Question:

The region is bounded by: 1. Hyperbola: $y = \dfrac{1}{x}$ 2. Line $L_1$: $5x - 4y - 1 = 0 \implies y = \dfrac{5x - 1}{4}$ 3. Line $L_2$: $4x + 4y - 17 = 0 \implies y = \dfrac{17 - 4x}{4}$ Let us find the intersection points: - Between $L_1$ and $L_2$: Adding $5x - 4y - 1 = 0$ and $4x + 4y - 17 = 0$ gives $9x - 18 = 0 \implies x = 2$. Thus $y = 9/4$. Point is $(2, 9/4)$. - Between $L_1$ and $y = 1/x$: $5x - 4/x - 1 = 0 \implies 5x^2 - x - 4 = 0 \implies (5x+4)(x-1) = 0$. Since $x > 0$, we have $x = 1 \implies y = 1$. Point is $(1, 1)$. - Between $L_2$ and $y = 1/x$: $4x + 4/x - 17 = 0 \implies 4x^2 - 17x + 4 = 0 \implies (4x-1)(x-4) = 0$. Since $x=1/4$ violates $5x-4y-1>0$, we take $x=4 \implies y = 1/4$. Point is $(4, 1/4)$. The upper boundary is $L_1$ on $[1, 2]$ and $L_2$ on $[2, 4]$. The lower boundary is the hyperbola $y = 1/x$ on $[1, 4]$. $$\text{Area} = \int_{1}^{2} \left(\dfrac{5x - 1}{4}\right) dx + \int_{2}^{4} \left(\dfrac{17 - 4x}{4}\right) dx - \int_{1}^{4} \dfrac{1}{x} dx$$ Calculate the integrals: 1. $\int_{1}^{2} \left(\dfrac{5x - 1}{4}\right) dx = \dfrac{1}{4}\left[ \dfrac{5x^2}{2} - x \right]_1^2 = \dfrac{1}{4}\left( (10 - 2) - \left(\dfrac{5}{2} - 1\right) \right) = \dfrac{13}{8}$ 2. $\int_{2}^{4} \left(\dfrac{17 - 4x}{4}\right) dx = \dfrac{1}{4}\left[ 17x - 2x^2 \right]_2^4 = \dfrac{1}{4}\left( (68 - 32) - (34 - 8) \right) = \dfrac{5}{2}$ 3. $\int_{1}^{4} \dfrac{1}{x} dx = \log_e 4$ Total Area: $$\text{Area} = \dfrac{13}{8} + \dfrac{5}{2} - \log_e 4 = \dfrac{33}{8} - \log_e 4$$