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)\Bigg|_{1}^{4} \\[2ex]
&=\left(\frac{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{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{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]

Revision as of 15:13, 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 {2t^{\frac {3}{2}}}{3}}+{\frac {2t^{\frac {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}}}