A community for students.
Here's the question you clicked on:
 0 viewing
twirlere
 2 years ago
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 xaxis over the given interval.
twirlere
 2 years ago
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 xaxis over the given interval.

This Question is Open

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2First step is to find to find the right end points, ∆x, which is defined as: \(\Delta x = \frac{ ba }{ n }\)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2I'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: \(\frac{ 3i }{ n }\).

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2And the summation in the definition of the definite integral is defined as: \[\sum_{i=1}^{n}f(x_i)\Delta x\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2In your case, you have: \[\sum_{i=1}^{n}f \left( \frac{ 3i }{ n } \right)\left( \frac{ 3 }{ n } \right)\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Are 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: \(f(x_i)\Delta x\)

twirlere
 2 years ago
Best ResponseYou've already chosen the best response.0okay, I'm understanding...

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2The 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
 2 years ago
Best ResponseYou've already chosen the best response.2Now, 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
 2 years ago
Best ResponseYou've already chosen the best response.2In 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
 2 years ago
Best ResponseYou've already chosen the best response.2Here 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
 2 years ago
Best ResponseYou've already chosen the best response.2I 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 :)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2I hope you used \[[(\frac{ 3i }{ n })^2+1](\frac{ 3 }{ n })\] because i i mistypoed a 2

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1really makes you appreciate the fundamental theorem of calculus, doesn't it?

twirlere
 2 years ago
Best ResponseYou've already chosen the best response.0that's what I used... and no :( lol

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1no? compare with \[F(x)=\frac{x^3}{3}+x\] and then \(F(3)F(0)\)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2I think that yourr result might be a tad bit off: \(\left[\frac{ 9i^2 }{ n^2 } +1\right] \frac{ 3 }{ n }\)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Oh wait, nevermid, you were right. Now distribute. Haha.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Also, remember the sum rules from I think algebra or precalculus? Where you can separate the sum: \[\sum_{i=1}^{n}(a \pm b) = \sum_{i=1}^{n}a \pm \sum_{i=1}^{n}b\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Great. 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
 2 years ago
Best ResponseYou've already chosen the best response.2Changing 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?

twirlere
 2 years ago
Best ResponseYou've already chosen the best response.0\[(9(n+1)(2n+1)/2n^2) + 3\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Good 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
 2 years ago
Best ResponseYou've already chosen the best response.2Combined it all into one fraction*

twirlere
 2 years ago
Best ResponseYou've already chosen the best response.0umm... I have trouble finding the limit

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Well, 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
 2 years ago
Best ResponseYou've already chosen the best response.0\[\Large \frac{27}{n^3}~{\frac{n(n+1)(2n+1)}{6}}+3\] \[\Large {\frac{9n(n+1)(2n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9n(2n^2+3n+1)+6n^3}{2n^3}}\] \[\Large {\frac{9(2+\cancel{\frac 3n+\frac1n)}^0+6}{2}}\] \[\Large 9+3\]

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0unless a typed it wrong of course .... latex is nice but consumes alot of brain power

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2@amistre64 how did you do the "slash" across? :P

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0lol, ``` \cancel{arg} ```

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0and i put an ^0 on it to indicate it canceld

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0theres also an alternative cancel that goes against the other diagonal, /bcancel i think ... never ued it tho :)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2awe snaps!! this just got real!! Thanks :)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Do you have a latex site for these codes? I have one but it's so basic. Lol.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0whenever 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
 2 years ago
Best ResponseYou've already chosen the best response.0try 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.