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

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<math>
<math>


\int_{0}^{1}\left(3+x\sqrt{x}\right)dx = \int_{0}^{1}\left(3+x^{1}{x}^{\frac{1}{2}}\right)dx
\begin{align}
\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


\end{align}
</math>
</math>

Revision as of 20:49, 23 August 2022