5.4 Indefinite Integrals and the Net Change Theorem/21: Difference between revisions

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<math>\int_{0}^{2}(6x^{2}-4x+5) dx</math> = <math>\frac{6x^{2+1}}{2+1}-\frac{4x^{1+1}}{1+1}+{5x}\bigg|_{0}^{2}</math>
<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>
&= <math>\frac{6x^{3}}{3}-\frac{4x^{2}}{2}+{5x}\bigg|_{0}^{2}</math> = <math>2x^{3}-2x^{2}+{5x}\bigg|_{0}^{2}</math>
&= <math>[2(2)^{3}-2(2)^{2}+{5(2)}]-[2(0)^{3}-2(0)^{2}+{5(0)}\bigg|_{0}^{2}]</math> = <math>16-18+0</math>
&= <math>18</math>

Latest revision as of 19:40, 21 September 2022

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