## twirlere Group Title y=x^2+1, [0,3] Use the limit process to find the area of the region between the graph of the function and the x-axis over the given interval. one year ago one year ago

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1. abb0t Group Title

First step is to find to find the right end points, ∆x, which is defined as: $$\Delta x = \frac{ b-a }{ n }$$

2. twirlere Group Title

3/n

3. abb0t Group Title

I'll leave it to you to verify your subintervals,, but as you can see,, the right endpoint on the $$i^{th}$$ interval is going to be: $$\frac{ 3i }{ n }$$.

4. abb0t Group Title

And the summation in the definition of the definite integral is defined as: $\sum_{i=1}^{n}f(x_i)\Delta x$

5. abb0t Group Title

In your case, you have: $\sum_{i=1}^{n}f \left( \frac{ 3i }{ n } \right)\left( \frac{ 3 }{ n } \right)$

6. abb0t Group Title

Are you with me so far? You're plugging in your right end point for "x". Hence: $\sum_{i=1}^{n}\left[ \left( \frac{ 2i }{ n } \right)^2+1 \right]\left( \frac{ 2 }{ n } \right)$ Remember: $$f(x_i)\Delta x$$

7. twirlere Group Title

okay, I'm understanding...

8. abb0t Group Title

The rest is easy algebra that you should be able to do. Combine like terms and distribute. I'll leave that to you to work on.

9. abb0t Group Title

Now, remember that the definition is defined as: $\lim_{n \rightarrow \infty}\sum_{i=1}^{n}f(x_i)\Delta x$ So you're going to be taking the limit of this!

10. abb0t Group Title

In other words, you're going to have to use some of the formulas you used in algebra of summation notation to eliminate the actual summation and get a formular in terms of just "n". Because that's what you're taking the limit. You cannot have 'i' in the equation. Also, NOTE: "n" is a constant in this care. The only thing that changes in this is "i". So you can factor OUT "n".

11. abb0t Group Title

Here are the general formulas that you SHOULD know (sidenote: these are only true if i=1) $\sum_{n}^{i=1}c = cn$ $\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }$ $\sum_{i=1}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }$ $\sum_{i=1}^{n}i^3=\left[ \frac{ n(n+1) }{ 2 } \right]^2$

12. abb0t Group Title

Are you still with me?

13. twirlere Group Title

yes

14. abb0t Group Title

I don't want to do all of yor work, especially not the algebra portion. So I'm going to ask, what did you get as your fuction? The one earlier: $$f(x_i) \Delta x$$? After combining the fraction into one single and distributing :)

15. twirlere Group Title

((9i^2/n^2)+1)(3/n)

16. abb0t Group Title

I hope you used $[(\frac{ 3i }{ n })^2+1](\frac{ 3 }{ n })$ because i i mistypoed a 2

17. satellite73 Group Title

really makes you appreciate the fundamental theorem of calculus, doesn't it?

18. twirlere Group Title

that's what I used... and no :( lol

19. satellite73 Group Title

no? compare with $F(x)=\frac{x^3}{3}+x$ and then $$F(3)-F(0)$$

20. abb0t Group Title

I think that yourr result might be a tad bit off: $$\left[\frac{ 9i^2 }{ n^2 } +1\right] \frac{ 3 }{ n }$$

21. abb0t Group Title

Oh wait, nevermid, you were right. Now distribute. Haha.

22. abb0t Group Title

Also, remember the sum rules from I think algebra or pre-calculus? Where you can separate the sum: $\sum_{i=1}^{n}(a \pm b) = \sum_{i=1}^{n}a \pm \sum_{i=1}^{n}b$

23. twirlere Group Title

(27i^2/n^3)+(3/n)

24. abb0t Group Title

Great. So now, you can separete them! Good job. Now, remember what I told you, how n is a constant, and the only thing changing is "i"? Well, You can actually factor out the constant, n, in this case. Use the properties above:$\frac{ 27 }{ n^3 } \sum_{i=1}^{n}i^2 + \frac{ 3 }{ n } \sum_{i=1}^{n}$ Use the properties i provided above to eliminate "i" and put it in terms of "n". Since you have $$i^2$$ you'll be using the third formula, and for the second, use the first formula.

25. abb0t Group Title

Changing them you will get: $$\frac{ 27 }{ n^3 }\left( \frac{ n(n+1)(2n+1) }{ 6 } \right) + \frac{ 3 }{ n }(n)$$. Can you do that? Cross cancel out the terms?

26. twirlere Group Title

$(9(n+1)(2n+1)/2n^2) + 3$

27. abb0t Group Title

Good job! Now, I think it might be easier to take the limit of this function if you combined it all into one (at least for me it's easier to see what cancels out). But all that's left to do is simply take the limit of that function! You're final result is your answer.

28. abb0t Group Title

Combined it all into one fraction*

29. twirlere Group Title

umm... I have trouble finding the limit

30. abb0t Group Title

Well, first off, did we get the same fraction, I did it fairly quick so I could be wrong so let's check that: $$\frac{ 21n^2+27n+9 }{ 2n^2 }$$ are we on the same page?

31. amistre64 Group Title

$\Large \frac{27}{n^3}~{\frac{n(n+1)(2n+1)}{6}}+3$ $\Large {\frac{9n(n+1)(2n+1)+6n^3}{2n^3}}$ $\Large {\frac{9n(2n^2+3n+1)+6n^3}{2n^3}}$ $\Large {\frac{9(2+\cancel{\frac 3n+\frac1n)}^0+6}{2}}$ $\Large 9+3$

32. amistre64 Group Title

unless a typed it wrong of course .... latex is nice but consumes alot of brain power

33. abb0t Group Title

@amistre64 how did you do the "slash" across? :P

34. amistre64 Group Title

lol,  \cancel{arg} 

35. amistre64 Group Title

and i put an ^0 on it to indicate it canceld

36. amistre64 Group Title

theres also an alternative cancel that goes against the other diagonal, /bcancel i think ... never ued it tho :)

37. abb0t Group Title

awe snaps!! this just got real!! Thanks :)

38. abb0t Group Title

Do you have a latex site for these codes? I have one but it's so basic. Lol.

39. amistre64 Group Title

whenever a see a neat latex written up on here, i right click it and "show math as tex commands" thatll copy the coding without the wrappers to inspect

40. amistre64 Group Title

try the Latex Practicing subject under "find more subjects" the math code they use here is not the complete setup, but it is pretty impressive nonetheless

41. abb0t Group Title

NEAT!