paralian
- 14
- 0
[SOLVED] Riemann sum
Important stuff:
\sum i^2 = \frac{n(n+1)(2n+1)}{6}
\sum i = \frac{n(n+1)}{2}
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.)
lim \sum t^2 + 6t - 4 \Delta t
\Delta t = \frac{5 - (-2)}{n}
t = i \Delta t = \frac{7i}{n}
etc, etc.
= \frac{343}{3} + 147 + 28
Which is about 233.3...
When I did the same thing by just doing the integral, I got -54.66...
Evaluate \int^{-2}_{5} t^2 + 6t - 4 dt 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:
\sum i^2 = \frac{n(n+1)(2n+1)}{6}
\sum i = \frac{n(n+1)}{2}
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.)
lim \sum t^2 + 6t - 4 \Delta t
\Delta t = \frac{5 - (-2)}{n}
t = i \Delta t = \frac{7i}{n}
etc, etc.
= \frac{343}{3} + 147 + 28
Which is about 233.3...
When I did the same thing by just doing the integral, I got -54.66...