Let e be the base of natural logarithm and let f : 1,2,3,4 1,e,e^2,e^3 and g : 1,e,e^2,e^3…

Question

Let $e$ be the base of natural logarithm and let $f : \{1,2,3,4\} 	o \{1,e,e^2,e^3\}$ and $g : \{1,e,e^2,e^3\} 	o \left\{1,\dfrac12,\dfrac13,\dfrac14ight\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x) = \left[f^{-1}\{g^{-1}(1/2)\}ight]^x$, then the area of the region $R = \{(x,y) : x^2 \le y \le \phi(x),\ 0 \le x \le 1\}$ is:

Accepted Answer

Since $f$ is strictly decreasing: $f(1)=e^3, f(2)=e^2, f(3)=e, f(4)=1$, so $f^{-1}(e^2)=2$.

Since $g$ is strictly increasing and $\tfrac14<\tfrac13<\tfrac12<1$: $g(1)=\tfrac14, g(e)=\tfrac13, g(e^2)=\tfrac12, g(e^3)=1$, so $g^{-1}(1/2)=e^2$.

Hence $f^{-1}(g^{-1}(1/2)) = f^{-1}(e^2) = 2$, so $\phi(x) = 2^x$, and $R = \{x^2 \le y \le 2^x,\ 0\le x\le1\}$.

$$A = \int_0^1(2^x - x^2)\,dx = \left[\frac{2^x}{\ln2}\right]_0^1 - \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{\ln2}-\frac{1}{3} = \frac{3-\log_e2}{3\log_e2}$$

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