Question

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = –x^3 and y = –x.

Answer 1

@terenzreignz Could you help me with this?

Answer 2

Okay, now, we need to have visuals of this graph...

Answer 3

Wait. Annoying haha

Answer 4

Tricky. There are two regions.

Answer 5

Yes, two congruent/symmetric regions

Answer 6

We could simply find the area of one and multiply by two

Answer 7

Are we to approximate both regions?

Answer 8

Yes

Answer 9

Well, if you say so.
In that case, we integrate from 0 to 1.

Answer 10

We'd integrate \[\Large \int\limits_0^1 x-x^3 dx \]

Answer 11

But only approximate.
And n = 4
So you need to divide the interval [0,1]
into four parts.

Answer 12

Would it be smarter to do them separately?

Answer 13

I see it faster to do one.

Answer 14

I have already setup the integrals

Answer 15

Just wasn't sure what the midpoint rule was.

Answer 16

If it means to divide it into 4 intervals then I should be fine.

Answer 17

Four equal intervals.
[0 , 0.25]
[0.25 , 0.5]
[0.5 , 0.75]
[0.75 , 1]

Get the midpoints of these intervals.

Answer 18

0.125
.375
.625
.875

Answer 19

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.

Answer 20

Then multiply by two

Answer 21

Yes :D
So... everything done, then? ^^

Answer 22

Let me finalize the results

Answer 23

I got 1.03125

Answer 24

Which I should multiply by 2?

Answer 25

2.0625 which is very inaccurate or wrong

Answer 26

Since the actual area is .5

Answer 27

well that's why we have integrals

Answer 28

So using that method it's right.

Answer 29

I should think so.

Answer 30

It will become more accurate the larger the value of n.
And at n = infinity, you get the integral

Answer 31

8/8

Answer 32

Wait... never mind... I got the problem...

Answer 33

Sorry, I oversimplified.

Answer 34

not 8/8?

Answer 35

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

Answer 36

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

Answer 37

So we're actually taking the sum of these values:
\[\Large \sum_{n=1}^4 f(x_n)\Delta x_n\]

Answer 38

Where \(\large x_n = 0.25n - 0.125\)

Answer 39

Anyway, see how that sum looks like an integral somewhat? haha

Answer 40

Yup

Answer 41

I should've seen that mistake honestly it was simply h*w and I didn't notice you forgot w so my bad

Answer 42

oh well...
I'm still super-awesome
HAHA
:>

Answer 43

Yes, yes you are