8767 #63
Posted: Thu Oct 23, 2014 12:24 am
Can anyone explain the answer to #63 on 8767?
Let $$f$$ be a continuous, strictly decreasing, real-valued function such that $$\int_0^\infty f(x)dx$$ is finite and $$f(0)=1$$. In terms of $$f^{-1}$$ (the inverse function of $$f$$), $$\int_0^\infty f(x)dx$$ is...
(a) less than $$\int_1^{\infty} f^{-1}(y)dy$$
(b) greater than $$\int_0^{1} f^{-1}(y)dy$$
(c) equal to $$\int_1^{\infty} f^{-1}(y)dy$$
(d) equal to $$\int_0^{1} f^{-1}(y)dy$$
(e) equal to $$\int_0^{\infty} f^{-1}(y)dy$$
The answer is D.
Let $$f$$ be a continuous, strictly decreasing, real-valued function such that $$\int_0^\infty f(x)dx$$ is finite and $$f(0)=1$$. In terms of $$f^{-1}$$ (the inverse function of $$f$$), $$\int_0^\infty f(x)dx$$ is...
(a) less than $$\int_1^{\infty} f^{-1}(y)dy$$
(b) greater than $$\int_0^{1} f^{-1}(y)dy$$
(c) equal to $$\int_1^{\infty} f^{-1}(y)dy$$
(d) equal to $$\int_0^{1} f^{-1}(y)dy$$
(e) equal to $$\int_0^{\infty} f^{-1}(y)dy$$
The answer is D.