GATE 2024 ME – Question 42
If the value of the double integral
$$\int_{x=3}^{4}\int_{y=1}^{2}\frac{dy\,dx}{(x + y)^2}$$
is $\log_e(a/24)$, then $a$ is ___________ (*answer in integer*).
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Correct answer: 25
Explanation
The inner integral is $\int_{1}^{2}\frac{dy}{(x + y)^2} = \frac{1}{x + 1} - \frac{1}{x + 2}$. Then $\int_{3}^{4}\left[\frac{1}{x + 1} - \frac{1}{x + 2}\right]dx = \left[\ln\frac{x + 1}{x + 2}\right]_3^4 = \ln\frac{5}{6} - \ln\frac{4}{5} = \ln\frac{25}{24}$. So $a = 25$.