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


\end{align}
\end{align}
</math>
</math>

Revision as of 15:11, 21 September 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\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] \\[2ex] &=\left[\frac{16}{3}+\frac{64}{5}]-[\frac{2}{3}+\frac{2}{5} \right] &=\frac{256}{15} \end{align} }