6.1 Areas Between Curves/17

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Revision as of 23:56, 18 September 2022 by Josuem95981@students.laalliance.org (talk | contribs) (Created page with "<math> \begin{align} 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[ \fra...")
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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}}}