6.1 Areas Between Curves/17: Difference between revisions
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y = \sqrt{x},\ y=\frac{1}{2}x,\ x=9 \\ | y = \sqrt{x},\ y=\frac{1}{2}x,\ x=9 \\ | ||
\sqrt{x}=\frac{1}{2}x\ \rightarrow \ \sqrt{x}-\frac{1}{2}x=0\ \rightarrow \ \sqrt{4}\ -\frac{1}{2}(4) = 2-2=0,\ x=4 \\ | |||
\end{align} | \end{align} | ||
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A= \int_{0}^{4} \left[\sqrt{x} - \frac{1}{2}x \right]\mathrm{d}x + \int_{4}^{9} \left[ \frac{1}{2}x - \sqrt{x} \right]\mathrm{d}x \\ | A= \int_{0}^{4} \left[\sqrt{x} - \frac{1}{2}x \right]\mathrm{d}x + \int_{4}^{9} \left[ \frac{1}{2}x - \sqrt{x} \right]\mathrm{d}x \\ | ||
= \ \left[\frac{2}{3}x^\frac{3}{2} - \frac{1}{4}x^2 \right]_{0}^{4} \ + \ \left[ \frac{1}{4}x^2 - \frac{2}{3}x^\frac{3}{2} \right]_{4}^{9} \\ | |||
=\left[ \frac{2}{3} \left(4\right)^\frac{3}{2} - \frac{1}{4} \left(4\right)^2 \right] - \left[0\right] + \left[\frac{1}{4}\left(9\right)^2 - \frac{2}{3}\left(9\right)^\frac{3}{2} \right] - \left[\frac{1}{4}\left(4\right)^2 - \frac{2}{3}\left(4\right)^\frac{3}{2} \right] = \left[ \frac{16}{3} - 4 \right] - \left[0\right] + \left[\frac{81}{4} - 18\right] - \left[4 - \frac{16}{3}\right] \\ | |||
= \frac{4}{3} + \frac{9}{4} + \frac{4}{3} = \frac{8}{3} + \frac{9}{4} = \frac{32}{12} + \frac{27}{12} \\ | |||
= \frac{59}{12} | |||
\end{align} | \end{align} | ||
</math> | </math> |
Revision as of 00:01, 19 September 2022