Question

If f (x): [1, 3] → [-1,1] satisfies $\int_{1}^{3} f(x)dx = 0$, then the maximum value of $\int_{1}^{3} \frac{f(x)}{x}dx$ is

• A. ln($\frac{3}{2}$)
• B. ln($\frac{4}{3}$)
• C. $\frac{8}{3}$
• D. ln($\frac{9}{2}$)

Answer 1

Answer: (B)

ln($\frac{4}{3}$)

Answer 2

Let $I = \int_{1}^{3} \frac{f(x)}{x}dx$. We are given that $f(x): [1, 3] \rightarrow [-1, 1]$ and $\int_{1}^{3} f(x)dx = 0$.
We want to maximize $I$. The integrand is $\frac{f(x)}{x}$. The term $\frac{1}{x}$ is a positive and decreasing function on the interval $[1, 3]$. To maximize the integral, we should make $f(x)$ as large as possible (i.e., $f(x)=1$) where $\frac{1}{x}$ is large (i.e., for small values of $x$), and $f(x)$ as small as possible (i.e., $f(x)=-1$) where $\frac{1}{x}$ is small (i.e., for large values of $x$).

This suggests that the maximizing function $f(x)$ should be of the form:
$f(x) = 1$ for $x \in [1, a]$
$f(x) = -1$ for $x \in (a, 3]$
for some value $a \in [1, 3]$.

We must satisfy the condition $\int_{1}^{3} f(x)dx = 0$.
$\int_{1}^{3} f(x)dx = \int_{1}^{a} 1 dx + \int_{a}^{3} (-1) dx = [x]_{1}^{a} + [-x]_{a}^{3}$
$= (a - 1) + (-3 - (-a)) = (a - 1) + (a - 3) = 2a - 4$.
Setting the integral to 0, we get $2a - 4 = 0$, which implies $a = 2$.
Since $a=2$ is in the interval $[1, 3]$, this function is valid.
The function is $f_0(x) = \begin{cases} 1 & \text{if } 1 \le x \le 2 \\ -1 & \text{if } 2 < x \le 3 \end{cases}$.
This function satisfies $f_0(x) \in [-1, 1]$ for all $x \in [1, 3]$ and $\int_{1}^{3} f_0(x)dx = 0$.

Now, let's calculate the value of the integral $I$ for this function $f_0(x)$:
$I_0 = \int_{1}^{3} \frac{f_0(x)}{x}dx = \int_{1}^{2} \frac{1}{x}dx + \int_{2}^{3} \frac{-1}{x}dx$
$I_0 = [\ln|x|]_{1}^{2} - [\ln|x|]_{2}^{3}$
$I_0 = (\ln 2 - \ln 1) - (\ln 3 - \ln 2)$
$I_0 = (\ln 2 - 0) - (\ln 3 - \ln 2)$
$I_0 = \ln 2 - \ln 3 + \ln 2 = 2 \ln 2 - \ln 3 = \ln(2^2) - \ln 3 = \ln 4 - \ln 3 = \ln(\frac{4}{3})$.