The area bounded by the curve xy - 3x - 2y - | vedclass

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Question

(C) Given the curve equation: $xy - 3x - 2y - 10 = 0$. Rearranging for $y$: $y(x - 2) = 3x + 10 \Rightarrow y = \frac{3x + 10}{x - 2}$. We can rewrite

Vedclass · Accepted Answer

$16 \log 2 + 3$

Vedclass · Answer

(C) Given the curve equation: $xy - 3x - 2y - 10 = 0$. Rearranging for $y$: $y(x - 2) = 3x + 10 \Rightarrow y = \frac{3x + 10}{x - 2}$. We can rewrite the expression as: $y = \frac{3(x - 2) + 16}{x - 2} = 3 + \frac{16}{x - 2}$. The required area $A$ is given by the integral: $A = \int_{3}^{4} y \, dx = \int_{3}^{4} \left( 3 + \frac{16}{x - 2} \right) dx$. Integrating term by term: $A = [3x + 16 \log |x - 2|]_{3}^{4}$. Evaluating at the limits: $A = (3(4) + 16 \log |4 - 2|) - (3(3) + 16 \log |3 - 2|)$. $A = (12 + 16 \log 2) - (9 + 16 \log 1)$. Since $\log 1 = 0$,we get $A = 12 + 16 \log 2 - 9 = 3 + 16 \log 2$ square units.

The area bounded by the curve $xy - 3x - 2y - 10 = 0$,the $x$-axis,and the lines $x = 3$ and $x = 4$ is:

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