Question

Intervals and definite integrals

Answer 1

The questions is attached

Answer 2

average value of an integral is:
\[\frac{ 1 }{ b-a }\int\limits_{a}^{b}f(x)dx\]

Answer 3

so in this case a=2 and b=11? and f(x)=x^2-1?

Answer 4

Yes, so you would solve that first to get your average value.

Answer 5

\[\frac{ 1 }{ 9 } \int\limits_{2}^{11}x^2-1= 48?\]

Answer 6

yep

Answer 7

ok that makes sense, thanks. now what do they mean by f(c) and finding c values?

Answer 8

Well, as of now, you have f(x) = x^2 - 1. Except they want f(c), so that means you have to replace x with c, giving you f(c) = c^2 - 1

Answer 9

and the value of c is 48?

Answer 10

Well, you would have to integrate c^2 - 1 and then apply the limits at the end and solve for c.

Answer 11

so thats \[\frac{ c^{3} }{ 3 }-x\]

Answer 12

The question asks: for what c is f(c) equal to the computed average.

Answer 13

So we are looking for a c such that f(c) = 48

Answer 14

In other words:
\[c^2 - 1 = 48\]

Answer 15

That equation has 2 solutions, \[\pm 7\] but only 1 of them is in the interval [2,11]

Answer 16

So the answer would be c = 7

Answer 17

Can you do the second part now?

Answer 18

oh so then solve for c which leads to 7

Answer 19

Yes, that is what you do.

Answer 20

so then for interval [-1,2] ?

Answer 21

It is the same process, can you try it?

Answer 22

is it \[\int\limits_{-1}^{2} c^2-1?\]

Answer 23

what is "it"?

Answer 24

thinking 0?

Answer 25

What are you thinking is 0?

Answer 26

the integral of c between -2 and 1
or not?

Answer 27

The integral\[\int\limits_{-1}^{2}c^2 - 1 dc = 0\]

Answer 28

That is correct

Answer 29

So what is the average value of the function between -1 and 2?

Answer 30

zero?
undefined ?

Answer 31

So the formula is:
\[\frac{1}{b-a}\int\limits_{a}^{b}f(c)dc\]

Answer 32

oh thats right

Answer 33

We just evaluated the integral part, so now what is the average?

Answer 34

0 again??

Answer 35

yes

Answer 36

Now we know the average value of the function between -1 and 2

Answer 37

We now need to find the value of c such that f(c) is the average, or f(c) = 0.

Answer 38

thank you

Answer 39

Did you find the answer?

Answer 40

for which part?

Answer 41

the last part is 0 ????

Answer 42

The average value of f(x) between -1 and 2 is 0, but that is not the answer to the question

Answer 43

is it +or- sort(1)?

Answer 44

sqrt

Answer 45

Yes

Answer 46

so thats c=+-1 is the answer?

Answer 47

Yes

Answer 48

thx

Answer 49

You are welcome