Question

If dy/dx=1/x, then the average rate of change of y with respect to x on the closed interval [1, 4] isa) -1/4b) (1/2)*ln(2)c) (2/3)*ln(2)d) 2/5e) 2

Answer 1

Answer: C) $\frac{2}{3} \ln(2)$

Step-by-step explanation:

Here, $\frac{dy}{dx} = \frac{1}{x}$

⇒ $dy = \frac{1}{x}dx$

⇒  $\int dy =\int \frac{1}{x}dx$

⇒  $y = \ln x + C$

⇒  $f(x) = \ln x + C$

Where C is any constant.

The average rate of change of y with respect to x on the closed interval [1, 4] = $\frac{f(4)-f(1)}{4-1}$

= $\frac{(\ln 4 + C) -(\ln 1 + C)}{3}$

= $\frac{\ln 4 + C - \ln 1 - C}{3}$

= $\frac{\ln 4 - ln 1}{3}$

= $\frac{\ln 2^2 - 0}{3}$

= $\frac{2 \ln 2}{3}$

⇒ Option C is correct.

Answer 2

Answer: C) $\dfrac{2\ln(2)}{3}$

Step-by-step explanation:

Given: $\dfrac{dy}{dx}=\dfrac{1}{x}$

$\Rightarrow dy=\dfrac{1}{x}dx$

Integrating on both the sides , we get

$\int \, dy =\int {\dfrac{1}{x}} \, dx \\\\\Rightarrow\ y=\ln(x)+C$

Let $f(x)=\ln(x)+C$

Now, the average rate of change of y with respect to x on the closed interval [1, 4] is given by :-

$R=\dfrac{f(4)-f(1)}{4-1}\\\\\Rightarrow R=\dfrac{\ln(4)+C-(\ln(1)+C)}{3}\\\\\Rightarrow R=\dfrac{\ln(4)-\ln(1))}{3}\\\\\Rightarrow R=\dfrac{\ln(2^2)-0}{3}\\\\\Rightarrow R=\dfrac{2\ln(2)}{3}$