5.4 Indefinite Integrals and the Net Change Theorem/21: Difference between revisions
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<math>\begin{align}\int_{0}^{2}(6x^{2}-4x+5) dx = \frac{6x^{2+1}}{2+1}-\frac{4x^{1+1}}{1+1}+{5x}\bigg|_{0}^{2} | <math>\begin{align}\int_{0}^{2}(6x^{2}-4x+5)dx &=\frac{6x^{2+1}}{2+1}-\frac{4x^{1+1}}{1+1}+{5x}\bigg|_{0}^{2}=\frac{6x^{3}}{3}-\frac{4x^{2}}{2}+{5x}\bigg|_{0}^{2}=2x^{3}-2x^{2}+{5x}\bigg|_{0}^{2}\\[2ex]&=[2(2)^{3}-2(2)^{2}+{5(2)}]-[2(0)^{3}-2(0)^{2}+{5(0)}]= 16-18+0\\[2ex]&=18\end{align}</math> | ||
=\frac{6x^{3}}{3}-\frac{4x^{2}}{2}+{5x}\bigg|_{0}^{2} = 2x^{3}-2x^{2}+{5x}\bigg|_{0}^{2} | |||
= [2(2)^{3}-2(2)^{2}+{5(2)}]-[2(0)^{3}-2(0)^{2}+{5(0)}] = 16-18+0 | |||
= 18\end{align}</math> | |||
Latest revision as of 19:40, 21 September 2022
![{\displaystyle {\begin{aligned}\int _{0}^{2}(6x^{2}-4x+5)dx&={\frac {6x^{2+1}}{2+1}}-{\frac {4x^{1+1}}{1+1}}+{5x}{\bigg |}_{0}^{2}={\frac {6x^{3}}{3}}-{\frac {4x^{2}}{2}}+{5x}{\bigg |}_{0}^{2}=2x^{3}-2x^{2}+{5x}{\bigg |}_{0}^{2}\\[2ex]&=[2(2)^{3}-2(2)^{2}+{5(2)}]-[2(0)^{3}-2(0)^{2}+{5(0)}]=16-18+0\\[2ex]&=18\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a2820a431bf7a715917f46dab4ee01aeaffb3669)