Calculating the area from 1280 subintervals

[Nemo1]

Hi Community,

I have this question and I stuck on the last question:

(b) How many subintervals do you need to estimate the area to within 0.1units\(\displaystyle ^2\)

View attachment 5069

I thought it was 40 subintervals but I am thinking that my answer is divisions along the $x$ axis.

The other questions I am confident I have been able to answer.

Any help would be appreciated.

Cheers Nemo.

 

[Deveno]

This is rather hard to do without the proviso that the sub-division be *regular* (i.e., into equal parts). The reason being, there are an infinite number of ways to sub-divide an interval into sub-intervals, and although the "error" in each way can be calculated by comparing the upper and lower sums, the rates of convergence will vary (so one may come up with different numbers of sub-intervals, depending on *how* the sub-division is being carried out).

Fortunately, you are in luck, on the interval $[-1,3]$ the function given by $f(x) = -x^2 + 10x + 24$ is increasing (how would you verify this?) *and* positive, and so the upper sums will be the right-hand sums, and the lower sums will be the left-hand sums. That is:

$U(f,n) = \sum\limits_{k = 1}^n \dfrac{4}{n}f\left(-1 + \dfrac{4k}{n}\right)$

$L(f,n) = \sum\limits_{k = 1}^n \dfrac{4}{n}f\left(-1 + \dfrac{4(k-1)}{n}\right)$

and you want $U(f,n) - L(f,n) = \epsilon < 0.1$

The algebra from here on out is going to be pretty nasty, but I can help you if you get stuck.

 

[Nemo1]

Hi Deveno,

Thanks for the response,

Your formula below I have to admit is beyond anything I have yet learnt.

$U(f,n) = \sum\limits_{k = 1}^n \dfrac{4}{n}f\left(-1 + \dfrac{4k}{n}\right)$

$L(f,n) = \sum\limits_{k = 1}^n \dfrac{4}{n}f\left(-1 + \dfrac{4(k-1)}{n}\right)$

and you want $U(f,n) - L(f,n) = \epsilon < 0.1$

As I review my training material provided they have the following method but I am unsure of how to use it.

This is the section provided.
View attachment 5073

And a graph of the equation:
View attachment 5074

I am happy to give a go to understanding your method, but find it a little daunting to be honest :(

Many thanks for your time, I really appreciate it!

Cheers Nemo.

 

[MarkFL]

It appears they are using the formula:

\(\displaystyle \frac{b-a}{n}|f(b)-f(a)|\le E\)

Can you identify $a,\,b,\,f(b),\,f(a),\,E$ for the given problem?

 

[Nemo1]

Hi Mark,

I started off by hand drawing a graph of \(\displaystyle y=-x^2+10x+24 \) and it looks near enough to:

View attachment 5075

I then drew dotted lines to get my upper and lower limit rectangles from $x=-1$ to $x=3$

I also drew a table to show my $x & y$ values:

x	y
-1	13
-0.75	15.9375
-0.5	18.75
-0.25	21.4375
0	24
0.25	26.4375
0.5	28.75
0.75	30.9375
1	33
1.25	34.9375
1.5	36.75
1.75	38.4375
2	40
2.25	41.4375
2.5	42.75
2.75	43.9375
3	45

Subdividing the interval \(\displaystyle [-1,3]\) into 4 equal intervals and using the table above the areas I got are the following:

Area Upper = \(\displaystyle 1*24+1*33+1*40+1*45=142units^2\)

Area Lower = \(\displaystyle 1*13+1*24+1*33+1*40=110units^2\)

I then created the definite integral:

\(\displaystyle \int_{-1}^{3} \,(-x^2+10x+24) dx\)

Which I solved to equal:

\(\displaystyle \frac{380}{3}\) or \(\displaystyle \approx 126.67\)

As per your question?

It appears they are using the formula:

\(\displaystyle \frac{b-a}{n}|f(b)-f(a)|\le E\)

Can you identify $a,\,b,\,f(b),\,f(a),\,E$ for the given problem?

I can see that $E = 0.1$

I am guessing that $n = 3$ which is the denominator from \(\displaystyle \frac{380}{3}\)

As for $b - a$ I am wondering if it is $a=-1$ and $b=3$ ?

To then get:

\(\displaystyle \frac{3-(-1)}{3}\) which is equal to \(\displaystyle \frac{4}{3}\)

Many thanks for your time and effort.

Cheers Nemo.

 

[Deveno]

Well, I admit it *does* look a bit terrifying. :P

The book there uses a nifty trick: that the bottom of the difference in areas rectangle in a given subdivision, is the top of the difference in areas in the previous subdivision. You would discover, if you were to use *my* formulas, that the net effect of this is to make a lot of terms "cancel out".

Let's look at your book's method, first.

We're going to have $n$ subdivisions of an interval that is 4 units long ($3 - (-1) = 4$). So the width of the "difference of sums rectangle" is going to be:

$\dfrac{4}{n}$.

Because of this unique "cancellation" feature (since $f$ is increasing), we can just take $f(3) - f(-1)$ for our "difference of sums rectangle" height.

Now, $f(3) = -(3^2) + 10\cdot 3 + 24 = -9 + 30 + 24 = 45$, and $f(-1) = -((-1)^2) + 10\cdot(-1) + 24 = -1 - 10 + 24 = -13$, so our rectangle has area:

$\dfrac{128}{n}$

********************

Now let's look at the method I propose (which is what you would use for "practically" calculating an integral).

We're going to chop up our interval of integration into $n$ segments, each of which has a base of $\dfrac{4}{n}$ length. The lower sums $L(f,n)$ will have a height of the value of $f$ at the right-hand endpoints of each sub-interval. Let's come up with an explicit way of saying what these endpoints are:

Well, the first one is easy, it's $-1$. The second one will be $-1 + \dfrac{4}{n}$. The third one will be $-1 + \dfrac{8}{n} = -1 + \dfrac{4}{n} + \dfrac{4}{n}$.

It's not hard to see that the $k$-th left endpoint will be:

$-1 + \dfrac{4(k-1)}{n}$.

So our $k$-th rectangle will have a base of $\dfrac{4}{n}$ and a height of $f\left(-1 + \dfrac{4k-4}{n}\right)$. As we add these all up $k$ goes up by $1$ each time, so we get the (lower) sum:

$L(f,n) = \sum\limits_{k=1}^n\dfrac{4}{n}f\left(-1 + \dfrac{4k-4}{n}\right)$

Let's compute this, even though it will get kind of nasty-looking for a bit.

First, let's compute the $f$-values (this is the hard part). Since $f(x) - x^2 + 10x + 24$, we have:

$f\left(-1 + \dfrac{4k-4}{n}\right) = -\left(-1 + \dfrac{4k-4}{n}\right)^2 + 10\left(-1 + \dfrac{4k-4}{n}\right) + 24$

$= -\left(1 - \dfrac{8k - 8}{n} + \dfrac{16k^2 - 32k + 16}{n^2}\right) - 10 + \dfrac{40k - 40}{n} + 24$

$= -1 + \dfrac{8k - 8}{n} - \dfrac{16k^2 - 32k + 16}{n^2} + \dfrac{40k - 40}{n} + 14$

$= 13 + \dfrac{48k - 48}{n} - \dfrac{16k^2 - 32k + 16}{n^2}$.

Oh my, that's ugly, isn't it?

We get one small break, since $\dfrac{4}{n}$ occurs in every term of our sum (over $k$), when we add all the terms together, we can use the distributive law to pull that out in front of the summation sign, like so:

$L(f,n) = \sum\limits_{k=1}^n\dfrac{4}{n}f\left(-1 + \dfrac{4k-4}{n}\right)$

$= \dfrac{4}{n}\sum\limits_{k=1}^n f\left(-1 + \dfrac{4k-4}{n}\right)$

Now for our upper sums, we use the right-hand endpoints of our sub-intervals, which results is a slightly different-looking sum:

$U(f,n) = \sum\limits_{k=1} \dfrac{4}{n} f\left(-1 + \dfrac{4k}{n}\right)$.

Again, calculating the $k$-th $f$-value is kind of scary:

$f\left(-1 + \dfrac{4k}{n}\right) = -\left(-1 + \dfrac{4k}{n}\right)^2 + 10\left(-1 + \dfrac{4k}{n}\right) + 24$

$= -\left(1 - \dfrac{8k}{n} + \dfrac{16k^2}{n^2}\right) - 10 + \dfrac{40k}{n} + 24$

$= -1 + \dfrac{8k}{n} - \dfrac{16k^2}{n^2} + \dfrac{40k}{n} + 14$

$= 13 + \dfrac{48k}{n} - \dfrac{16k^2}{n^2}$

To calculate $U(f,n) - L(f,n)$ let's keep the $\dfrac{4}{n}$ "out in front", and subtract the $f$-values "term by term" (that is, for each $k$). So the $k$-th *difference* in $f$ values is:

$13 + \dfrac{48k}{n} - \dfrac{16k^2}{n^2} - \left(13 + \dfrac{48k - 48}{n} - \dfrac{16k^2 - 32k + 16}{n^2}\right)$

$= \dfrac{48}{n} + \dfrac{-32k + 16}{n^2} = \dfrac{48}{n} + \dfrac{16}{n^2} - \dfrac{32k}{n^2}$

So, $U(f,n) - L(f,n) = \dfrac{4}{n}\sum\limits_{k=1}^n \left(\dfrac{48}{n} + \dfrac{16}{n^2} - \dfrac{32k}{n^2}\right)$

$=\dfrac{4}{n}\left(\sum\limits_{k=1}^n \dfrac{48}{n} + \sum\limits_{k=1}^n \dfrac{16}{n^2} - \sum\limits_{k=1}^n \dfrac{32k}{n^2}\right)$

Now the summands in the first two summations don't involve $k$, so we just add each summand $n$ times, which is the same as multiplying by $n$. So:

$\dfrac{4}{n}\left(\sum\limits_{k=1}^n \dfrac{48}{n} + \sum\limits_{k=1}^n \dfrac{16}{n^2} - \sum\limits_{k=1}^n \dfrac{32k}{n^2}\right) = \dfrac{4}{n}\left(48 + \dfrac{16}{n} - \dfrac{32}{n^2}\sum\limits_{k=1}^n k\right)$

$=\dfrac{192}{n} + \dfrac{64}{n^2} - \dfrac{128}{n^3}\sum\limits_{k=1}^n k$

You may or may not know this, but the last summation, is:

$\sum\limits_{k=1}^n k = \dfrac{n(n+1)}{2}$.

So, we arrive at:

$U(f,n) - L(f,n) = \dfrac{192}{n} + \dfrac{64}{n^2} - \dfrac{64(n+1)}{n^2}$

$= \dfrac{192}{n} + \dfrac{64}{n^2} - \dfrac{64}{n} - \dfrac{64}{n^2} = \dfrac{128}{n}$.

(whew!), as we obtained before.

You still have to solve:

$\dfrac{128}{n} < \dfrac{1}{10}$ for $n$, mind you.

 

[Nemo1]

Hi Deveno,

Thank you for your detailed response.

In

We're going to have $n$ subdivisions of an interval that is 4 units long ($3 - (-1) = 4$). So the width of the "difference of sums rectangle" is going to be:

$\dfrac{4}{n}$.

Because of this unique "cancellation" feature (since $f$ is increasing), we can just take $f(3) - f(-1)$ for our "difference of sums rectangle" height.

Now, $f(3) = -(3^2) + 10\cdot 3 + 24 = -9 + 30 + 24 = 45$, and $f(-1) = -((-1)^2) + 10\cdot(-1) + 24 = -1 - 10 + 24 = -13$, so our rectangle has area:

$\dfrac{128}{n}$

I see how using the formula given by Mark of:

\(\displaystyle \frac{b-a}{n}\) becomes \(\displaystyle \frac{3-(-1)}{n}\)

And if I plug that into the worked example from my book by using the $y$ values when $x$ equals $-1$ & $3$

I get:

\(\displaystyle 4(45-13) \le 0.1\)

Simplified:

\(\displaystyle 128 \le 0.1\)

Its from here I am lost, is 128 the amount of intervals to give me the area within \(\displaystyle \pm 0.1\)units$^2$?

I have had a detailed read and re read of your other method and I find myself even more confused.

I really appreciate your time you have given in your explanations.

Many thanks

Nemo :)

 

[MarkFL]

Hi Deveno,

Thank you for your detailed response.

In
I see how using the formula given by Mark of:

\(\displaystyle \frac{b-a}{n}\) becomes \(\displaystyle \frac{3-(-1)}{n}\)

And if I plug that into the worked example from my book by using the $y$ values when $x$ equals $-1$ & $3$

I get:

\(\displaystyle 4(45-13) \le 0.1\)

Simplified:

\(\displaystyle 128 \le 0.1\)

Its from here I am lost, is 128 the amount of intervals to give me the area within \(\displaystyle \pm 0.1\)units$^2$?

I have had a detailed read and re read of your other method and I find myself even more confused.

I really appreciate your time you have given in your explanations.

Many thanks

Nemo :)

You have dropped $n$ from the inequality...what you should wind up with is:

\(\displaystyle \frac{128}{n}\le\frac{1}{10}\)

And now you can solve for $n$...:)

 

[Nemo1]

Hi Mark and Deveno,

It was staring me in the face and I missed it :(

So we have:

\(\displaystyle \frac{128}{n} \le 0.1\)

To solve we can multiply both sides by $n$ to get:

\(\displaystyle 128=0.1n\)

Then divide both sides by $0.1$

\(\displaystyle \frac{128}{0.1} = \frac{0.1n}{0.1}\)

Simplify:

\(\displaystyle n=1280\)

So $1280$ divisions between $-1$ & $3$ will get the area to be within $0.1units$$^2$

Thanks for your persistence

To better understand the question how would I calculate the area from the $1280$ divisions.

As I know the correct area as defined by:

\(\displaystyle \int_{-1}^{3}(-x^2+10x+24) \,dx\) = \(\displaystyle \frac{380}{3}\) \(\displaystyle \approx 126.67\)

My thought leads me to the fact that if I can figure out the area from $1280$ divisions then subtract that from:

\(\displaystyle \frac{380}{3}\) \(\displaystyle \approx 126.67\)

To prove that I am indeed within $0.1units$$^2$

It also has occurred to me that:

\(\displaystyle 1280\le\infty\) could also be an accurate statement.

I am curious if there is a number less than $1280$ that could still achieve the requirement.

The more I play around with this question the more I am learning :)

Many thanks again :)

Cheers Nemo

 

[Deveno]

Hi Mark and Deveno,

It was staring me in the face and I missed it :(

So we have:

\(\displaystyle \frac{128}{n} \le 0.1\)

To solve we can multiply both sides by $n$ to get:

\(\displaystyle 128=0.1n\)

Then divide both sides by $0.1$

\(\displaystyle \frac{128}{0.1} = \frac{0.1n}{0.1}\)

Simplify:

\(\displaystyle n=1280\)

So $1280$ divisions between $-1$ & $3$ will get the area to be within $0.1units$$^2$

Thanks for your persistence

To better understand the question how would I calculate the area from the $1280$ divisions.

As I know the correct area as defined by:

\(\displaystyle \int_{-1}^{3}(-x^2+10x+24) \,dx\) = \(\displaystyle \frac{380}{3}\) \(\displaystyle \approx 126.67\)

My thought leads me to the fact that if I can figure out the area from $1280$ divisions then subtract that from:

\(\displaystyle \frac{380}{3}\) \(\displaystyle \approx 126.67\)

To prove that I am indeed within $0.1units$$^2$

It also has occurred to me that:

\(\displaystyle 1280\le\infty\) could also be an accurate statement.

I am curious if there is a number less than $1280$ that could still achieve the requirement.

The more I play around with this question the more I am learning :)

Many thanks again :)

Cheers Nemo

Well...it depends on how you "sub-divide". On portions of the curve where it is "flatter" (the slope (deriviative) is near $0$), we could use "larger" sub-intervals, whereas on portions where the curve is "steeper" we'd want pretty small sub-intervals. As bizarre as this seems, it is the beginnings of a more complicated kind of integration than the kind you are studying (Riemann integration), called Lebesgue integration.

But to be explicit, since we have established:

$U(f,n) - L(f,n) = \dfrac{128}{n}$ it follows that:

$\lim\limits_{n \to \infty} U(f,n) - L(f,n) = \lim\limits_{n \to \infty} \dfrac{128}{n} = 0$, hence:

$\lim\limits_{n \to \infty} U(f,n) = \lim\limits_{n \to \infty} L(f,n) = \int\limits_{-1}^3 f$

So, for example, since:

$U(f,n) = \dfrac{4}{n}\sum\limits_{k=1}^n \left(13 + \dfrac{48k}{n} - \dfrac{16k^2}{n^2}\right)$

$= \dfrac{4}{n}\left(13n + \dfrac{48}{n}\sum\limits_{k = 1}^n k - \dfrac{16}{n^2}\sum\limits_{k=1}^n k^2\right)$

$=\dfrac{4}{n}\left(13n + \dfrac{48}{n}\dfrac{n(n+1)}{2} - \dfrac{16}{n^2}\dfrac{n(n+1)(2n+1)}{6}\right)$

$= 52 + \dfrac{96(n+1)}{n} - \dfrac{32(n+1)(2n+1)}{3n^2}$

$= 52 + 96 + \dfrac{96}{n} - \dfrac{64}{3} - \dfrac{96n + 32}{n^2}$

$= \dfrac{380}{3} + \dfrac{1}{n}\left(96 - \dfrac{96n + 32}{n}\right)$

It follows that:

$\int\limits_{-1}^3 f = \lim\limits_{n \to \infty} U(f,n) = \dfrac{380}{3}$.

Now, if you *want* to, you can plug in $1,280$ for $n$ in:

$\dfrac{1}{n}\left(96 - \dfrac{96n + 32}{n}\right)$

and verify you get a number $< 0.1$, but there is no need.

You could then do the same for the lower sums, which would be an equal amount of hair-pulling...