Use a left and midpoint Rienman sum to approximate

[flyingpig]

Homework Statement

Use a left and midpoint Rienman sum to approximate \int_{0}^{3} x^3 dx

The Attempt at a Solution

Left

\Delta x = \frac{3}{n}

Now here is the problem, for left points, what is my 'a'? I can't carve up my intervals and this isn't a right-end approximation so I can't take a= 0 for my x_i

 

[Sethric]

What formula are you using that has an 'a'? Without seeing it, I can't tell you what it is.

 

[flyingpig]

x_i = a + i\Delta x

 

[Sethric]

What is x_0 in this case?

 

[Sethric]

For clarification, what is the left endpoint of the interval over which you are approximating? That's what your 'a' is supposed to be here. Your function shouldn't affect your choice of a.

 

[flyingpig]

What do you mean...?

 

[Mark44]

a is 0.

x₀ = 0 + 0*3/n
x₁ = 0 + 1*3/n = 3/n
x₂ = 0 + 3*3/n = 6/n
.
.
.
x_n = 0 + n*3/n = 3

 

[flyingpig]

Is the answer for the left end and right end going to be the same for this question?

 

[Mark44]

No, of course not.

 

[flyingpig]

But that's what I got...

\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\Delta x

With \Delta x the same for all the f(x_i)

Like i isn't given I know (I drew a picture) that you don't evaluate f(x_i) at the first x₁, you do it for x₂

 

[Mark44]

You're getting confused between the partition and the x_i values. Suppose you partition the interval [0, 3] into 10 subintervals of equal length. Your partition will have 11 numbers in it:
x₀ = 0 + 0 * 3/10 = 0
x₁ = 0 + 1 * 3/10 = 3/10
x₂ = 0 + 2 * 3/10 = 6/10
x₃ = 0 + 3 * 3/10 = 9/10
.
.
.
x₁₀ = 0 + 10 * 3/10 = 30/10 = 3

You can estimate the value of the integral by this sum:
\sum_{i=1}^{10} f(c_i)\Delta x

where c_i is any number in the i-th subinterval. Notice that I removed the limit part.

For example, c₀ could be the left endpoint (x₀), the right endpoint (x₁), the midpoint ( (x₀ + x₁)/2), or any other point in the first interval.

If you take a larger value of n, you'll probably get a closer estimate.

 

[flyingpig]

No, the value of n is not given

 

[Mark44]

Then you pick a value of n.

 

[flyingpig]

You aren't suppose to, you are suppose to use the summation properties at the end like

\sum_{i=1}^n i = \frac{n(n+1)}{2} to evaluate the sum. I could do it for right-end sums, but I can't do it for left end

 

[Sethric]

Then leave it at an unknown n. Therefore, every partition has a width of 3/n. Show us your work for the right side. They should be extremely similar - differing only by which side of each partition you plug into f.

 

[Mark44]

Use a left and midpoint Rienman sum to approximate ...

I think you're supposed to pick a value for n to come up with an actual estimate.

 

[flyingpig]

\lim_{n\to \infty}\sum_{i=1}^{n} f(x_i) \Delta x

\Delta x = \frac{3}{n}

x_i = 0 + i\frac{3}{n}

f(x_i) = \frac{9i^2}{n^2}

\frac{3}{n} \lim_{n\to \infty}\sum_{i=1}^{n} \frac{9i^2}{n^2}

Sum this up and it turns out to be exactly as the integral for a right-end approximate

 

[Mark44]

\lim_{n\to \infty}\sum_{i=1}^{n} f(x_i) \Delta x

\Delta x = \frac{3}{n}

x_i = 0 + i\frac{3}{n}

f(x_i) = \frac{9i^2}{n^2}

\frac{3}{n} \lim_{n\to \infty}\sum_{i=1}^{n} \frac{9i^2}{n^2}

Sum this up and it turns out to be exactly as the integral for a right-end approximate

I'm reasonably certain this is NOT what they're looking for. The tipoff is that they have asked you to approximate the integral by a Riemann sum.

 

[flyingpig]

This was an example question for exam. I'll pull one from my midterm and I'll pull out the whole solutions [PLAIN]http://img21.imageshack.us/img21/6295/unledel.png

 

[Sethric]

Compute and approximate are different. You said approximate. Hence Mark's advice.

 

[Sethric]

Also, I can't help but notice your sum used x^2, while the problem stated x^3

 

[flyingpig]

Oh is it too late to ask how to "compute" for leftend approximate now...?

 

[Sethric]

For your right hand side, you took the right endpoint of each section in the partition, and plugged it into the formula, correct? For the left hand side, we just plug in the left endpoint:

\Delta x = \frac{3}{n}

\sum_{i=1}^n \frac{3}{n} f(x_{i-1})

That i-1 index is the key here.

What is x_{i-1} in terms of n and i?

 

[Mark44]

Oh is it too late to ask how to "compute" for leftend approximate now...?

Pick a number n.
Divide the interval [0, 3] into that many subintervals.
Evaluate f at the left end of each subinterval.
Multiply by \Delta x.
Add the n terms.

 

[flyingpig]

So n has to be defined...? But this doesn't apply to MidPoint right?

 

[Mark44]

So n has to be defined...? But this doesn't apply to MidPoint right?

Yes, n has to be defined. You haven't done this before?

Yes, you need to pick an n to do a midpoint approximation.

Pick a number n.
Divide the interval [0, 3] into that many subintervals.
Evaluate f at the [STRIKE]left end[/STRIKE] midpoint of each subinterval.
Multiply by \Delta x.
Add the n terms.

 

[Sethric]

If your professor meant for you to compute, then by all means let n go to infinity. If he wanted an approximation, do as Mark says.

 

[flyingpig]

Yes, n has to be defined. You haven't done this before?

Yes, you need to pick an n to do a midpoint approximation.

Pick a number n.
Divide the interval [0, 3] into that many subintervals.
Evaluate f at the [STRIKE]left end[/STRIKE] midpoint of each subinterval.
Multiply by \Delta x.
Add the n terms.

No I have done approximation before, in fact I am really good at them. While everyone (including my TAs) just memorize the formulas, I go and use pictures to help me.

 

[flyingpig]

If your professor meant for you to compute, then by all means let n go to infinity. If he wanted an approximation, do as Mark says.

But the question said "right Reinman"

 

[Sethric]

Right Riemann just means to create rectangles based on the height of f(x) on the right hand side of each section in the partition. We really only take the limit of sums when we are dealing with whether something is integrable.

 

[Mark44]

But the question said "right Reinman"

What's your point? All this means is that you use the right endpoint of each subinterval.

Pick a number n.
Divide the interval [0, 3] into that many subintervals.
Evaluate f at the [STRIKE]left end [/STRIKE] [STRIKE]midpoint[/STRIKE] right end of each subinterval.
Multiply by \Delta x.
Add the n terms.



BTW, it's Riemann.

 

[flyingpig]

What if you want to use left endpoint

 

[Sethric]

Try this:

Riemann Sum

Particularly the difference between right, midpoint, and left.

 

[Mark44]

What if you want to use left endpoint

See post #24.

I really hope that you have finished this problem by now. Approximating an integral by a Riemann sum is really nothing more than adding up the areas of a number of (n) rectangles. The rectangles are all \Delta x (= (b - a)/n) in width. For a left Riemann sum, you use the left endpoint of the subinterval to find the height of the rectangle. For a right Riemann sum, you use the right endpoint of the subinterval to find the height of the rectangle. For a midpoint Riemann sum, you use the midpoint of the subinterval to find the height of the rectangle.

To get your approximation, add the areas of the n rectangles. That's all there is to it.