Hi, I was wondering if anyone can help: Find a positive value of k such that the average value of f(x) = 1/ (kx^2) over the interval [k,1] is 2. Thanks,
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Originally Posted by Wintaker99 Hi, I was wondering if anyone can help: Find a positive value of k such that the average value of f(x) = 1/ (kx^2) over the interval [k,1] is 2. Thanks, The average of f(x) over [k,1] is: Av(f) = [integral_{x=k to 1} f(x) dx]/(1-k) = [(1-k)/k^2]/(1-k) = 1/k^2, so if Av(f) = 2, we have 1/k^2=2, and so k=1/sqrt(2). RonL
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