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^{1+\frac{1}{2}}\right)dx = \int_{0}^{1}\left(3+x^{\frac{3}{2}}\right)dx \\[2ex] | \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 \\[2ex] | ||
&= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg|_{0}^{1} = 3x+\frac{x^{\tfrac{5}{2}}}{\frac{5}{2}}\bigg|_{0}^{1} = 3x+\frac{2x^{\frac{5}{2}}}{5}\bigg|_{0}^{1} \\ | &= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg|_{0}^{1} = 3x+\frac{x^{\tfrac{5}{2}}}{\frac{5}{2}}\bigg|_{0}^{1} = 3x+\frac{2x^{\frac{5}{2}}}{5}\bigg|_{0}^{1} \\[2ex] | ||
&= \left[3(1)+\frac{2(1)^{5/2}}{5}\right]-\left[3(0)+\frac{2(0)^{5/2}}{5}\right] \\ | |||
&= \left[3(1)+\frac{2(1)^{5/2}}{5}\right]-\left[3(0)+\frac{2(0)^{5/2}}{5}\right] \\[2ex] | |||
&= 3+\frac{2}{5} = \frac{15}{5}+\frac{2}{5} = \frac{17}{5} | &= 3+\frac{2}{5} = \frac{15}{5}+\frac{2}{5} = \frac{17}{5} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 21:57, 23 August 2022