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Math Help - AREA under the curve

  1. #1
    Senior Member pankaj's Avatar
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    AREA under the curve

    1) For x\in[-\sqrt{2},1), the equation (y-1)x^2-3(y+1)x+2(y-1)=0 defines a monotonic function y=f(x).Find area bounded by y=f^{-1}(x),x=-17+12\sqrt{2},x=1 and the x-axis.


    2) Let f(x) be continuous and bijective such that f(0)=0.If \forall t\in R,area bounded by y=f(x),x=a-t,x=a and x-axis is equal to area bounded by y=f(x),x=a+t,x=a and x-axis,then prove that
    \int_{-\lambda}^{\lambda} f^{-1}(x)=2a\lambda

    3) If f(x)=\sin x,\forall x\in[0,\frac{\pi}{2}], f(x)+f(\pi-x)=2 \forall x\in\big(\frac{\pi}{2},\pi], f(x)=f(2\pi-x),\forall x\in\big(\frac{3\pi}{2},2\pi\big],then find the area enclosed by y=f(x) and the x-axis.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by pankaj View Post
    1) For x\in[-\sqrt{2},1), the equation (y-1)x^2-3(y+1)x+2(y-1)=0 defines a monotonic function y=f(x).Find area bounded by y=f^{-1}(x),x=-17+12\sqrt{2},x=1 and the x-axis.
    Rewrite the equation in the form y=f(x), sketch the curve and on that sketch shade the area definded. This will allow you to express the area as an integral of f(x).

    CB
    Last edited by CaptainBlack; November 28th 2010 at 06:53 AM.
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