anonymous
  • anonymous
Intervals and definite integrals
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
The questions is attached
Psymon
  • Psymon
average value of an integral is: \[\frac{ 1 }{ b-a }\int\limits_{a}^{b}f(x)dx\]

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anonymous
  • anonymous
so in this case a=2 and b=11? and f(x)=x^2-1?
Psymon
  • Psymon
Yes, so you would solve that first to get your average value.
anonymous
  • anonymous
\[\frac{ 1 }{ 9 } \int\limits_{2}^{11}x^2-1= 48?\]
anonymous
  • anonymous
anonymous
  • anonymous
ok that makes sense, thanks. now what do they mean by f(c) and finding c values?
Psymon
  • 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
  • anonymous
and the value of c is 48?
Psymon
  • Psymon
Well, you would have to integrate c^2 - 1 and then apply the limits at the end and solve for c.
anonymous
  • anonymous
so thats \[\frac{ c^{3} }{ 3 }-x\]
anonymous
  • anonymous
The question asks: for what c is f(c) equal to the computed average.
anonymous
  • anonymous
So we are looking for a c such that f(c) = 48
anonymous
  • anonymous
In other words: \[c^2 - 1 = 48\]
anonymous
  • anonymous
That equation has 2 solutions, \[\pm 7\] but only 1 of them is in the interval [2,11]
anonymous
  • anonymous
So the answer would be c = 7
anonymous
  • anonymous
Can you do the second part now?
anonymous
  • anonymous
oh so then solve for c which leads to 7
anonymous
  • anonymous
Yes, that is what you do.
anonymous
  • anonymous
so then for interval [-1,2] ?
anonymous
  • anonymous
It is the same process, can you try it?
anonymous
  • anonymous
is it \[\int\limits_{-1}^{2} c^2-1?\]
anonymous
  • anonymous
what is "it"?
anonymous
  • anonymous
thinking 0?
anonymous
  • anonymous
What are you thinking is 0?
anonymous
  • anonymous
the integral of c between -2 and 1 or not?
anonymous
  • anonymous
The integral\[\int\limits_{-1}^{2}c^2 - 1 dc = 0\]
anonymous
  • anonymous
That is correct
anonymous
  • anonymous
So what is the average value of the function between -1 and 2?
anonymous
  • anonymous
zero? undefined ?
anonymous
  • anonymous
So the formula is: \[\frac{1}{b-a}\int\limits_{a}^{b}f(c)dc\]
anonymous
  • anonymous
oh thats right
anonymous
  • anonymous
We just evaluated the integral part, so now what is the average?
anonymous
  • anonymous
0 again??
anonymous
  • anonymous
yes
anonymous
  • anonymous
Now we know the average value of the function between -1 and 2
anonymous
  • anonymous
We now need to find the value of c such that f(c) is the average, or f(c) = 0.
anonymous
  • anonymous
thank you
anonymous
  • anonymous
Did you find the answer?
anonymous
  • anonymous
for which part?
anonymous
  • anonymous
the last part is 0 ????
anonymous
  • anonymous
The average value of f(x) between -1 and 2 is 0, but that is not the answer to the question
anonymous
  • anonymous
is it +or- sort(1)?
anonymous
  • anonymous
sqrt
anonymous
  • anonymous
Yes
anonymous
  • anonymous
so thats c=+-1 is the answer?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
thx
anonymous
  • anonymous
You are welcome

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