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

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
m (Protected "5.4 Indefinite Integrals and the Net Change Theorem/33" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite)))
 
(8 intermediate revisions by the same user not shown)
Line 3: Line 3:




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


&= 2\sqrt{5\times4}-2\sqrt{5\times1} \\[2ex]
&= [2\sqrt{5}\sqrt{4}]-[2\sqrt{5}{\sqrt{1}}] = 4\sqrt{5}-2\sqrt{5} \\[2ex]
 
&= 2\sqrt{20}-2\sqrt{5} = 4\sqrt{5}-2\sqrt{5} = 2\sqrt{5}
&= 2\sqrt{5}
\end{align}
\end{align}
</math>
</math>

Latest revision as of 19:41, 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={\sqrt {5}}\int _{1}^{4}x^{-{\frac {1}{2}}}dx\\[2ex]&=2{\sqrt {5}}x^{\frac {1}{2}}{\bigg |}_{1}^{4}\\[2ex]&=[2{\sqrt {5}}{\sqrt {4}}]-[2{\sqrt {5}}{\sqrt {1}}]=4{\sqrt {5}}-2{\sqrt {5}}\\[2ex]&=2{\sqrt {5}}\end{aligned}}}