5.4 Indefinite Integrals and the Net Change Theorem/27: Difference between revisions
Jump to navigation
Jump to search
No edit summary Tags: Manual revert Reverted |
No edit summary Tag: Manual revert |
||
Line 6: | Line 6: | ||
&=\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 15:10, 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} }