Find b, a and f(x) for the definite integral

[Painguy]

Homework Statement

Three terms are used in a left hand sum to approximate the integral of ∫a to b f(x)dx

((2+0*(4/3))^2 * 4/3) + ((1+1*(4/3))^2 * 4/3) + ((2+2*(4/3))^2 * 4/3)

find a possible value of b and a, and f(x

Homework Equations

Ʃ Δx(f(a+Δxi))

The Attempt at a Solution

based on the the given summation I figured f(x)=x^2

and I want to say a is 2 and b is 4 because of (2+0*(4/3))^2 and because (b-a)/n=Δx. so that would mean n = 3

but the part that says ((1+1*(4/3))^2 * 4/3) is confusing me.

 

[Simon Bridge]

Well you have three terms in the equation - so write out (expand) the summation in symbols to three terms. Write out the numerical equation directly undernieth this. Compare them.

and I want to say a is 2 and b is 4 because of (2+0*(4/3))^2 and because (b-a)/n=Δx. so that would mean n = 3

n=3 because there are three terms ... but you have made a fence-post error in computing $\Delta x$.

 

[Painguy]

Well you have three terms in the equation - so write out (expand) the summation in symbols to three terms. Write out the numerical equation directly undernieth this. Compare them.


n=3 because there are three terms ... but you have made a fence-post error in computing $\Delta x$.

I don't understand what you mean. I have 3 terms? (i, Δx, and a) right?
i know what Δx is, but I'm stumped on what to put for a, and how to get it.

 

[Simon Bridge]

Not what I meant.
I am somewhat restricted by trying not to do the problem for you ;)

You have an expression with a bunch of numbers in it - that expression has three terms in it like this: term1 + term2 + term3

You also have an expression with symbols in it after a big Greek letter sigma - it is a summation. You have to match this one up with the one with all the numbers.

You should expand that summation into the corresponding three terms ... that will give you something to compare the first expression to. Then you will see where you have made your mistake and you'll see how to find the correct delta-x and b.

Do you not know what a "fence-post error" is?
Look it up. hint: $n \neq 3$.

 

[SammyS]

I don't understand what you mean. I have 3 terms? (i, Δx, and a) right?
i know what Δx is, but I'm stumped on what to put for a, and how to get it.

Inspecting the expression,

((2+0*(4/3))^2 * 4/3) + ((1+1*(4/3))^2 * 4/3) + ((2+2*(4/3))^2 * 4/3) ,

it's quite easy to determine a, n, and Δx .

What do you get for those values?

 

[Simon Bridge]

Recap - we have:$$\int_a^b f(x)dx \approx \sum_{i=0}^n f(a+i\Delta x)\Delta x =\left (2+0\cdot\frac{4}{3}\right )^2 \frac{4}{3} + \left (1+1\cdot\frac{4}{3}\right )^2\frac{4}{3} + \left ( 2 + 2\cdot\frac{4}{3}\right )^2\frac{4}{3}=T_1 + T_2 + T_3$$... the task is to find $a$ and $\Delta x$ and $b$.
painguy had left off the indexes for the sum - which are an important clue.
I still think that expanding the sum symbolically will be useful to understanding the problem.

You can relate $b$ to $a$, $\Delta x$ and $n$, and find it that way but you don't really need to. When painguy did this, he got the wrong value for n.

I do have a bit of a problem with that second term [1] - which is also confusing painguy -I think it is supposed to be $$T_2= \left (2+1\cdot\frac{4}{3}\right )^2\frac{4}{3} $$

Apart from that it is a matter of reading off the values.

----------------------------

[1] ... otherwise the three terms are inconsistent.

 

[SammyS]

Recap - we have:$$\int_a^b f(x)dx \approx \sum_{i=0}^n f(a+i\Delta x)\Delta x =\left (2+0\cdot\frac{4}{3}\right )^2 \frac{4}{3} + \left (1+1\cdot\frac{4}{3}\right )^2\frac{4}{3} + \left ( 2 + 2\cdot\frac{4}{3}\right )^2\frac{4}{3}=T_1 + T_2 + T_3$$... .

I think I see where one problem could be.

I usually see the Riemann sum written with n being the number of intervals into which [a,b] is partitioned.

So for a left hand sum, I would write \displaystyle \int_a^b f(x)dx \approx \sum_{i=0}^{n-1} f(a+i\Delta x)\Delta x \ .

And for a right hand sum \displaystyle \int_a^b f(x)dx \approx \sum_{i=1}^{n} f(a+i\Delta x)\Delta x \ .

So I would say n=3.

Painguy still has b wrong.

As for the second term -- I think that's just a typo.

 

[Simon Bridge]

I usually see the Riemann sum written with n being the number of intervals into which [a,b] is partitioned.

Fair enough. However - PainGuy says:

(b-a)/n=Δx

... which, at face value, suggests that n is the number of $\Delta x$'s between a and b. Which is not 3.

I maintain this is a fence-post error either way.

I suspect that the confusion here is the source of the matter.
I agree about the second term being a typo. I was hoping to get PainGuy to expand the sum and see that as the most likely explanation: there is nothing like the confidence you get from showing the question is wrong ;)

Now we need to hear back from OP :)

 

[SammyS]

Fair enough. However - PainGuy says: ... which, at face value, suggests that n is the number of $\Delta x$'s between a and b. Which is not 3.

It looks like 3 Δx's to me.

 

[Simon Bridge]

1st one's is multiplied by zero ... am I talking at cross purposes again?

I've been doing a+(n-1)Δx = b having the sum slightly overrun. But of course the 4/3 makes more sense the way you tell it ... and I owe OP an apology :( <sigh>

It would be more:
if $\Delta x = (b-a)/n$ then do the algebra first (solve for b), then plug in the numbers.

You know what it is? I've been doing too many MATLAB simulations where I'm sampling a function over n discrete points, including start an finish, which means I need n-1 intervals.
I'm going to go have a stiff drink!

 

[SammyS]

...

I'm going to go have a stiff drink!

I'll drink to that !