The area of the region R = {(x, y) : 0 le y l | vedclass

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Question

(B) The region $R$ is defined by $0 \le y \le \frac{27}{x}$ and $1 \le x \le 9$. The area $A$ is given by the integral: $A = \int_{1}^{9} \frac{27}{x}

Vedclass · Accepted Answer

$54 \log_e 3 - \frac{52}{3}$

Vedclass · Answer

(B) The region $R$ is defined by $0 \le y \le \frac{27}{x}$ and $1 \le x \le 9$. The area $A$ is given by the integral: $A = \int_{1}^{9} \frac{27}{x} dx$ $A = 27 [\ln |x|]_{1}^{9}$ $A = 27 (\ln 9 - \ln 1)$ Since $\ln 9 = \ln(3^2) = 2 \ln 3$ and $\ln 1 = 0$: $A = 27 (2 \ln 3) = 54 \ln 3$. Note: If the region is bounded by $y \le \frac{27}{x}$,$y \ge 0$,$x \ge 1$,and $x \le 9$,the area is $54 \ln 3$. Given the options provided,the calculation $54 \log_e 3 - \frac{52}{3}$ suggests a more complex boundary condition (e.g.,subtracting a triangular area). Assuming the standard interpretation of such problems,the result is $54 \ln 3$.

The area of the region $R = \{(x, y) : 0 \le y \le \frac{27}{x}, 1 \le x \le 9\}$ is equal to:

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