Question

Use the form of the definition of the integral as the limit of a partial sum AND the formulas for the sum of powers of positive integers to evaluate the following integrals

Answer 1

This problem is from Calculus I

Answer 2

\[\large{\int_{-2}^{4}(x^2-6)dx}\]

\[\Delta x=\frac{4-(-2)}{n}=\frac{6}{n}\]

\[\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\Delta x=\lim_{n\to\infty}\sum_{i=1}^{n}\left(\left(-2+i\frac{6}{n}\right)^2-6\right)\frac{6}{n}\]




\[=\lim_{n\to\infty}\sum_{i=1}^{n}\left(4-\frac{24}{n}i+i^2\frac{36}{n^2}-6\right)\frac{6}{n}\]


\[=\lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{24}{n}-\frac{144}{n^2}i+i^2\frac{216}{n^3}-\frac{36}{n}\right)\]

\[=\lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{-12}{n}-\frac{144}{n^2}i+i^2\frac{216}{n^3}\right)\]


\[=\lim_{n\to\infty}\left(\frac{-12}{n}\sum_{i=1}^{n}1-\frac{144}{n^2}\sum_{i=1}^{n}i+\frac{216}{n^3}\sum_{i=1}^{n}i^2\right)\]

\[=\lim_{n\to\infty}\left(\frac{-12}{n}n-\frac{144}{n^2}\frac{n(n+1)}{2}+\frac{216}{n^3}\frac{n(n+1)(2n+1)}{6}\right)\]

\[=-12-\frac{144}{2}+\frac{216\times2}{6}=-12\]

part b is very similar

Answer 3

I hope u found your answer!

Answer 4

CAN I HAVE A MEDAL?! PLEASE?!!!!!!!!!!

Answer 5

So you're trying to understand this i business?

If we busted up this function into rectangles,
the `height` of the first rectangle would be \(\large\rm (-2)^2-6\) right?
(If we're using left end-points).

If instead we're using right end points,
our first rectangle height could be written as \(\large\rm (-2+1\Delta x)^2-6\)

The width of the rectangle is \(\large\rm \Delta x\), so we've moved from -2 to the next x-location \(\large\rm -2+\Delta x\) to get the right edge for the height.

Answer 6

Ok. I kind of get that...

Answer 7

So the height of our second rectangle would be \(\large\rm (-2+2\Delta x)^2-6\)

You'll probably see where the i is coming from, with the way the are counting up, if you're able to understand that first post I made.

Answer 8

Wait, is this going to help me understand where \(i\) comes from? lol

Answer 9

From current knowledge: I understand that in \(\Sigma\) notation, \(i\) represents "index" or the starting point in terms of sub-intervals or rectangles.

Answer 10

The area of the first rectangle will be the height \(\large\rm (-2+\Delta x)^2-6\) times the width,

\(\large\rm A_1=[(-2+\Delta x)^2-6]\Delta x\)

Answer 11

Well if you can understand what EACH rectangle looks like,
it will make the summation a bit more clear. :)

Answer 12

\[\sum_{i=a}^{n}\]\[a=\text{starting sub-interval/rectangle value}\]

Answer 13

I just don't see how it applies into the actual \(f(x_{i}*)\)?

Answer 14

Or is it because it is \(\large{f(x_{i}*)}\) and not \(\large{f(x_{i})}\)?

Answer 15

So ummm... xi* means we can choose any point in the interval.
Since we're taking a limit, it won't affect the outcome in any way.

Whether we use right end points, left end points, mid points,
it doesn't matter, all turns out the same in the end.

So I was using right end points (because that's how Zarkon set up the summation (at least that's how I'm reading it).

Answer 16

Okay, I see what you mean... but why does \(i\) show up in the summation?

Answer 17

Be the turtle!!
Slow and steady wins the race! lol

One sec, we'll get to that XD

So here is the area of our first rectangle,
\(\large\rm A_1=[(-2+\Delta x)^2-6]\Delta x\)

To get to our next height, we travel another width to the right,
\(\large\rm A_2=[(-2+2\Delta x)^2-6]\Delta x\)

Answer 18

Here is the area of our third rectangle,
\(\large\rm A_3=[(-2+3\Delta x)^2-6]\Delta x\)

Answer 19

Sorry, mom is rushing me to go to bed but I want to understand this problem >_<

Answer 20

Here is the area of our ith rectangle,
\(\large\rm A_i=[(-2+i\Delta x)^2-6]\Delta x\)

Answer 21

STOP WITH THE CUSTOM TEXT :(
It's so tiny and difficult to read :((

Answer 22

There is a bunch of stuff I'm really hating about this site... ugh

Answer 23

grumble grumble <.<

Answer 24

Maybe your zoom is too small? lol I can see it :o

Answer 25

Why are there rainbow lines between the posts?
This site is so stupid :((

Answer 26

Because Jay wanted it that way LOL.

Answer 27

Is Jay 7 years old?
This site is so weird...

Answer 28

He does what he wants. LOL

Answer 29

It feels like a 7 year old girl designed this site:
Dolphins... rainbows... custom emotes...

Answer 30

Dolphins are nice. >:O

Also, custom emotes was in his extension for OS as well.

Answer 31

I hate that more than anything :(
When I type ":(" I want to SEE ":("
not some ridiculous picture.

Answer 32

What's wrong with the emoji >:O

Answer 33

Maybe time to head to PeerAnswer :( I dunno... not digging this.

Answer 34

It's just like Facebook... well Facebook overdoes it but you see the similarity

Answer 35

OK, BACK TO THE QUESTION.

Answer 36

** I can tell Jay that he should include an option for on/off emoji if you would like that.

Answer 37

ANYWAY... what is that graph about?

Answer 38

Tell him to stop channeling his inner 7 year old girl :c

Answer 39

Zep, that's mean. >_>

Answer 40

Our ith rectangle.

Answer 41

Hmm, ok. I see O:

Answer 42

Ok, I need to sleep :<

I will see you some other time ? @Zepdrix

Answer 43

:p

Answer 44

Is that a yes? -_-

Answer 45

On OpenStudy ... yes.
Here? ... . . . we will see -_-

Answer 46

OpenStudy is a sinking ship.