5.3 The Fundamental Theorem of Calculus/19: Difference between revisions
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<math>\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}\end{align}</math> | <math>\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{(2)^4}{4}-\frac{2(2)^2}{2}\right]-\left[\frac{(-1)^4}{4}-\frac{2(-1)^2}{2}\right]\end{align}</math> | ||
Revision as of 19:37, 25 August 2022
![{\displaystyle {\begin{aligned}\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 {(2)^{4}}{4}}-{\frac {2(2)^{2}}{2}}\right]-\left[{\frac {(-1)^{4}}{4}}-{\frac {2(-1)^{2}}{2}}\right]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/21215d7fa169abd255d61dfd4b8d3c35cb07200d)