Intervals and definite integrals

- anonymous

Intervals and definite integrals

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- anonymous

##### 1 Attachment

- anonymous

The questions is attached

- Psymon

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

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- anonymous

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

- Psymon

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

- anonymous

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

- anonymous

yep

##### 1 Attachment

- anonymous

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

- Psymon

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

- anonymous

and the value of c is 48?

- Psymon

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

So the answer would be c = 7

- anonymous

Can you do the second part now?

- anonymous

oh so then solve for c which leads to 7

- anonymous

Yes, that is what you do.

- anonymous

so then for interval [-1,2] ?

- anonymous

It is the same process, can you try it?

- anonymous

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

- anonymous

what is "it"?

- anonymous

thinking 0?

- anonymous

What are you thinking is 0?

- anonymous

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

- anonymous

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

- anonymous

That is correct

- anonymous

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

- anonymous

zero?
undefined ?

- anonymous

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

- anonymous

oh thats right

- anonymous

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

- anonymous

0 again??

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

thank you

- anonymous

Did you find the answer?

- anonymous

for which part?

- anonymous

the last part is 0 ????

- anonymous

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

- anonymous

is it +or- sort(1)?

- anonymous

sqrt

- anonymous

Yes

- anonymous

so thats c=+-1 is the answer?

- anonymous

Yes

- anonymous

thx

- anonymous

You are welcome

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