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Sagot :
In this problem
the distance traveled between the seconds t=1 and t=3 is given by
[tex]\int_1^3(t^2-t+6)dt=\frac{50}{3}\text{ m}[/tex]The answer is
50/3 meters
or 16.67 meters
Explanation of integrals
In this problem we have
[tex]\int_1^3(t^2-t+6)dt=\int_1^3t^2dt-\int_1^3tdt+\int_1^36dt[/tex][tex]\begin{gathered} \int_1^3t^2dt=\frac{t^3}{3} \\ Evaluate\text{ at 3 and 1} \\ \frac{(3)^3}{3}-\frac{1^3}{3}=\frac{27}{3}-\frac{1}{3}=\frac{26}{3} \end{gathered}[/tex][tex]\begin{gathered} -\int_1^3tdt=-\frac{t^2}{2} \\ evaluate\text{ at 3 and 1} \\ -\frac{3^2}{2}+\frac{1^2}{2}=-\frac{9}{2}+\frac{1}{2}=-4 \end{gathered}[/tex][tex]\begin{gathered} \int_1^36dt=6t \\ evaluate\text{ at 3 and 1} \\ 6(3)-6(1)=12 \end{gathered}[/tex]substitute
[tex]\int_1^3t^2dt-\int_1^3tdt+\int_1^36dt=\frac{26}{3}-4+12=\frac{50}{3}[/tex]
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