The function is continuous on the interval [10, 20] with some of its values given in the table above. Estimate the average value of the function with a Trapezoidal Sum Approximation, using the intervals between those given points.
x 10 12 15 19 20
f(x) –2 –5 –9 –12 –16

- anonymous

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- jim_thompson5910

how far did you get?

- anonymous

Im stuck on how to start it

- anonymous

what trips me up is that it asks to use intervals between those values

Looking for something else?

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

## More answers

- jim_thompson5910

I would start it by plotting the points on an xy grid (see attached)

##### 1 Attachment

- anonymous

what would you do after plotting the points

- jim_thompson5910

then draw trapezoids

##### 1 Attachment

- jim_thompson5910

tell me what you get for the area of each trapezoid

- jim_thompson5910

|dw:1443395568574:dw|
area of trapezoid = h*(B1+B2)/2
notice how the B1 and B2 are parallel

- anonymous

17.5 for the first trapezoid

- jim_thompson5910

incorrect

- anonymous

7

- jim_thompson5910

yes for the one on the very left

- anonymous

21 for the second

- jim_thompson5910

good

- anonymous

42 for the third

- jim_thompson5910

yep

- anonymous

and 14 for the last

- jim_thompson5910

very good

- jim_thompson5910

now add up those areas to get ???

- anonymous

84

- anonymous

what would be the next step

- jim_thompson5910

|dw:1443396060396:dw|

- jim_thompson5910

imagine we have something like this function
|dw:1443396074151:dw|
some curve below the x axis

- jim_thompson5910

points A through E lie on this curve
|dw:1443396096733:dw|

- jim_thompson5910

we can't get the exact area between the curve and x axis, but we can get the approximate area by using trapezoids
|dw:1443396136613:dw|

- jim_thompson5910

it's not perfect, but it's the best we can do

- jim_thompson5910

what we do is divide the approximate area by the distance between the two endpoints (b-a = 20-10 = 10)

- anonymous

so the answer would be -8.4?

- jim_thompson5910

you may have seen the formula
\[\Large A = \frac{1}{b-a}\int_{a}^{b}f(x)dx\]
A = average value of the function f(x) on the interval x = a to x = b

- jim_thompson5910

so you just do 84/10 = 8.4

- jim_thompson5910

sorry yeah the area is actually negative because we're below the x axis
-84/10 = -8.4

- anonymous

wow thank you so much! your explanation and steps were extremely helpful!

- jim_thompson5910

|dw:1443396331117:dw|

- jim_thompson5910

no problem

Looking for something else?

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