## nathanruff Group Title Find the exact area bounded by f(x)=x^2 and the x-axis between x=1 and x=2 using n partitions. one year ago one year ago

1. ByteMe Group Title

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 Group Title

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 Group Title

|dw:1353405044723:dw|

4. nathanruff Group Title

Okay. I understand that.

5. ByteMe Group Title

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

6. nathanruff Group Title

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 Group Title

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

8. ByteMe Group Title

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 Group Title

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

10. ByteMe Group Title

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

11. ByteMe Group Title

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

12. ByteMe Group Title

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

13. ByteMe Group Title

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

14. nathanruff Group Title

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 Group Title

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

16. nathanruff Group Title

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

17. ByteMe Group Title

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

18. ByteMe Group Title