5.4 Indefinite Integrals and the Net Change Theorem/33: Difference between revisions
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(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>") |
m (Protected "5.4 Indefinite Integrals and the Net Change Theorem/33" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))) |
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\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4} | \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} | &= 2\sqrt{5}x^{\frac{1}{2}}\bigg|_{1}^{4} \\[2ex] | ||
&= 2\sqrt{5 | &= [2\sqrt{5}\sqrt{4}]-[2\sqrt{5}{\sqrt{1}}] = 4\sqrt{5}-2\sqrt{5} \\[2ex] | ||
&= 2\sqrt{5} | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 19:41, 21 September 2022
![{\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6ed4dda3c0671d278be3d931ee7adb2ef8514117)