5.4 Indefinite Integrals and the Net Change Theorem/27: Difference between revisions

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&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg|_{1}^{4} \\[2ex]
&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg|_{1}^{4} \\[2ex]


&=\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}
&=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]


&=\frac{256}{15}
&=\frac{256}{15}

Revision as of 15:09, 21 September 2022

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \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] &=\frac{256}{15} \end{align} }