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

Revision as of 20:37, 6 September 2022

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

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 <math>
-\int\limits_{1}^{4}(5-2t+3t^2)dt
+\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>
-<math>
-\int5dt-\int2tdt+\int3t^2dt=5t-t^2+t^3
-</math>
-<math>
-(5t-t^2+t^3)\bigg|_{1}^{4}=5\cdot4-4^2+4^3-(5\cdot1-1^2+1^3)=63
 </math>