Question

Some Calculus Integration
Help Please!!

Answer 1

@hartnn can you help?

Answer 2

@freckles can you help?

Answer 3

I have an answer, but I just need confirmation.

Answer 4

@mathmale can you help please?

Answer 5

My answer for the left-hand approximation is 2,115.

Answer 6

I'll help, subject to your sharing all that you already know about the first problem. I'm sure you understand some of it already.

Which rule for approx. areas under curves do you wish to use? Because the problem specifies use of "left endpoints," we won't consider the Trapezoidal or Simpson Rule(s).

Answer 7

What is the interval width?
Which x-value comes first, and how many x-values will we actually need?
What's the height of the first approximating rectangle?

Answer 8

The interval width is 15

Answer 9

0 comes first and the height of the first rectangle 0

Answer 10

@mathmale

Answer 11

Yes. The first x-value, for left endpoints, is 0, and the corresponding y-value, which is the height of the approx. rectangle is 0. Good.
and yes, the interval width is 15. What's a good name to give to this 15? Review example problems for this info.

Answer 12

a good name would be "n" right?

Answer 13

n = 15?

Answer 14

No. n represents the # of approx. rectangles.
My question about the interval width should give you a strong clue. Try again. 15 is represented by the symbol ... ?

Answer 15

Delta x

Answer 16

Perfect.

How many approx rect. will there be? How many y values are given you in the table?

Answer 17

Hint: Tricky question.

Answer 18

Okay since this is left-hand approximation i wouldn't use the last one so there are 8

Answer 19

V ery good. Now, what's the area of the first approx. rect? Show your work.
Also, what's that of the 2nd?

Answer 20

(height of first rect.)(width of 1st rect) = (area of first rect) = ?
area of 2nd rect = ?

Answer 21

The first is 0 because base = 15 and the height =0

Answer 22

Right.
2nd?

Answer 23

the second is 15*15 so it would be 225

Answer 24

Right.
But...let's reread the instructions. They specify that n=4, right? Not 8.

How will this affect your calculation of the area of the first rect?
...of the second rect?

Answer 25

Would i multiply it by 4?

Answer 26

No. You could do a really quick sketch showing that there are only 4 approx. rects., not 8.
Then, again, calc the area of the 1st and 2nd rect. Hint: How wide are the approx. rects. if n=4 instead of n=8?

Answer 27

Oh okay the base would now be 30 so the second rectangle would now be 450

Answer 28

right. first is zero times 30; second is 18 times 30. Agree with the latter calculation?

Answer 29

Height of 2nd rec. is 18. Width is 30. Area is .... ?

Answer 30

yes so the second rectangle would be 540 i realized my mistake after i drew the picture

Answer 31

Now please find the areas of the 3rd and 4th approximating rect.

Answer 32

I won't be checking your work, so be really careful.

Answer 33

okay

Answer 34

The are of the third would be 570 and the area of the fourth would be 720

Answer 35

Now I add up all of the areas and that is the total approximation?

Answer 36

I'm going to take your word for that.
How would you now come up with a single figure that represents the approx area under this curve when n=4? Explain.

Answer 37

0 + ? + ? + ? = ?

Answer 38

1830

Answer 39

comfortable with this approach? I believe we've nailed it (but as before, I'm not checking your calculations).

Answer 40

Okay i can do the rest thanks except canyou help with question 2

Answer 41

Unless you have questions about this, I'd like to move on to using the Trap Rule. Use of the right-hand rule follows precisely the same steps as we used for the left-hand rule, except that you begin x at 12 and work backward. n=4, so the interval width delta x is still 30. first approx rect. has a height of 120. OK with that?

Answer 42

Music?

Answer 43

Sorry

Answer 44

Yah so for the trapezoid rule the equation i use for the area is \[A = b *\frac{ h _{1} +h _{2} }{ 2 }\]

Answer 45

that's for a single trapez. The Trap Rule is based upon that but is quite different otherwise. Do you have in front of you a reference that shows the formula for the Trap Rule?

Answer 46

Yes it is T = (f(a) + 2f(x1) + 2f(x2)...+2(f(xn-1) + f(b)

Answer 47

That's part of the deal, but there's more to the deal than that.

Answer 48

\[A=\frac{ \delta~x }{ 2}[(your~ expression)\]

Answer 49

Where the interval [a,b] is divided into "n" equal sub-intervals [xi-1,xi]

Answer 50

Please note carefully: n=4 here. You must use n+1, or 4+1, or 5, x values and their corresponding y-values. Agreed? Yes, you're right on dividing up the interval [0,12].

Answer 51

Let's actually identify the 5 necessary y-values. What are they?

List them here, please:

Answer 52

Oh and in the beginning of the equation there is a \[\frac{ b-a }{ 2n }\]

Answer 53

the five necessary y-values are 0,18,19,24,12

Answer 54

b-a/2n = 15

Answer 55

Regarding (b-a)/n: That's delta x.

Regardng (b-a)/(2n): Multiply your 5-term expression by that.

Answer 56

In the end, your approach and mine should deliver precisely the same result.

Answer 57

Okay i will calculate my answer now

Answer 58

Wait please

Answer 59

oh okay

Answer 60

I've checked out the y-values and find that I have a dramatically different response. I claim that the set of y values is {0, 30, 60, 90, 120}.

Answer 61

Please accept that (after checking) or argue why this is not correct. ;)

Answer 62

Those are x-values though right?

Answer 63

Just kidding

Answer 64

Nope. You won't even need the x-values. Only the y-values. Really. Not kidding. ;)

Answer 65

Okay sothose are the values i use when calculating the traprule

Answer 66

I dont have and equation to do the trap rule though, or do i just use the given base and heights

Answer 67

Yes.

One more thing. You have the coefficients right in your expression: They are:

1, 2, 2, 2, 1.

compare this to your result...See what I mean?

Answer 68

delta x= (b-a)/n, yes.

The desired approx. area, calc. with n=4, is

\[A=\frac{ \delta~x }{ 2 }[1*f(x _{0})+2*f(x _{1})+2*f(x _{2}) + (2 ~more~ terms).\]

Answer 69

Type in the last 2 terms, please./

Answer 70

\[2*f(x _{3}) + f(120)\]

Answer 71

Good.

Answer 72

What is delta x in this case, given that n=4?

Answer 73

15

Answer 74

120/2*4

Answer 75

(b-a)/2 = delta x
Is (12-0)/4 equal to 15?

Answer 76

i thought we were using the [a,b] from the integral

Answer 77

I'm willing to go along with whatever format you wish to use.

Your format (b-a)/(2n) is different from mine, but is usable.

I would begin my expression for Area as follows:

\[A = \frac{ \delta~x }{ 2 }\]

Answer 78

okay so delta x would be 3 then

Answer 79

whereas you're beginning yours with

Answer 80

\[A =\frac{ (b-a) }{ 2n }\]

Answer 81

Go ahead and use whichever form you prefer.

Answer 82

Yes, I figure that delta x is 3, and thus I would start out my expression for A as \[A=\frac{ 3 }{ 2 }[f(x _{0}) + .... \]

Answer 83

okay

Answer 84

I was wrong earlier: we do indeed need to use 5 x-values, which are

{0, 3, 6, 9, 12}. Do you agree with these values?

Answer 85

What i was doing was using the a and b values from the integral making a=0 and b=120

Answer 86

yes i agree with those values

Answer 87

Caution: Don't do that. [a,b] refers to [0,12], and not the y-values at all.

Answer 88

We are approx. an integral integrated from x=0 to x=12, not to x=120.

Answer 89

f(x) = y right?

Answer 90

Yes, y=f(x).

But in this approximation problem, we identify and use 5 specific values for y.

Answer 91

Nothing to calculate, since all of the x and y values are given you in a table.

Answer 92

So if f(x) = y then why are we using the f(x) values for the interval from [a,b]

Answer 93

Recall: x is the independent variable, and here ranges from 0 to 12; y is the dependent, or function, value, and ranges from 0 to 120.

Does that answer your question?

Answer 94

\[A=\frac{ \delta~x }{ 2 }[1*f(x _{0})+2*f(x _{1})+2*f(x _{2}) + (2 ~more~ terms).\]

Answer 95

Why is y the one that is labeled x in the table?

Answer 96

\[A=\frac{ 3 }{ 2 }[1*0+2*30)+2*f(x _{2}) + (2 ~more~ terms).\]