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

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<math>
<math>
\begin{align}
\begin{align}
\int_{-1}^{2}(x-2|x|)dx &= \int_{-1}^{0}(x-2(-x))dx + \int_{0}^{2}(x-2(x))dx = \int_{-1}^{0}3x\,dx - \int_{0}^{2}x\,dx \\[2ex]




\int\limits_{-1}^{2}(x-2|x|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[2ex]
&= \left(\frac{3x^2}{2} \right)\bigg|_{-1}^{0} - \left(\frac{x^2}{2} \right)\bigg|_{0}^{2} \\[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} \\[2ex]
&= \left[\frac{3(0)^2}{2}-\frac{3(-1)^2}{2}\right]-\left[\frac{(2)^2}{2} - \frac{(0)^2}{2}\right] \\[2ex]


&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[2ex]
&= \left[-\frac{3}{2}\right]-\left[\frac{4}{2}\right] \\[2ex]


&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[2ex]
&= -\frac{7}{2} \\[2ex]
\end{align}
\end{align}
</math>
</math>

Latest revision as of 19:42, 21 September 2022