5.4 Indefinite Integrals and the Net Change Theorem/27: Difference between revisions
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&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex] | &=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex] | ||
&=\left[\frac{16}{3}+\frac{64}{5}]-[\frac{2}{3}+\frac{2}{5} \right] | &=\left[\frac{16}{3}+\frac{64}{5}\right]-\left[\frac{2}{3}+\frac{2}{5} \right] \\[2ex] | ||
&=\frac{256}{15} | &=\frac{256}{15} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 15:12, 21 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{1}^{4}{\sqrt {t}}(1+t)dt&=\int _{1}^{4}\left(t^{\frac {1}{2}}+t^{\frac {3}{2}}\right)dt\\[2ex]&=\left({\frac {2(t)^{3/2}}{3}}+{\frac {2(t)^{5/2}}{5}}\right){\Bigg |}_{1}^{4}\\[2ex]&=\left[{\frac {2(4)^{3/2}}{3}}+{\frac {2(4)^{5/2}}{5}}\right]-\left[{\frac {2(1)^{3/2}}{3}}+{\frac {2(1)^{5/2}}{5}}\right]\\[2ex]&=\left[{\frac {16}{3}}+{\frac {64}{5}}\right]-\left[{\frac {2}{3}}+{\frac {2}{5}}\right]\\[2ex]&={\frac {256}{15}}\end{aligned}}}