5.3 The Fundamental Theorem of Calculus/28: Difference between revisions

From Burton Tech. Points Wiki
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{2x^{frac{5}{2}}}{5}}
&= 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