5.3 The Fundamental Theorem of Calculus/23: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 4: | Line 4: | ||
\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg|_{0}^{1} \\[2ex] | \int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg|_{0}^{1} \\[2ex] | ||
&=\frac{5\sqrt[5]{1^9}}{9}-\frac{5 \sqrt[5]{0^9}}{9} \\[2ex] | &=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex] | ||
&=\cfrac{5}{9} | &=\cfrac{5}{9} | ||
Revision as of 20:50, 6 September 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg|_{0}^{1} \\[2ex] &=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex] &=\cfrac{5}{9} \end{align} }