5.4 Indefinite Integrals and the Net Change Theorem/39: Difference between revisions
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<math>\int_{1}^{64}\frac{1+\sqrt[3]{x}}\sqrt{x}dx</math> | <math>\int_{1}^{64}\frac{1+\sqrt[3]{x}}\sqrt{x}dx</math> | ||
= <math>\int_{1}^{64}\frac{1}{x^{1/2}}</math> + <math>\int_{1}^{64}\frac{x^{1/3}}{x^{1/2}}</math> | = <math>\int_{1}^{64}\frac{1}{x^{1/2}}</math> + <math>\int_{1}^{64}\frac{x^{1/3}}{x^{1/2}}\\[3ex]</math> | ||
= <math>\int_{1}^{64}x^{-1/2}+x^{\frac{1}{3}-{\frac{1}{2}}}</math> = <math>\int_{1}^{64}x^{-\frac{1}{2}}+x^{-\frac{1}{6}}</math> | = <math>\int_{1}^{64}x^{-1/2}+x^{\frac{1}{3}-{\frac{1}{2}}}</math> = <math>\int_{1}^{64}x^{-\frac{1}{2}}+x^{-\frac{1}{6}}</math> | ||
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Add one to the exponents and divide by the new exponent | Add one to the exponents and divide by the new exponent | ||
= <math>\int_{1}^{64}\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+ \frac{x^{\frac{5}{6}}}{\frac{5}{6}}</math> = <math>\int_{1}^{64}2x^\frac{1}{2} + \frac{6}{5}x^\frac{5}{6} | = <math>\int_{1}^{64}\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+ \frac{x^{\frac{5}{6}}}{\frac{5}{6}}</math> = <math>\int_{1}^{64}2x^\frac{1}{2} + \frac{6}{5}x^\frac{5}{6}</math> | ||
=<math>2(x)^\frac{1}{2} + \frac{6}{5}(x)^\frac{5}{6}\bigg|_{1}^{64}</math> | =<math>2(x)^\frac{1}{2} + \frac{6}{5}(x)^\frac{5}{6}\bigg|_{1}^{64}</math> | ||
Revision as of 19:25, 30 August 2022
= + Failed to parse (syntax error): {\displaystyle \int_{1}^{64}\frac{x^{1/3}}{x^{1/2}}\\[3ex]}
![{\displaystyle \int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cdb02981593f318bab17caae838949d55b200cb6)
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Add one to the exponents and divide by the new exponent
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