anonymous
  • anonymous
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = –x^3 and y = –x.
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@terenzreignz Could you help me with this?
terenzreignz
  • terenzreignz
Okay, now, we need to have visuals of this graph...
terenzreignz
  • terenzreignz
Wait. Annoying haha

Looking for something else?

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

More answers

terenzreignz
  • terenzreignz
Tricky. There are two regions.
anonymous
  • anonymous
Yes, two congruent/symmetric regions
anonymous
  • anonymous
We could simply find the area of one and multiply by two
terenzreignz
  • terenzreignz
Are we to approximate both regions?
1 Attachment
anonymous
  • anonymous
Yes
terenzreignz
  • terenzreignz
Well, if you say so. In that case, we integrate from 0 to 1.
terenzreignz
  • terenzreignz
We'd integrate \[\Large \int\limits_0^1 x-x^3 dx \]
terenzreignz
  • terenzreignz
But only approximate. And n = 4 So you need to divide the interval [0,1] into four parts.
anonymous
  • anonymous
Would it be smarter to do them separately?
anonymous
  • anonymous
I see it faster to do one.
anonymous
  • anonymous
I have already setup the integrals
anonymous
  • anonymous
Just wasn't sure what the midpoint rule was.
anonymous
  • anonymous
If it means to divide it into 4 intervals then I should be fine.
terenzreignz
  • terenzreignz
Four equal intervals. [0 , 0.25] [0.25 , 0.5] [0.5 , 0.75] [0.75 , 1] Get the midpoints of these intervals.
anonymous
  • anonymous
0.125 .375 .625 .875
terenzreignz
  • terenzreignz
Right. Now evaluate the function \(\large x - x^3\) at those values, add up the results, and you get your approximated area for the region.
anonymous
  • anonymous
Then multiply by two
terenzreignz
  • terenzreignz
Yes :D So... everything done, then? ^^
anonymous
  • anonymous
Let me finalize the results
anonymous
  • anonymous
I got 1.03125
anonymous
  • anonymous
Which I should multiply by 2?
anonymous
  • anonymous
2.0625 which is very inaccurate or wrong
anonymous
  • anonymous
Since the actual area is .5
terenzreignz
  • terenzreignz
well that's why we have integrals
anonymous
  • anonymous
So using that method it's right.
terenzreignz
  • terenzreignz
I should think so.
terenzreignz
  • terenzreignz
It will become more accurate the larger the value of n. And at n = infinity, you get the integral
anonymous
  • anonymous
8/8
terenzreignz
  • terenzreignz
Wait... never mind... I got the problem...
terenzreignz
  • terenzreignz
Sorry, I oversimplified.
anonymous
  • anonymous
not 8/8?
terenzreignz
  • terenzreignz
We don't just simply get the function \(x - x^3\) at the midpoints 0.125 0.375 0.625 0.875 We also multiply the entire sum by 0.25, which is the length of a single partition I'm sorry ^^ I literally haven't done this sort of primitive integration in YEARS
terenzreignz
  • terenzreignz
After all, we're getting areas of the rectangular strips... and we've only been taking the length. We need to multiply them by the width, which happens to be 0.25
terenzreignz
  • terenzreignz
So we're actually taking the sum of these values: \[\Large \sum_{n=1}^4 f(x_n)\Delta x_n\]
terenzreignz
  • terenzreignz
Where \(\large x_n = 0.25n - 0.125\)
terenzreignz
  • terenzreignz
Anyway, see how that sum looks like an integral somewhat? haha
anonymous
  • anonymous
Yup
anonymous
  • anonymous
I should've seen that mistake honestly it was simply h*w and I didn't notice you forgot w so my bad
terenzreignz
  • terenzreignz
oh well... I'm still super-awesome HAHA :>
anonymous
  • anonymous
Yes, yes you are

Looking for something else?

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