5.4 Indefinite Integrals and the Net Change Theorem/31: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 6: | Line 6: | ||
\int_{0}^{1}x\left(\sqrt[3]{x}+\sqrt[4]{x}\right)dx &=\int_{0}^{1}x\left(x^{\frac{1}{3}}+x^{\frac{1}{4}}\right)dx=\int_{0}^{1}\left(x^{\frac{4}{3}}+x^{\frac{5}{4}}\right)dx \\[2ex] | \int_{0}^{1}x\left(\sqrt[3]{x}+\sqrt[4]{x}\right)dx &=\int_{0}^{1}x\left(x^{\frac{1}{3}}+x^{\frac{1}{4}}\right)dx=\int_{0}^{1}\left(x^{\frac{4}{3}}+x^{\frac{5}{4}}\right)dx \\[2ex] | ||
&= \left(\frac{3x^{\frac{7}{3}}{7}+\frac{4x^{\frac{9}{4}}{9}\right)\bigg|_{0}^{1} | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 20:14, 20 September 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int_{0}^{1}x\left(\sqrt[3]{x}+\sqrt[4]{x}\right)dx &=\int_{0}^{1}x\left(x^{\frac{1}{3}}+x^{\frac{1}{4}}\right)dx=\int_{0}^{1}\left(x^{\frac{4}{3}}+x^{\frac{5}{4}}\right)dx \\[2ex] &= \left(\frac{3x^{\frac{7}{3}}{7}+\frac{4x^{\frac{9}{4}}{9}\right)\bigg|_{0}^{1} \end{align} }