5.3 The Fundamental Theorem of Calculus/28: Difference between revisions
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\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} = 3x+\frac{2x^{frac{5}{2}}{5}} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 20:52, 23 August 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \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 \\ &= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} = 3x+\frac{2x^{frac{5}{2}}{5}} \end{align} }