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Math Help - Area Under Curve

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    Area Under Curve

    x=2-t^2, y= t(t+1)

    Find the exact area of the region bounded by the curve and the coordinate axes.

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    Quote Originally Posted by Air View Post
    x=2-t^2, y= t(t+1)

    Find the exact area of the region bounded by the curve and the coordinate axes.

    To find the x-intercepts we set y =0 and solve for t.

    0=t(t+1) \\\ t=0 \\\ t=-1

    The area under the curve can be found using

    \int_{x_0}^{x_1}ydx

    Now we need to calculate dx so

    dx=-2t dt

    so we get


    \int_{1}^{2}ydx=\int_{-1}^{0}-2t^3-2t^2dt=-\frac{1}{6}

    Note that the area is negative because it is below the x-axis

    Here is a graph. Good luck.

    Area Under Curve-graph.jpg
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    Now that I have read your question again, did you mean the part of the graph below the curve and inbetween the x and y axis. If so then

    \int_{0}^{1}ydx=\int_{-\sqrt{2}}^{-1}-2t^3-2t^2dt

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  4. #4
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    I don't know. The answer in mark scheme is...

     <br />
2 + \frac43 \sqrt2<br />
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  5. #5
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    Okay this shaded part is the area they want...

    Area Under Curve-graph.jpg

    \int_{0}^{2}ydx=-\int_{2}^{0}ydx=-\int_{0}^{\sqrt{2}}t(t+1)(-2t)dt=\int_{0}^{\sqrt{2}}2t^3+2t^2dt=2+\frac{4}{3}  \sqrt{2}
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