- #1

- 1,753

- 143

Code:

```
x 20 25 30 35 40 45 50
f(x) 42 38 31 29 35 48 60
```

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- Thread starter tony873004
- Start date

- #1

- 1,753

- 143

Code:

```
x 20 25 30 35 40 45 50
f(x) 42 38 31 29 35 48 60
```

- #2

- 1

- 0

Dont forget that to find this it is the intergral divided by the interval.

you also have your midpoint formula wrongly set up

delta x = (b - a)/n

so (50 - 20)/6

then you find the midpoints between the 7 f(x) values, note you used the x values in this step instead of the f(x) values

Midpoint sum = (50 -20)/60 * (.5(42 + 38) + .5(38 + 31) ...)

Now that you have the midpoint sum, or in other words the intergral of the function, divide that by the interval of 30. The anwser of 38.3333 is correct.

- #3

- 1,753

- 143

Your 1st post... Welcome!

Thanks for the explanation.

(b-a)/n is what I have, but I used 7 and you used 6. I guess that's because they're being paired up, so its only 6 pairings?

But then you use (50-20)/60 instead of (50-20)/6. Where did the 60 come from?

When I punch this into the calculator, I get

(50-20)/60 * (.5*(42+38)+.5*(38+31)+.5*(31+29)+.5*(29+35)+.5*(35+48)+.5*(48+60))

which equals 116.

Edit ***

I forgot to divide by 30. If I use (50-20)/6 instead of 60, and divide by 30 I get 38.66667, instead of 38.33333. I'm guessing the 60 was a typo? I'm still off by 1/3

Thanks for the explanation.

(b-a)/n is what I have, but I used 7 and you used 6. I guess that's because they're being paired up, so its only 6 pairings?

But then you use (50-20)/60 instead of (50-20)/6. Where did the 60 come from?

When I punch this into the calculator, I get

(50-20)/60 * (.5*(42+38)+.5*(38+31)+.5*(31+29)+.5*(29+35)+.5*(35+48)+.5*(48+60))

which equals 116.

Edit ***

I forgot to divide by 30. If I use (50-20)/6 instead of 60, and divide by 30 I get 38.66667, instead of 38.33333. I'm guessing the 60 was a typo? I'm still off by 1/3

Last edited:

- #4

- 1

- 0

The formula for the average value is

(1/b-a)http://upload.wikimedia.org/math/4/0/c/40c4f3fd4e90d666f43f275de6264a1e.png

and the midpoint rule is the same as the midpoint sum except that dt is factored out of the equation.

dt = (b-a/n)

midpont sum = f(t_{0})dt +...f(t_{n})dt and with the

average value = (1/b-a)(midpont sum)

(1/b-a)http://upload.wikimedia.org/math/4/0/c/40c4f3fd4e90d666f43f275de6264a1e.png

and the midpoint rule is the same as the midpoint sum except that dt is factored out of the equation.

dt = (b-a/n)

midpont sum = f(t

average value = (1/b-a)(midpont sum)

Last edited:

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