6.1 Areas Between Curves/17: Difference between revisions

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\begin{align}
\begin{align}


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 \\
& \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 \\






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}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] \\  
&=\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{4}{3} + \frac{9}{4} + \frac{4}{3} = \frac{8}{3} + \frac{9}{4} = \frac{32}{12} + \frac{27}{12} \\
= \frac{59}{12}
&= \frac{59}{12}


\end{align}
\end{align}
</math>
</math>

Revision as of 00:23, 19 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}&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\\&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{aligned}}}