
    Then simply solve both equations for y:
    y = 1-x

    y=\frac{r}{x}
    The graphs intersect at \lbrace{(x,y): \left(\frac{1+\sqrt{1-4r}}{2},\frac{1-\sqrt{1-4r}}{2}\right),\left(\frac{1-\sqrt{1-4r}}{2},\frac{1+\sqrt{1-4r}}{2}\right)\rbrace}

    So you need to evaluate:

    \int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} (1-x)\,dx-\int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} \frac{r}{x}\,dx

    Hahahah... I might be wrong

    On second thought it might be better to solve for r instead...

    x = 0.65, y = .2 seems like a good spot.

    (0.65)(0.2)=r \Rightarrow r = 0.13
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  3. #3
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    r is a constant

    but varies from 0 to 0.25, that means that r can be any number between [0,0.25]
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    r is just a constant, isn't it?
    Then simply solve both equations for y:
    y = 1-x

    y=\frac{r}{x}
    The graphs intersect at \lbrace{(x,y): \left(\frac{1+\sqrt{1-4r}}{2},\frac{1-\sqrt{1-4r}}{2}\right),\left(\frac{1-\sqrt{1-4r}}{2},\frac{1+\sqrt{1-4r}}{2}\right)\rbrace}

    So you need to evaluate:

    \int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} (1-x)\,dx-\int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} \frac{r}{x}\,dx
    I see no problem here. The final answer will depend on r.

    -Dan
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