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

Revision as of 20:38, 6 September 2022

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

@@ Line 1: / Line 1: @@
 <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\cdot4-4^2+4^3-(5\cdot1-1^2+1^3)=63
 </math>