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>") |
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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 = 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 21:37, 3 September 2022