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

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\begin{align}
\begin{align}


\int_{1}^{4}\sqrt{t}(1+t)dt =\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt
\int_{1}^{4}\sqrt{t}(1+t)dt &=\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt \\[2ex]


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


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

Latest revision as of 19:40, 21 September 2022