# Quick help with average value function problem please

## Main Question or Discussion Point

I am in a time crunch and I am stumped........

I need to find the average value function of f(x) = (3x+5)^2 on [1, 2]

arildno
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What would you suppose to be the method of finding it out?

.

ok..i typed the problem wrong....its (3x+5)^5

I know the formula is 1/b-a int from 1 to 2 f(X)dx

but We havent gone over this along with 4 other different problems I have left....I don't know how to handle the 5th power?

arildno
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You'd know how to handle the integral $$\int_{1}^{2}t^{5}dt$$

Right?

i think I could figure it out...we haven't gone over this material because we ran out of time...but I have figured out problems similar to that.....

is this a substitution problem?

arildno
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Allright:
If I asked you to find:
$$\frac{d}{dt}\frac{1}{6}t^{6}$$
I hope you agree with me that we have:
$$\frac{d}{dt}\frac{1}{6}t^{6}=t^{5}$$

Hence, to solve the problem:
1) Make a variable substitution (you've done this type of stuff, right?): t=3x+5
2) since t=8 when x=1 and t=11 when x=2, we get the integral:
$$\frac{1}{3}\int_{8}^{11}t^{5}dt$$
which is the average value you're seeking.

no....I haven't done this stuff, but with what you just posted, I might be able to figure it out....thanks, I'll try it again.........

so I don't use the formula I posted above?..I am just looking in the book, its a different type of problem(polynomial), but it gets a constant as the answer and it uses a formula to get it....

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arildno
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c19dale said:
ok..i typed the problem wrong....its (3x+5)^5

I know the formula is 1/b-a int from 1 to 2 f(X)dx
This is certainly the formula I've used:
Since 2-1=1, we have:
$$\hat{f}=\frac{1}{1}\int_{1}^{2}(3x+5)^{5}dx=\int_{1}^{2}(3x+5)^{5}dx=\int_{8}^{11}\frac{1}{3}t^{5}dt$$

thats what I got after you explained it before, but does it simplify from there or is that the answer? ..never mind...i see...sub back in the value of t and solve....thanks, I got it...