5.3 The Fundamental Theorem of Calculus/28: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 4: | Line 4: | ||
\int_{0}^{1}\left(3+x\sqrt{x}\right)dx &= \int_{0}^{1}\left(3+x^{1}{x}^{\frac{1}{2}}\right)dx | \int_{0}^{1}\left(3+x\sqrt{x}\right)dx &= \int_{0}^{1}\left(3+x^{1}{x}^{\frac{1}{2}}\right)dx | ||
= \int_{0}^{1}\left(3+x^{1+\frac{1}{2}}\right)dx = \int_{0}^{1}\left(3+x^{\frac{3}{2}}\right)dx \\ | = \int_{0}^{1}\left(3+x^{1+\frac{1}{2}}\right)dx = \int_{0}^{1}\left(3+x^{\frac{3}{2}}\right)dx \\ | ||
&= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} = | &= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} = | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
<!-- | |||
3x+\frac{2x^{frac{5}{2}}}{5}} | |||
--> | |||
Revision as of 20:53, 23 August 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{1}\left(3+x{\sqrt {x}}\right)dx&=\int _{0}^{1}\left(3+x^{1}{x}^{\frac {1}{2}}\right)dx=\int _{0}^{1}\left(3+x^{1+{\frac {1}{2}}}\right)dx=\int _{0}^{1}\left(3+x^{\frac {3}{2}}\right)dx\\&=3x+{\frac {x^{{\frac {3}{2}}+1}}{{\frac {3}{2}}+1}}=\end{aligned}}}