5.3 The Fundamental Theorem of Calculus/19: Difference between revisions
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&=\left[\frac{(2)^4}{4}-\frac{2(2)^2}{2}\right]-\left[\frac{(-1)^4}{4}-\frac{2(-1)^2}{2}\right]\\[2ex] | &=\left[\frac{(2)^4}{4}-\frac{2(2)^2}{2}\right]-\left[\frac{(-1)^4}{4}-\frac{2(-1)^2}{2}\right]\\[2ex] | ||
&=[0]-\left[\frac{1}{4}-1\right]=\frac{3}{4} | &=[0]-\left[\frac{1}{4}-1\right] | ||
&=\frac{3}{4} | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 20:35, 6 September 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align}\int_{-1}^{2}(x^3-2x)dx &= \frac{x^4}{4}-\frac{2x^2}{2}\Bigg|_{-1}^{2}\\[2ex] &=\left[\frac{(2)^4}{4}-\frac{2(2)^2}{2}\right]-\left[\frac{(-1)^4}{4}-\frac{2(-1)^2}{2}\right]\\[2ex] &=[0]-\left[\frac{1}{4}-1\right] &=\frac{3}{4} \end{align} }