5.4 Indefinite Integrals and the Net Change Theorem/25: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 4: | Line 4: | ||
\int_{-2}^{2}({3u+1})^2 du = \int {3u^2+6u+1} {du} \\[2ex] | \int_{-2}^{2}({3u+1})^2 du = \int {3u^2+6u+1} {du} \\[2ex] | ||
= {3u^3+3u^2+u}\bigg|_{-2}^{2} = {3\cdot 2^3 + \cdot 2^2 +2 - 3\cdot -2^3 + 3 \cdot-2^2 -2} = {52} | &= {3u^3+3u^2+u}\bigg|_{-2}^{2} = {3\cdot 2^3 + \cdot 2^2 +2 - 3\cdot -2^3 + 3 \cdot-2^2 -2} = {52} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 17:40, 7 September 2022
![{\displaystyle {\begin{aligned}\int _{-2}^{2}({3u+1})^{2}du=\int {3u^{2}+6u+1}{du}\\[2ex]&={3u^{3}+3u^{2}+u}{\bigg |}_{-2}^{2}={3\cdot 2^{3}+\cdot 2^{2}+2-3\cdot -2^{3}+3\cdot -2^{2}-2}={52}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0975dfe73f5d6da993638d5874267b95dac33cc6)