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A cannonball is shot into the air. Its velocity is given as a function $f(t) m/s$, where $t$ measured in seconds since $1:00$ PM. We know that $f (t)$ takes the following values:

t 0 7.5 15 22.5 30 37.5 45 52.5 60

f(t) 10.0 6.46 5.00 3.88 2.93 2.09 1.34 0.646 0

For the two parts below, let $L_n$ be the Riemann sum for $I$ using $n$ subintervals and left endpoints, $R_n$ be the Riemann sum for $I$ using $n$ subintervals and right endpoints, and $M_n$ be the Riemann sum for $ I$ using $n$ subintervals and midpoints.

Write out the terms in $L_4$, $M_4$,$R_4$.

Can somebody help me how the question defines the Riemann sums?

mathnoob123
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  • Hint: break the interval into four pieces and apply the quadrature rule - the Riemann sum that is. – Sean Roberson Jul 04 '17 at 20:24
  • Hint: $L_4=15(f(0)+f(15)+f(30)+f(45)).$ – mfl Jul 04 '17 at 20:24
  • @mfl I am not sure how you got that. – mathnoob123 Jul 04 '17 at 20:25
  • Divide $[0,60]$ in four intervals of the same lenght: $[0,15], [15,30], [30,45]$ and $[45,60].$ The left points are $0,15,30$ and $45.$ Thus you have $L_4.$ – mfl Jul 04 '17 at 20:27
  • What if the question was n=6 or n=9? How to do in that case? and why was 15 used to multiply the sum? – mathnoob123 Jul 04 '17 at 20:28
  • In the same way. For $n=6$ you have to divide in $6$ intervals: $[0,10],\cdots, [50,60].$ – mfl Jul 04 '17 at 20:31
  • Oh so the procedure would be same but in this question, n=6 wouldn't be applicable because $f(10)$ etc. are not provided. (This means the numerical value cannot be estimated but the sum for $n=6$ would exist) Moreover can you aslo tell why was 15 used for multiplication? – mathnoob123 Jul 04 '17 at 20:37

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