fundamental theorem of calculus part 1-

[ku1005]

hi...i understand te methods of computing definite/indefinite integrals...but when it comes to using the fundamental theorem part 1...my understanding lacks....so i have gone back to where I dont quite follow:

Prove that :

the Integral a to b x dx = (b^2-a^2)/2 where b > a

Because the method I used, which i know isnt correct is that

1) you divide the region a - b into n subintervals...therfore

each delta x will be (b-a)/n

2) now the height at each xi* will be (choosing left endpoints of intervals) f(x)=

x

but i stop now since there is no point saying any more since this approach gets me no where.

I no you require an expression where n tends to infinite....but I dont understand where the 2 comes from???

would someone plz be able to point me in the correct direction...because with an understanding of this Q...it allows me to proceed with alot more questions!

thanks heaps
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[ku1005]

also im quite sure that the sum of i = n(n+1)/2 has to be used.....

 

[galactus]

You're looking at the Riemann sum form what I can gather from your post.


You know we're dividing the interval [a,b] into n equal parts, then each part will have length:

\(\displaystyle \L\\{\Delta}x=\frac{b-a}{n}\)

Using the right endpoint, we have:

\(\displaystyle \L\\x_{k}=a+k{\Delta}x=a+k\cdot\frac{b-a}{n}\)

Therefore, the kth rectangle has area:

\(\displaystyle \L\\f(x_{k}){\Delta}x=x_{k}{\Delta}x=\left(a+k\cdot\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right)\)

Do the algebra and expand, we find the sum of these rectangles is:

\(\displaystyle \L\\\sum_{k=1}^{n}\left[\frac{a^{2}k}{n^{2}}-\frac{a^{2}}{n}-\frac{2abk}{n^{2}}+\frac{ab}{n}+\frac{b^{2}k}{n^{2}}\right]\)

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

As you know, the sum of the first k integers is \(\displaystyle \L\\\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\)

So, we have:

\(\displaystyle \L\\\frac{a^{2}}{n^{2}}\left(\frac{n(n+1)}{2}\right)-\frac{a^{2}}{n}(n)-\frac{2ab}{n^{2}}\left(\frac{n(n+1)}{2}\right)+\frac{ab}{n}(n)+\frac{b^{2}}{n^{2}}\left(\frac{n(n+1)}{2}\right)\)

Doing all this monstrous algebra, we arrive at:

\(\displaystyle \L\\\frac{a^{2}}{2n}-\frac{a^{2}}{2}-\frac{ab}{n}+\frac{b^{2}}{2n}+\frac{b^{2}}{2}\)

Now take the limit of the above:

\(\displaystyle \L\\\lim_{n\to\infty}\left[\frac{a^{2}}{2n}-\frac{a^{2}}{2}-\frac{ab}{n}+\frac{b^{2}}{2n}+\frac{b^{2}}{2}\right]\)

As you can see, as n approaches infinity, you're left with:

\(\displaystyle \L\\\frac{b^{2}}{2}-\frac{a^{2}}{2}\)

Which is the answer if you do it the easy way:

\(\displaystyle \L\\\int_{a}^{b}xdx=\frac{b^{2}}{2}-\frac{a^{2}}{2}\)

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

This is a good representation of what an integral is. You find the area of the rectangles and add them all up. As the number of rectangles, n, approaches infinity, you have the area under the curve...the integral.



EDIT: I just noticed you stated left endpoints. I used the right.

For left, use \(\displaystyle \L\\a+(k-1){\Delta}x\) instead of \(\displaystyle \L\\a+k{\Delta}x\)

 

[ku1005]

thanks heaps....u managed to just teach it better to me then my uni lecturer did...and she had an hour lol...cheers