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?(Nod)

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- Oct 13th 2010, 07:55 PMhuybinhsFind the average value
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?(Nod) - Oct 13th 2010, 08:13 PMraphw
You missed that: $\displaystyle \overline{f(x)} = \int_a^b f(x)x dx$

- Oct 13th 2010, 08:17 PMhuybinhs
no, f(ave) = 1/(b-a) * integral of fx from b to a

- Oct 13th 2010, 08:22 PMEducated
- Oct 13th 2010, 08:24 PMhuybinhs
no b/c we also have du = -dt/2 so => 1/2 * (-1/2) = -1.4 infront, right?

- Oct 13th 2010, 08:25 PMraphw
Sorry, you are right, I was in statistics. Your solution is indeed: $\displaystyle \frac{\ln 5}{4}$

- Oct 13th 2010, 08:27 PMhuybinhs
Thanks for double check my work ;)

- Oct 13th 2010, 08:29 PMEducated
So what did you get for your new integral from -1 to 1?

- Oct 13th 2010, 08:32 PMhuybinhs
I did not change the limit because I replace 1 + x^2 = u after that ;)