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:
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
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:
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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