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

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<math>
<math>
\int\limits_{-1}^{2}(x-2|x|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx
\begin{align}


</math>


\int\limits_{-1}^{2}(x-2|x|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx


<math>
\begin{align}
&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg|_{0}^{2}  
&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg|_{0}^{2}  


\end{align}
</math>
<math>
\begin{align}
&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0
&= 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)
&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right)
\end{align}
\end{align}
</math>
</math>

Revision as of 18:58, 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=\int \limits _{-1}^{0}(x-2(-x))dx+\int \limits _{0}^{2}(x-2(x))dx&=\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}}}