Riemann sum

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[SOLVED] Riemann sum

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....
 

Answers and Replies

HallsofIvy
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The difficulty is that you haven't shown HOW you got either of those values. Your [itex]\Delta x[/itex] should be - since you are integrating from 5 down to -2. You have also neglected the "-" on "- 4" in the integrand.
 
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The difficulty is that you haven't shown HOW you got either of those values. Your [itex]\Delta x[/itex] should be - since you are integrating from 5 down to -2. You have also neglected the "-" on "- 4" in the integrand.
Sorry, I messed up the thing I meant for it to be from -2 to 5 but I put the numbers in the wrong places when I was typing it up :P

Ok :) This is the "etc, etc." Where did I miss a "-" on the -4 ?

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

[tex]lim \sum ( ( \frac{7i}{n})^2 + 6 \frac{7i}{n} - 4 ) \frac{7}{n}[/tex]

[tex]lim \sum ( \frac{7i}{n})^2 * \frac{7}{n} + 6 \frac{7i}{n} * \frac{7}{n} - 4 * \frac{7}{n}[/tex]

[tex]lim \sum \frac{7^2}{n^2} * i^2 * \frac{7}{n} + 6 \frac{7}{n} * i * \frac{7}{n} - 4 * \frac{7}{n}[/tex]

[tex]lim \frac{7^2}{n^2} * \frac{n(n+1)(2n+1)}{6} * \frac{7}{n} + 6 \frac{7}{n} * \frac{n(n+1)}{2} * \frac{7}{n} - 4n * \frac{7}{n}[/tex]

[tex]lim \frac{7^3}{6} * \frac{2n^2 + 3n + 1}{n^2} + 21 * 7 * \frac{n+1}{n} - 4 * 7[/tex]

[tex]\frac{7^3}{3} + 21 * 7 - 4 * 7[/tex]

If you can follow the algebra there :)

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Sorry all for not realizing this part of the forum wasn't for homework. If anyone has the ability to move it to the right part of the forum, it would be shiny.
 
Last edited:

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