Find the average value of the function on the given interval: h(u) = (3-2u)^-1 ; [-1,1] Let t = 3-2u => du = -dt/2 => -1/4 integral of 1/t dt = -1/4 ln|t| = -1.4 * ln(1/(3-2u)) from -1 to 1 then I got result = 0.402, correct?
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You missed that: $\displaystyle \overline{f(x)} = \int_a^b f(x)x dx$
no, f(ave) = 1/(b-a) * integral of fx from b to a
Originally Posted by huybinhs Let t = 3-2u => du = -dt/2 -1/4 integral of 1/t dt = -1/4 ln|t| I think it should be $\displaystyle \displaystyle \frac{-1}{2} \int \frac{1}{t}dt = \frac{-1}{2} \ln |t| $
no b/c we also have du = -dt/2 so => 1/2 * (-1/2) = -1.4 infront, right?
Sorry, you are right, I was in statistics. Your solution is indeed: $\displaystyle \frac{\ln 5}{4}$
Thanks for double check my work
So what did you get for your new integral from -1 to 1?
I did not change the limit because I replace 1 + x^2 = u after that
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