5.3 The Fundamental Theorem of Calculus/19: Difference between revisions
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\begin{align}\int_{-1}^{2}(x^3-2x)dx &= \frac{x^4}{4}-\frac{2x^2}{2}\Bigg|_{-1}^{2}\\[2ex] | \begin{align}\int_{-1}^{2}(x^3-2x)dx &= \frac{x^4}{4}-\frac{2x^2}{2}\Bigg|_{-1}^{2}\\[2ex] | ||
&=\frac{(2)^4}{4}-\frac{2(2)^2}{2}-\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]=0-\left[-\frac{3}{4}\right]=\frac{3}{4} | &=0-\left[\frac{1}{4}-1\right]=0-\left[-\frac{3}{4}\right]=\frac{3}{4} | ||
Revision as of 20:34, 6 September 2022
![{\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]=0-\left[-{\frac {3}{4}}\right]={\frac {3}{4}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b77b18bf97275b62138325f1be457a6340b0552d)