Evaluating a Riemann Sum for $\int^{-2}_{5} t^2 + 6t - 4 dt$

In summary, the limit of the Riemann sum is 233.3, and the solution is (where I write "lim" I mean limit as n-->infinity).
  • #1
paralian
14
0
[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...
 
Physics news on Phys.org
  • #2
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.
 
  • #3
HallsofIvy said:
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 :)

------

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:

1. What is a Riemann Sum?

A Riemann Sum is a method used in calculus to approximate the area under a curve. It involves dividing the area into smaller rectangles and finding the sum of the areas of these rectangles.

2. How do you evaluate a Riemann Sum for a definite integral?

To evaluate a Riemann Sum for a definite integral, you first divide the interval of integration into smaller subintervals. Then, you choose a point within each subinterval and use the function to calculate the height of a rectangle. Finally, you multiply the height by the width of the subinterval and sum up all the rectangles to get an approximation of the area under the curve.

3. What does the notation of $\int^{-2}_{5} t^2 + 6t - 4 dt$ mean?

The notation indicates that we are finding the area under the curve of the function $t^2 + 6t - 4$ between the limits of -2 and 5 on the x-axis.

4. What is the significance of using more rectangles in a Riemann Sum?

Using more rectangles in a Riemann Sum increases the accuracy of the approximation of the area under the curve. As the number of rectangles approaches infinity, the Riemann Sum becomes a more accurate representation of the actual area under the curve.

5. Can a Riemann Sum be used for any type of function?

Yes, a Riemann Sum can be used for any continuous function, regardless of its shape. However, the accuracy of the approximation may vary depending on the function and the number of rectangles used.

Similar threads

Replies
2
Views
1K
Replies
16
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
  • Calculus
Replies
2
Views
2K
Replies
1
Views
2K
Replies
1
Views
909
Replies
5
Views
2K
Replies
3
Views
1K
Back
Top