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{x^{\frac{5}{2}}}{\frac{5}{2}} = 3x+ | &= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} = 3x+\frac{x^{\frac{5}{2}}}{\frac{5}{2}} = 3x+\frac{2x^{5/2}}{5} | ||
\end{align} | \end{align} |
Revision as of 21:39, 23 August 2022