5.4 Indefinite Integrals and the Net Change Theorem/33
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Revision as of 21:37, 3 September 2022 by Kattieh70488@students.laalliance.org (talk | contribs) (Created page with "<math> \begin{align} \int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}(\frac{\sqrt{5}}{\sqrt{x}})dx = 5^\frac{1}{2}\int_{1}^{4}x^\frac{-1}{2}dx\\[2ex] &= \sqrt{5}\times2x^\frac{1}{2}\bigg|_{1}^{4} = \sqrt{5}\times2\sqrt{x}\bigg|_{1}^{4} = 2\sqrt{5x}\bigg|_{1}^{4} \\[2ex] &= 2\sqrt{5\times4}-2\sqrt{5\times1}\bigg|_{1}^{4} \\[2ex] &= 2\sqrt{20}-2\sqrt{5}\bigg|_{1}^{4} = 4\sqrt{5}-2\sqrt{5}\bigg|_{1}^{4} = 2\sqrt{5} \end{align} </math>")
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{1}^{4}{\sqrt {\frac {5}{x}}}dy&=\int _{1}^{4}({\frac {\sqrt {5}}{\sqrt {x}}})dx=5^{\frac {1}{2}}\int _{1}^{4}x^{\frac {-1}{2}}dx\\[2ex]&={\sqrt {5}}\times 2x^{\frac {1}{2}}{\bigg |}_{1}^{4}={\sqrt {5}}\times 2{\sqrt {x}}{\bigg |}_{1}^{4}=2{\sqrt {5x}}{\bigg |}_{1}^{4}\\[2ex]&=2{\sqrt {5\times 4}}-2{\sqrt {5\times 1}}{\bigg |}_{1}^{4}\\[2ex]&=2{\sqrt {20}}-2{\sqrt {5}}{\bigg |}_{1}^{4}=4{\sqrt {5}}-2{\sqrt {5}}{\bigg |}_{1}^{4}=2{\sqrt {5}}\end{aligned}}}