Quick help with average value function problem please

[c19dale]

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]


help please..

 

[arildno]

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

 

[c19dale]

.

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

 

[arildno]

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

Right?

 

[c19dale]

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...

 

[c19dale]

is this a substitution problem?

 

[arildno]

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.

 

[c19dale]

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...

 

[c19dale]

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...

 

[arildno]

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

 

[c19dale]

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...