Question

The area (in sq. units) bounded by the curves $y=\frac{8}{x}$, $y=2x$ and $x=4$ is

• A. $12-8\log 2$
• B. $12+8\log 2$
• C. $12-8\log 4$
• D. $12+8\log 4$

Hint:

Area Between Curves:

• Compute $\int (f(x) - g(x)) dx$ where $f(x)>g(x)$.
• Carefully identify limits of intersection and dominance.

Answer 1

The Correct Option is A

Intersection: $\frac{8}{x} = 2x \Rightarrow x^2 = 4 \Rightarrow x = 2$. Between $x=2$ and $x=4$, $y=2x$ lies above $y=\frac{8}{x}$.
\[ \text{Area} = \int_2^4 (2x - \frac{8}{x}) dx = \left[ x^2 - 8 \ln|x| \right]_2^4 = (16 - 8 \ln 4) - (4 - 8 \ln 2) \] \[ = 12 - 8(\ln 4 - \ln 2) = 12 - 8 \ln 2 \]

Answer 2

Step 1: Identify the curves and limits
The curves given are:
\( y = \frac{8}{x} \),
\( y = 2x \),
and the vertical line \( x = 4 \).
We need to find the area bounded by these curves.

Step 2: Find the points of intersection between \( y = \frac{8}{x} \) and \( y = 2x \)
Set \( \frac{8}{x} = 2x \)
Multiply both sides by \( x \) (assuming \( x > 0 \)):
8 = 2x^2
\( x^2 = 4 \) → \( x = 2 \) (considering positive x-axis)

Step 3: Set up the integral for the bounded area
From \( x=2 \) to \( x=4 \), the upper curve is \( y = \frac{8}{x} \) and the lower curve is \( y = 2x \).
So, the area \( A = \int_2^4 \left(\frac{8}{x} - 2x \right) dx \).

Step 4: Calculate the integral
\[ A = \int_2^4 \frac{8}{x} dx - \int_2^4 2x dx \]
Calculate each separately:
\[ \int_2^4 \frac{8}{x} dx = 8 \int_2^4 \frac{1}{x} dx = 8 [\ln x]_2^4 = 8 (\ln 4 - \ln 2) = 8 \ln 2 \]
\[ \int_2^4 2x dx = 2 \int_2^4 x dx = 2 \left[ \frac{x^2}{2} \right]_2^4 = [x^2]_2^4 = 16 - 4 = 12 \]

Step 5: Compute the final area
\[ A = 8 \ln 2 - 12 = 12 - 8 \ln 2 \]
(Note the subtraction order, area must be positive; since \( \frac{8}{x} \) is above \( 2x \) between 2 and 4, the subtraction order is correct.)

Step 6: Conclusion
The area bounded by the curves is \( 12 - 8 \log 2 \) square units.