5.4 Indefinite Integrals and the Net Change Theorem/27: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 8: | Line 8: | ||
&=\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} }