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 $y=\frac{8}{x}$.
\[ \text{Area} = \int_2^4 (2x - \frac{8}{x}) dx = \left[ x^2 - 8 \ln|x| \right]_2^4 = (16 - 8 \ln 4) - (4 - 8 \ln 2) \] \[ = 12 - 8(\ln 4 - \ln 2) = 12 - 8 \ln 2 \]