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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\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{aligned}}}