a region in the plane is bounded by the graph of $y = \\frac{1}{x}$, the $x$-axis, the line $x = m$, and the…

a region in the plane is bounded by the graph of $y = \\frac{1}{x}$, the $x$-axis, the line $x = m$, and the line $x = 2m$, $(m>0)$. the area of this region\na is independent of $m$.\nb increases as $m$ increases.\nc decreases as $m$ increases.\nd decreases as $m$ increases when $m\\frac{1}{2}$.\ne increases as $m$ increases when $m < \\frac{1}{2}$; decreases as $m$ increases when $m > \\frac{1}{2}$.

a region in the plane is bounded by the graph of $y = \\frac{1}{x}$, the $x$-axis, the line $x = m$, and the line $x = 2m$, $(m>0)$. the area of this region\na is independent of $m$.\nb increases as $m$ increases.\nc decreases as $m$ increases.\nd decreases as $m$ increases when $m\\frac{1}{2}$.\ne increases as $m$ increases when $m < \\frac{1}{2}$; decreases as $m$ increases when $m > \\frac{1}{2}$.

Answer

Explanation:

Step1: Recall area - under - curve formula

The area $A$ under the curve $y = f(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}f(x)dx$. Here, $f(x)=\frac{1}{x}$, $a = m$, and $b = 2m$. So, $A=\int_{m}^{2m}\frac{1}{x}dx$.

Step2: Evaluate the integral

We know that the antiderivative of $\frac{1}{x}$ is $\ln|x|$. Using the fundamental theorem of calculus $\int_{m}^{2m}\frac{1}{x}dx=\left[\ln x\right]_{m}^{2m}$.

Step3: Substitute the limits of integration

$\left[\ln x\right]_{m}^{2m}=\ln(2m)-\ln(m)$.

Step4: Use logarithm property

By the property $\ln a-\ln b=\ln\frac{a}{b}$, we have $\ln(2m)-\ln(m)=\ln\frac{2m}{m}=\ln 2$.

Answer:

A. is independent of $m$.