nathanruff 3 years ago Find the exact area bounded by f(x)=x^2 and the x-axis between x=1 and x=2 using n partitions.

1. ByteMe

n partitions imply a finite number... if you're going to find the EXACT area, you'll need an infinite number of very tiny partitions... ie, let $$\large n\rightarrow \infty$$

2. nathanruff

I ended up with this: $A = (\frac{ 1 }{ n } + \frac{ 2(1) }{ n ^{2} } + \frac{ 1^{2} }{ n ^{3} }) + (\frac{ 1 }{ n } + \frac{ 2(2) }{ n ^{2} } + \frac{ 2^{2} }{ n ^{3} }) + (\frac{ 1 }{ n } + \frac{ 2(3) }{ n ^{2} } + \frac{ 3^{2} }{ n ^{3} }) + (\frac{ 1 }{ n } + \frac{ 2(4) }{ n ^{2} } + \frac{ 4^{2} }{ n ^{3} }) +...+ (\frac{ 1 }{ n } + \frac{ 2(n) }{ n ^{2} } + \frac{ n^{2} }{ n ^{3} })$ Simplified: $A = \frac{ n }{ n } + \frac{ 2 }{ n ^{2} }(1+2+3+4+...+n)+\frac{ 1 }{ n ^{3} }(1^{2}+2^{2}+3^{2}+4^{2}+...+n ^{2})$ Is that correct? I know I have to take the limit of it eventually, but I know it simplifies more. That's what I'm unsure about.

3. ByteMe

|dw:1353405044723:dw|

4. nathanruff

Okay. I understand that.

5. ByteMe

$$\large f(1+\frac{i-1}{n})=(1+\frac{i-1}{n})^2$$ so the area of each rectangle is: $$\large A_i=(1+\frac{i-1}{n})^2 \cdot \frac{1}{n}$$

6. nathanruff

That matches with what I have. I'm just not sure if what I did next was right. Would you simplify that, or expand it (and distribute)?

7. ByteMe

yeah... sorry, i'm more visual so i had to draw that picture for me to see what i was doing...

8. ByteMe

yes.. expand that an add up all the $$\large A_i$$ from 1 to n then take the limit as n approaches infinity...

9. ByteMe

yuck... looks like you'll have to expand $$\large (n + i - 1)^2$$

10. ByteMe

lol... with a little help... http://www.wolframalpha.com/input/?i=expand+%28n%2Bi-1%29%5E2

11. ByteMe

whoa.. wait a minute... wolfram thinks i is the imaginary unit....

12. ByteMe

this is better... the index "i" is the "x" here: http://www.wolframalpha.com/input/?i=expand+%28n%2Bx-1%29%5E2

13. ByteMe

$$\large \lim_{n\rightarrow \infty} \sum_{i=1}^{n}A_i$$

14. nathanruff

I sort of understand what you're doing. The example in my textbook (Calculus 7E, Stewart) uses the formula for the sum of squares in simplifying this when taking my approach: $A= \frac{ n }{ n } + \frac{ 2 }{ n ^{2} } (1+2+3+4+...+n) + \frac{ 1 }{ n ^{3} }(1^{2}+2^{2}+3^{2}+4^{2}+...+n ^{2})$ Where.. $(1^{2}+2^{2}+3^{2}+4^{2}+...+n ^{2}) = \frac{ n(n+1)(2n+1) }{ 6 }$ Is there a way to simplify this? $(1+2+3+4+...+n)$

15. ByteMe

yes... that's n(n+1)/2

16. nathanruff

How did you do that? My textbook doesn't explain anything.

17. ByteMe

let the sum from 1 to n be x, and also, write it backwards, it's still x... |dw:1353407364634:dw|

18. ByteMe

19. ByteMe

so x, the sum, is n(n+1)/2.....

20. ByteMe

Stewart Calculus 7E ??? is this for AP calculus?

21. nathanruff

I'm taking Calculus I at a community college and my professor is supposedly the toughest calc instructor on campus. We can't use our calculators on any of the tests, there's no margin for error with him, and we have to memorize identities, theorems, and definitions.

22. ByteMe

sorry man.... I have to take off... btw... that's what i used for AP calculus last year.... got a 1... :(