*newbie post* Calc II, finding the average value.

[lovely_aly]

Alright, so I know how to do this. My question is more about simplification.

f(x) = ln(x), [1,3]

The antiderivative of ln(x) is x ln(x)-1 so, for my average value of the function (f ave) on the interval I have (3/2)ln(3) -1. The book problem then asks me to find c such that f ave = f(c). I'm confused about the order of operations of (3/2)ln(3)-1 and don't know how I should go about simplifying it so that I can solve for c.

Thanks for any help.

 

[galactus]

The antiderivative of ln(x) is \(\displaystyle xln(x)-x\), not \(\displaystyle xln(x)-1\)

\(\displaystyle f_{ave}=\frac{1}{3-1}\int_{1}^{3}ln(x)dx=\frac{3ln(3)-2}{2}\), but you are correct. I assume that is a typo then.

I reckon you want to find a value c in [1,3] such that \(\displaystyle f(c)=f_{ave}\)

\(\displaystyle ln(c)=\frac{3ln(3)-2}{2}\)

Solve for c.

 

[Deleted member 4993]

Remember

If

ln(x) = a

then

x = e^a

 

[lovely_aly]

thanks for your help.