5.4 Indefinite Integrals and the Net Change Theorem/43: Difference between revisions

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&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} (4) - 4\right)
&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} (4) - 4\right)


&= \left(\frac{1}{2}+1 \right) + \left(2-4 \right)
&= \frac{1}{2}+1


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

Revision as of 18:50, 30 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int \limits _{-1}^{2}(x-2|x|)dx\\[1ex]&=\int \limits _{-1}^{0}(x-2(-x))dx+\int \limits _{0}^{2}(x-2(x))dx\\[2ex]&=\left({\frac {1}{2}}{x^{2}}+x^{2}\right){\bigg |}_{-1}^{0}+\left({\frac {1}{2}}{x^{2}}-x^{2}\right){\bigg |}_{0}^{2}&=0-\left({\frac {1}{2}}(-1)^{2}+(-1)^{2}\right)+\left({\frac {1}{2}}(2)^{2}-(2)^{2}\right)-0&=\left({\frac {1}{2}}+1\right)+\left({\frac {1}{2}}(4)-4\right)\end{aligned}}}