5.3 The Fundamental Theorem of Calculus/21: Difference between revisions

Revision as of 20:39, 6 September 2022

$\int _{1}^{4}(5-2t+3t^{2})\,dt=5t-t^{2}+t^{3}{\bigg |}_{1}^{4}=5(4)-(4)^{2}+(4)^{3}-(5(1)-(1)^{2}+(1)^{3})=63$

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 <math>
-\int_{1}^{4}(5-2t+3t^2)\,dt=5t-t^2+t^3\bigg|_{1}^{4}=5\cdot4-4^2+4^3-(5\cdot1-1^2+1^3)=63
+\int_{1}^{4}(5-2t+3t^2)\,dt=5t-t^2+t^3\bigg|_{1}^{4}=5(4)-(4)^2+(4)^3-(5(1)-(1)^2+(1)^3)=63
 </math>