5.4 Indefinite Integrals and the Net Change Theorem/33: Difference between revisions

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\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]
\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]
&= \sqrt{5}\times2x^\frac{1}{2}\bigg|_{1}^{4} = \sqrt{5}\times2\sqrt{x}\bigg|_{1}^{4} = 2\sqrt{5x}\bigg|_{1}^{4} \\[2ex]

Revision as of 15:26, 21 September 2022

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}}\\[2ex]&=2{\sqrt {20}}-2{\sqrt {5}}=4{\sqrt {5}}-2{\sqrt {5}}=2{\sqrt {5}}\end{aligned}}}