How to Integrate Upper Bounds with Circumscribed Rectangles

[tmclary]

[SOLVED] Upper Bounds Integration

Homework Statement

Integrate y=4x from 2 to 5 using the limit with circumscribed rectangles.

Homework Equations

A=lim(n to inf.) Summation of f(xsubi) times delta (xsubi)

The Attempt at a Solution

A=lim(4/n)(4/n)(4)(2+3+4+...+(n+1))
=64/n^2((n^2+3n)/2))= 32lim((n+3)/n)) =32. But from integration the answer is obviously 48. What am I doing wrong? (Sorry about lack of typo skills-newbie)

 

[sutupidmath]

why do you think the answer is obviously 48?
Your answer isn't correct again thout..

 

[sutupidmath]

well i got 42 as my answer, either by directly integrating

\int_2^5 4xdx and also by using Rieman sums.

I'll try to post my work, on my next post.

 

[tmclary]

Sorry-wrong limits!

Sorry! The limits were 1 to 5, not 2 to 5!

 

[sutupidmath]

we want to calculate

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

now let us create n mini segments on the segment [2,5]

that is let the points be

x_0=2,x_1,x_2,...x_i_-_1,x_i,...,x_n=5

Now our concern is to determine what our function will be.
First let's notice certian facts:

\delta x_i=x_i-x_i_-_1 also let \epsilon_i=x_i

this way we have:

\epsilon_i=\delta x_i+x_i_-_1

also: \delta x_i=\frac{5-2}{n}=\frac{3}{n}

Now, for to determine our function let's try some values for i=1,2,3,...

f(x_1)=4\left(\frac{3}{n}+2\right),f(x_2)=4(\frac{6}{n}+2),f(x_3)=4(\frac{9}{n}+2),..., f(x_i)=4(\frac{3i}{n}+2)


Hence:

\int_2^54xdx=\lim_{n\to\infty}\sum_{i=1}^{n}4\left(\frac{3i}{n}+2\right)\frac{3}{n}=...=42

 

[sutupidmath]

Sorry! The limits were 1 to 5, not 2 to 5!

Well, then do the same thing as i did here, just take into consideration that you have the lower limit 1, in this case. I am not going to troube to go the same route again, i think you can do it now. If you can't ask again.

cheers!

 

[sutupidmath]

Well it doesn't change a lot by the way, the difference is that now you'll have

\delta x_i=\frac{4}{n} and

f(x_i)=4\left(\frac{4i}{n}+1\right)

and the answer will be 48.

 

[tmclary]

Thanks very much for your replies- I'm still stuck expanding the summation- will attempt another query when I have time, and can clarify.

 

[tmclary]

Got it- I wasn't adding the 1 to the 4/n. Thanks again for your answer.

 

[sutupidmath]

Got it- I wasn't adding the 1 to the 4/n. Thanks again for your answer.

I tried to post a detailed answer, including how the summation expanded and all that stuff, but after i typed it all, i don't know for some crappy reason it did not show up. Anyways, I'm glad you got it !