Here's the question you clicked on:
twirlere
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.
First step is to find to find the right end points, ∆x, which is defined as: \(\Delta x = \frac{ b-a }{ n }\)
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: \(\frac{ 3i }{ n }\).
And the summation in the definition of the definite integral is defined as: \[\sum_{i=1}^{n}f(x_i)\Delta x\]
In your case, you have: \[\sum_{i=1}^{n}f \left( \frac{ 3i }{ n } \right)\left( \frac{ 3 }{ n } \right)\]
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: \(f(x_i)\Delta x\)
okay, I'm understanding...
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.
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!
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".
Here are the general formulas that you SHOULD know (sidenote: these are only true if i=1) \[\sum_{n}^{i=1}c = cn\] \[\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }\] \[\sum_{i=1}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }\] \[\sum_{i=1}^{n}i^3=\left[ \frac{ n(n+1) }{ 2 } \right]^2\]
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 :)
I hope you used \[[(\frac{ 3i }{ n })^2+1](\frac{ 3 }{ n })\] because i i mistypoed a 2
really makes you appreciate the fundamental theorem of calculus, doesn't it?
that's what I used... and no :( lol
no? compare with \[F(x)=\frac{x^3}{3}+x\] and then \(F(3)-F(0)\)
I think that yourr result might be a tad bit off: \(\left[\frac{ 9i^2 }{ n^2 } +1\right] \frac{ 3 }{ n }\)
Oh wait, nevermid, you were right. Now distribute. Haha.
Also, remember the sum rules from I think algebra or pre-calculus? Where you can separate the sum: \[\sum_{i=1}^{n}(a \pm b) = \sum_{i=1}^{n}a \pm \sum_{i=1}^{n}b\]
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.
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?
\[(9(n+1)(2n+1)/2n^2) + 3\]
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.
Combined it all into one fraction*
umm... I have trouble finding the limit
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?
\[\Large \frac{27}{n^3}~{\frac{n(n+1)(2n+1)}{6}}+3\] \[\Large {\frac{9n(n+1)(2n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9n(2n^2+3n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9(2+\cancel{\frac 3n+\frac1n)}^0+6}{2}}\] \[\Large 9+3\]
unless a typed it wrong of course .... latex is nice but consumes alot of brain power
@amistre64 how did you do the "slash" across? :P
lol, ``` \cancel{arg} ```
and i put an ^0 on it to indicate it canceld
theres also an alternative cancel that goes against the other diagonal, /bcancel i think ... never ued it tho :)
awe snaps!! this just got real!! Thanks :)
Do you have a latex site for these codes? I have one but it's so basic. Lol.
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
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