Riemann sum..?

Question

I have the following exercise: Calculate like a limit of the sum of Riemann the following integral: 1~2 (x^2) (~ is integral)
I have the solution, but I can't understand what was done.

Answer 1

Do you mean:
[1, 2]∫x² dx?

The fastest method of doing this is of course to take the antiderivative of x² (which is x³/3) and find the difference between (2)³/3 and (1)³/3, which is 8/3 - 1/3 or 7/3. However, since you are specifically asked to calculate this using riemeann sums:

First note that as the mesh of the partitions approaches zero, the value of all riemann sums will converge to the same limit. It therefore suffices to find the limit as the mesh decreases of any particular subset of riemann sums. To make this as easy as possible to compute, we will consider only those sums involving regular partitions evaluated at the right-hand side. If there are n subintervals, then the partition is:

x_k = 1 + k/n

Each partition has width 1/n. Since this is a right-hand sum, the kth partition is evaluated at x_k. So our riemann sum is:

[k=1, n]∑((1 + k/n)² 1/n)

At this point I should explain what we are doing: we are dividing the interval [1, 2] into n subintervals, such that each subinterval is the same length. This implies that the kth subinterval is [x_(k-1), x_k], where x_k is defined using the formula given above. We then construct n rectangles each having as base the kth subinterval (which is always of length 1/n) and height the value of the function x² at the right-hand side of that interval. These rectangles form an approximation to the region under x² on the interval [1, 2]. Because of this, we can approximate the area under the function simply by adding up the areas of the rectangles. This sum is the riemann sum given above.

To find the exact area under the function, we take the limit of these riemann sums as the width of all the rectangles approaches zero. Since these rectangles all have the same width, this can be done simply by letting the number of rectangles approach infinity. Thus, [1, 2]∫x² dx is:

[n→∞]lim [k=1, n]∑((1 + k/n)² 1/n)

In order to evaluate this limit, we need a closed-form expression for that sum. First note that 1/n is a constant w.r.t. k, so we can bring it outside the sum:

[n→∞]lim 1/n [k=1, n]∑(1 + k/n)²

Expanding the part in the parentheses:

[n→∞]lim 1/n [k=1, n]∑(1 + 2k/n + k²/n²)

Breaking up the sum:

[n→∞]lim 1/n ([k=1, n]∑1 + 2/n[k=1, n]∑k + 1/n²[k=1, n]∑k²)

You should remember closed form expressions for each of these simple sums:

[n→∞]lim 1/n (n + 2/n(n(n+1)/2) + 1/n²(n(n+1)(2n+1)/6))

Simplifying:

[n→∞]lim 1/n (n + n+1 + (n+1)(2n+1)/(6n))
[n→∞]lim 1/n (n + n+1 + (2n²+3n+1)/(6n))
[n→∞]lim 1/n (2n+1 + n/3+1/2+1/n)
[n→∞]lim 1/n (7n/3+3/2+1/n)
[n→∞]lim 7/3 + 3/(2n)+1/n²

Finally, we evaluate the limit. As n→∞, both 3/(2n) and 1/n² approach zero, so we have:

[1, 2]∫x² dx = 7/3

Which is what we got at the beginning.

Answer 2

Ok, lets solve it.

You have the integral [I would also use the same sign for integral as you did]
1~2 (x^2) (~ is integral)

Now, we devide the interval [1,2] in some small parts, with equal distance between them.
a = x0 < x1 < x2 ... < xn = b.
a = 1 1.2 1.4 1.6 1.8 2=b
Also,
P = {[x0, x1), [x1, x2), ... [xn-1, xn]}

So the Riemann sum is

R = sum(i=1 to n) [f(y_i)*(x_i - x_(i-1))]

With x_i < y_i < x_(i-1). Lets say y_i is mean(or mid point) of x_i and x_(i-1).

So, the Riemann sum will become

R = sum(i=1 to n) [f((x_i - x_(i-1)/2)*(x_i - x_(i-1))]

Substituting n=6 we get

R = sum(i=1 to 6) [f((x_i - x_(i-1)/2)*(x_i - x_(i-1))]

(1.2100*0.2420)+(1.6900*0.3380)+(2.2500*0.4500)+(2.8900*0.5780)+(3.6100*0.7220)
[I might have made some mistake for copying this so, pleas be careful while interpreting]
R = 2.3300

Now, if you increase n from 6 to 60, the answer precises to
R = 2.3333
And it remains so if you any more increasse values of n.

[I am sorry but its too tidious to show some steps for higher values of n, so I just showed for n=6,I hope you understand]

All the best.

Answer 3

yoiu shoud dfine rectangles with the upper left corner on the graph of the function and rthe uopper right corner of the rect. on the grapph. the widt of the rectangle is h. write the sum of the surface for all rctangles and then take the limit of h->0.

Answer 4

Make the integration with lower limit 2 and upper limit 6. Ans is: [(5x^3)/3 + (3x^2)/2 + 2x] , put upper limit and lower limit values on this final result will be 402.67 unit square

Answer 5

Perhaps this will help:
http://en.wikipedia.org/wiki/Riemann_sum

Also, there may be other sources if you search on Google.

I have seen some sites that show examples which could help.

In fact, it is here http://mathworld.wolfram.com/RiemannIntegral.html
I believe that is the same solution you are asking for.

Answer 6

i havent learned the Riemann sum,but hope dis web can be helpful:
http://en.wikipedia.org/wiki/Riemann_sum
:)

Answer 7

Great article at http://mathworld.wolfram.com/RiemannSum.html


Doug