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Evaluate [tex]\int^{-2}_{5} t^2 + 6t - 4 dt[/tex] by writing it as the limit of a Riemann sum, and taking the limit of that sum using properties of sigma notation. (Do NOT use the fundamental theorem of calculus to evaluate this integral.)Important stuff:

[tex]\sum i^2 = \frac{n(n+1)(2n+1)}{6}[/tex]

[tex]\sum i = \frac{n(n+1)}{2}[/tex]

And the solution:(Where I write "lim" I mean limit as n-->infinity. Where I write the summation sign I mean from i=1 to n.)

[tex]lim \sum t^2 + 6t - 4 \Delta t[/tex]

[tex]\Delta t = \frac{5 - (-2)}{n}[/tex]

[tex]t = i \Delta t = \frac{7i}{n}[/tex]

etc, etc.

[tex]= \frac{343}{3} + 147 + 28[/tex]

Which is about 233.3...

When I did the same thing by just doing the integral, I got -54.66....

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# Riemann sum

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