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Math Help - Area under curve

  1. #1
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    Area under curve

    Find the area under the curve y=05x−15

    from x=6 to x=t and evaluate it for t=10 , t=100.
    Then find the total area under this curve for x6.

    The Integral is read as .5x^(-1.5)
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  2. #2
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    Hello, radioheadfan!

    (a) Find the area under the curve: . y \:=\:\tfrac{1}{2}x^{-\frac{3}{2}} .from x=6 to  x=t.
    A \;=\;\tfrac{1}{2}\int^t_6 x^{-\frac{3}{2}}dx \;=\;\tfrac{1}{2}\left(-\tfrac{2}{1}\right)x^{-\frac{1}{2}}\,\bigg]^t_6 \;=\;-\frac{1}{\sqrt{x}}\,\bigg]^t_6 \;=\;\left(-\frac{1}{\sqrt{t}}\right) - \left(-\frac{1}{\sqrt{6}}\right)<br />

    Therefore: . A \;=\;\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{t}}



    \text{(b) Evaluate it for: }t=10,\: t=100
    \text{For }t = 10\!:\;\;A \;=\;\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{10}}

    \text{For }t = 100\!:\;\;A \;=\;\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{100}} \;=\;\frac{1}{\sqrt{6}} - \frac{1}{10}



    (c) Find the total area under this curve for x \geq 6.
    This would be: . \lim_{t\to\infty}\left(\frac{1}{\sqrt{6}} - \frac{1}{\sqrt{t}}\right) \;\;=\;\;\frac{1}{\sqrt{6}} - 0 \;\;=\;\;\frac{1}{\sqrt{6}}

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