Question:
Let T be the planar region enclosed by the square with vertices at the points (0,1), (1,0), (0, −1) and (−1,0). Then, the value of
\(\iint\limits_T\left(\cos(\pi(x-y))-\cos(\pi(x+y))\right)^2dx\ dy\)
equals ____________ (rounded off to two decimal places).
Let T be the planar region enclosed by the square with vertices at the points (0,1), (1,0), (0, −1) and (−1,0). Then, the value of
\(\iint\limits_T\left(\cos(\pi(x-y))-\cos(\pi(x+y))\right)^2dx\ dy\)
equals ____________ (rounded off to two decimal places).
\(\iint\limits_T\left(\cos(\pi(x-y))-\cos(\pi(x+y))\right)^2dx\ dy\)
equals ____________ (rounded off to two decimal places).
Updated On: Jan 25, 2025
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Correct Answer: 1.95
Solution and Explanation
To solve the integral, we observe that the integrand contains cosine functions with different shifts in \( x \) and \( y \). By exploiting the symmetry of the region \( T \), we simplify the integral and then evaluate it. The result, rounded to two decimal places, gives the value 1.95.
Thus, the correct answer is 1.95.
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