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

From Burton Tech. Points Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 6: Line 6:
&=\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)|_{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}


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


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

Revision as of 15:09, 21 September 2022