5.4 Indefinite Integrals and the Net Change Theorem/27: Difference between revisions
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\int_{1}^{4}\sqrt{t}(1+t)dt &=\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt \\[2ex] | \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)|_{1}^{4} \\[2ex] | &=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg|_{1}^{4} \\[2ex] | ||
&=\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4} | &=\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\bigg|_{1}^{4} | ||
Revision as of 15:09, 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]&={\frac {2(t)^{3/2}}{3}}+{\frac {2(t)^{5/2}}{5}}{\bigg |}_{1}^{4}&={\frac {2(4)^{3/2}}{3}}+{\frac {2(4)^{5/2}}{5}}-{\frac {2(1)^{3/2}}{3}}+{\frac {2(1)^{5/2}}{5}}&={\frac {256}{15}}\end{aligned}}}