Area between curves

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January 22nd 2009, 09:30 AM
Dana_Scully

Area between curves

Is there something I'm doing wrong in this problem? I can't figure it out, which probably means it's blindingly obvious.

Find the area of the region enclosed by $y=\sqrt{x}, y=\frac{1}{2}x,$ and $x=16$ .

$\sqrt{x}=\frac{1}{2}x$ at $x=(0, 4)$

So $\displaystyle\int^{16}_4 \frac{1}{2}x - x^{\frac{1}{2}} dx$

$\left(\frac{x^2}{4} - \frac{2\sqrt{x^3}}{3}\right) \Bigg|_{4}^{16}$

$\left(64 - \frac{128}{3}\right) - \left(4 - \frac{16}{3}\right)$

Which equals $\frac{68}{3}$
January 22nd 2009, 09:39 AM
ThePerfectHacker

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Quote:

Originally Posted by Dana_Scully

Is there something I'm doing wrong in this problem? I can't figure it out, which probably means it's blindingly obvious.

Find the area of the region enclosed by $y=\sqrt{x}, y=\frac{1}{2}x,$ and $x=16$ .

$\sqrt{x}=\frac{1}{2}x$ at $x=(0, 4)$

So $\displaystyle\int^{16}_4 \frac{1}{2}x - x^{\frac{1}{2}} dx$

$\left(\frac{x^2}{4} - \frac{2\sqrt{x^3}}{3}\right) \Bigg|_{4}^{16}$

$\left(64 - \frac{128}{3}\right) - \left(4 - \frac{16}{3}\right)$

Which equals $\frac{68}{3}$

Look at the picutre.
January 22nd 2009, 09:54 AM
Dana_Scully

Thank you! I guess I thought that it meant only the part directly bounded by x=16...I'm new to this, so thanks.