anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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abb0t
  • abb0t
First step is to find to find the right end points, ∆x, which is defined as: \(\Delta x = \frac{ b-a }{ n }\)
anonymous
  • anonymous
3/n
abb0t
  • abb0t
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: \(\frac{ 3i }{ n }\).

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abb0t
  • abb0t
And the summation in the definition of the definite integral is defined as: \[\sum_{i=1}^{n}f(x_i)\Delta x\]
abb0t
  • abb0t
In your case, you have: \[\sum_{i=1}^{n}f \left( \frac{ 3i }{ n } \right)\left( \frac{ 3 }{ n } \right)\]
abb0t
  • abb0t
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: \(f(x_i)\Delta x\)
anonymous
  • anonymous
okay, I'm understanding...
abb0t
  • abb0t
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.
abb0t
  • abb0t
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!
abb0t
  • abb0t
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".
abb0t
  • abb0t
Here are the general formulas that you SHOULD know (sidenote: these are only true if i=1) \[\sum_{n}^{i=1}c = cn\] \[\sum_{i=1}^{n}i=\frac{ n(n+1) }{ 2 }\] \[\sum_{i=1}^{n}i^2=\frac{ n(n+1)(2n+1) }{ 6 }\] \[\sum_{i=1}^{n}i^3=\left[ \frac{ n(n+1) }{ 2 } \right]^2\]
abb0t
  • abb0t
Are you still with me?
anonymous
  • anonymous
yes
abb0t
  • abb0t
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 :)
anonymous
  • anonymous
((9i^2/n^2)+1)(3/n)
abb0t
  • abb0t
I hope you used \[[(\frac{ 3i }{ n })^2+1](\frac{ 3 }{ n })\] because i i mistypoed a 2
anonymous
  • anonymous
really makes you appreciate the fundamental theorem of calculus, doesn't it?
anonymous
  • anonymous
that's what I used... and no :( lol
anonymous
  • anonymous
no? compare with \[F(x)=\frac{x^3}{3}+x\] and then \(F(3)-F(0)\)
abb0t
  • abb0t
I think that yourr result might be a tad bit off: \(\left[\frac{ 9i^2 }{ n^2 } +1\right] \frac{ 3 }{ n }\)
abb0t
  • abb0t
Oh wait, nevermid, you were right. Now distribute. Haha.
abb0t
  • abb0t
Also, remember the sum rules from I think algebra or pre-calculus? Where you can separate the sum: \[\sum_{i=1}^{n}(a \pm b) = \sum_{i=1}^{n}a \pm \sum_{i=1}^{n}b\]
anonymous
  • anonymous
(27i^2/n^3)+(3/n)
abb0t
  • abb0t
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.
abb0t
  • abb0t
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?
anonymous
  • anonymous
\[(9(n+1)(2n+1)/2n^2) + 3\]
abb0t
  • abb0t
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.
abb0t
  • abb0t
Combined it all into one fraction*
anonymous
  • anonymous
umm... I have trouble finding the limit
abb0t
  • abb0t
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?
amistre64
  • amistre64
\[\Large \frac{27}{n^3}~{\frac{n(n+1)(2n+1)}{6}}+3\] \[\Large {\frac{9n(n+1)(2n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9n(2n^2+3n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9(2+\cancel{\frac 3n+\frac1n)}^0+6}{2}}\] \[\Large 9+3\]
amistre64
  • amistre64
unless a typed it wrong of course .... latex is nice but consumes alot of brain power
abb0t
  • abb0t
@amistre64 how did you do the "slash" across? :P
amistre64
  • amistre64
lol, ``` \cancel{arg} ```
amistre64
  • amistre64
and i put an ^0 on it to indicate it canceld
amistre64
  • amistre64
theres also an alternative cancel that goes against the other diagonal, /bcancel i think ... never ued it tho :)
abb0t
  • abb0t
awe snaps!! this just got real!! Thanks :)
abb0t
  • abb0t
Do you have a latex site for these codes? I have one but it's so basic. Lol.
amistre64
  • amistre64
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
amistre64
  • amistre64
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
abb0t
  • abb0t
NEAT!

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