Question

The area of the region \[ R=\{(x,y): xy\le 8,\; 1\le y\le x^2,\; x\ge 0\} \] is:

• A. \( \dfrac{2}{3}(20\log_e 2+9) \)
• B. \( \dfrac{1}{3}(40\log_e 2+27) \)
• C. \( \dfrac{1}{3}(49\log_e 2-15) \)
• D. \( \dfrac{2}{3}(24\log_e 2-7) \)

Hint:

When a region has two possible upper boundaries, always split the integral at their point of intersection.

Answer 1

The Correct Option is D

Concept: The region is bounded by the curves \( y=1 \), \( y=x^2 \), and \( y=\dfrac{8}{x} \). The upper boundary changes where \( x^2=\dfrac{8}{x} \).
Step 1: Find the point of intersection \[ x^2=\frac{8}{x}\Rightarrow x^3=8 \Rightarrow x=2 \]
Step 2: Set up the integrals For \( 1\le x\le 2 \), upper curve is \( y=x^2 \) For \( 2\le x\le 8 \), upper curve is \( y=\dfrac{8}{x} \) \[ \text{Area}=\int_1^2 (x^2-1)\,dx+\int_2^8\left(\frac{8}{x}-1\right)\,dx \]
Step 3: Evaluate the integrals \[ \int_1^2(x^2-1)\,dx=\left[\frac{x^3}{3}-x\right]_1^2=\frac{4}{3} \] \[ \int_2^8\left(\frac{8}{x}-1\right)\,dx =\left[8\ln x-x\right]_2^8 =16\ln 2-6 \]
Step 4: Add the results \[ \text{Area}=16\ln 2-\frac{14}{3} =\frac{2}{3}(24\ln 2-7) \]