Find an expression for the area under the graph as a limit

[MarkSA]

Hello,

We just started on Integrals, and as with derivatives, we're having to start with the longer and more confusing ways of solving them before getting to the shortcuts.

The problem is:
Find an expression for the area under the graph of f as a limit. Do not evaluate the limit.

f(x) = 1 + x^4, with 2 <= x <= 5

The notation they are looking for is lim n->infinity of [f(x1) * dx + f(x2) * dx.. etc]
Using sigma notation. I hope i've made that clear enough (i'm not entirely sure what to use myself)

Right now I have
lim n->infinity of sigma i=1 to n of:
3 + (3i/n)^4 * 3/n

I don't know if that is right. Is there a way to check that with a graphing calculator somehow? What I did was
dx = width = (b-a)/n = (5-2)/n = 3/n
xi = a + i*dx = 2 + (3i/n)
1 + 2 + (3i/n) * 3/n comes out to what I put above.

Can you confirm if i'm doing this correctly? The book only had one example problem so i'm not sure. Thanks.

 

[tkhunny]

Re: Find an expression for the area under the graph as a lim

we're having to start with the longer and more confusing ways of solving them before getting to the shortcuts.

This is, indeed, not how children learn to speak or to walk. Should all things be learned as little children?

h = 3
n = (5-2)/3 = 1
f(2)*(h)

h = 1
n = (5-2)/1 = 3
[f(2) + f(2+h) + f(2+2h)]*(h)

h = 1/2
n = (5-2)/(1/2) = 6
[f(2) + f(2+h) + f(2+2h) + f(2+3h) + f(2+4h) + f(2+5h)]*(h)

How are we doing?

h = 1/4
n = (5-2)/(1/4) = 12
[f(2) + f(2+h) + f(2+2h) + ... + f(2+(n-1)h)]*(h)

 

[MarkSA]

This is, indeed, not how children learn to speak or to walk. Should all things be learned as little children?

I can't really tell what that means but i'm guessing you're agreeing with learning the long and hard method first

I'm a little confused about the examples you gave, could you clarify some about them?

If I understand right you're using h as the number of subintervals, and n as the width of the subinterval (dx). dx = n = (b-a)/h in your example. Is this correct?

I don't really understand the f(2)*(h) for instance though. I should be adding the areas of each rectangle correct? Wouldn't that be f(x)*dx for each one, with x changing? Is the way you wrote it another way of solving the problem?

For instance say I have 3 subintervals, n = 3
f(xi)*dx is the area of each...
dx = width = (5-2)/3 = 1
And xi = a + i*dx or
xi = 2 + i
So, the area of these three rectangles should be:
A = (1)*[f(2 + 1) + f(2 + 2) + f(2 + 3)]
Is this correct?

 

[galactus]

You're supposed to use n subintervals?.

The heights of the approximating rectangles are the values of f at the points \(\displaystyle \L\\x_{1}, x_{2}, x_{3},............,x_{n}\)

So the approximating rectangles that make up the region have areas:

\(\displaystyle \L\\f(x_{1}){\Delta}x, f(x_{2}){\Delta}x, f(x_{3}){\Delta}x, ................., f(x_{n}){\Delta}x\)

The total area of the region is then:

\(\displaystyle \L\\f(x_{1}){\Delta}x+f(x_{2}){\Delta}x+f(x_{3}){\Delta}x+.....................+f(x_{n}){\Delta}x\)

Or in sigma notation:

\(\displaystyle \L\\\sum_{k=1}^{n}f(x_{k}){\Delta}x\)

So, we can write:

\(\displaystyle \L\\Area \;\ = \;\ \lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k}){\Delta}x\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, let's use the right endpoint method: \(\displaystyle \L\\x_{k}=a+k{\Delta}x\). Where a = 2, in this case, in the subinterval [2,5].

\(\displaystyle \L\\{\Delta}x=\frac{5-2}{n}=\frac{3}{n}\)

\(\displaystyle \L\\x_{k}=a+k{\Delta}x=2+\frac{3k}{n}\)

So, we have:

\(\displaystyle \L\\f(x_{k}){\Delta}x=\left[1+\left(2+\frac{3k}{n}\right)^{4}\right]\cdot\frac{3}{n}\)

Expand this out and the algebra is a mess.....but it will work:

\(\displaystyle \L\\\frac{243}{n^{5}}k^{4}+\frac{648}{n^{4}}k^{3}+\frac{648}{n^{3}}k^{2}+\frac{288}{n^{2}}k+\frac{51}{n}\)

Take note that the k terms can be written as sums:

\(\displaystyle \L\\\frac{243}{n^{5}}\sum_{k=1}^{n}k^{4}+\frac{648}{n^{4}}\sum_{k=1}^{n}k^{3}+\frac{648}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{288}{n^{2}}\sum_{k=1}^{n}k+\frac{51}{n}\sum_{k=1}^{n}1\)

We can continue to the area under the curve, but it's messy.

\(\displaystyle \L\\\sum_{k=1}^{n}k^{4}=\frac{n(n+1)(6n^{3}+9n^{2}+n-1)}{30}\)

\(\displaystyle \L\\\sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4}\)

\(\displaystyle \L\\\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}\)

\(\displaystyle \L\\\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\)

Now, sub these in for the sums above and get:

\(\displaystyle \L\\\lim_{n\to\infty}\left[\frac{243}{2n}+\frac{81}{n^{2}}-\frac{81}{10n^{4}}+\fbox{\frac{243}{5}}+\frac{324}{n}+\frac{162}{n^{2}}+\fbox{162}+\frac{324}{n}+\frac{108}{n^{2}}+\fbox{216}+\frac{144}{n}+\fbox{144}+\fbox{51}\right]\)

Take note that all the terms with n in the denominator approach 0, because n is unbounded ("approaches infinity").

So, we are left with \(\displaystyle \L\\\frac{243}{5}+162+216+144+51\)=\(\displaystyle \H\\\frac{3108}{5}\)

WHEW.....and that is the area under the curve using a Riemann sum.


EDIT: Now I see it says do not evaluate the limit. Oh well. See how it's done now?. Hope that helps.

 

[MarkSA]

Yes that was very helpful, thank you. Wow, it would take me about a century to evaluate the limit. Hopefully the homework problems are a little easier

One thing I had a question on that you mentioned:
"Now, let's use the left endpoint method: xk = a + k*dx, where a = 2 in this case"

You called this the left endpoint method. Is there a different version of it for right endpoints? Would doing it for right endpoints simply be: xk = a + (k+1)*dx?

How about midpoints, would the one for that be: xk = a + (2k + 1)/2*dx?

Thanks

 

[galactus]

The right endpoint is \(\displaystyle \L\\x_{k}=a+k{\Delta}x\)

The left endpoint is \(\displaystyle \L\\x_{k-1}=a+(k-1){\Delta}x\)

The mid point is \(\displaystyle \L\\x_{k}=a+(k-\frac{1}{2}){\Delta}x\)

Sorry, I believe I said I used the left endpoint, that was the right I used.