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.

- anonymous

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- abb0t

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

- anonymous

3/n

- 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: \(\frac{ 3i }{ n }\).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- abb0t

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

- abb0t

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

- 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

okay, I'm understanding...

- 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

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

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

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

Are you still with me?

- anonymous

yes

- 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

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

- abb0t

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

- anonymous

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

- anonymous

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

- anonymous

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

- abb0t

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

- abb0t

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

- 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

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

- 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

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

\[(9(n+1)(2n+1)/2n^2) + 3\]

- 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

Combined it all into one fraction*

- anonymous

umm... I have trouble finding the limit

- 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

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

- amistre64

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

- abb0t

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

- amistre64

lol,
```
\cancel{arg}
```

- amistre64

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

- amistre64

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

- abb0t

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

- abb0t

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

- 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

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

NEAT!

Looking for something else?

Not the answer you are looking for? Search for more explanations.