Question

The value of the integral \(I = \int_0^1\int_0^1 x\cos(xy)\,dx\,dy\) is

• A. \(1 - \cos(1)\)
• B. \(1 + \cos(1)\)
• C. \(1 - \sin(1)\)
• D. \(1 + \sin(1)\)

Hint:

Try swapping the order of integration to get a simple antiderivative.

Answer 1

The Correct Option is A

Step 1: Recognise that the order of integration can be swapped.
The function \(x\cos(xy)\) is continuous on the square \([0,1]\times[0,1]\), so by Fubini's theorem the iterated integral gives the same value in either order.
It is easier to integrate over \(y\) first.

Step 2: Integrate with respect to \(y\), holding \(x\) fixed.
\(\displaystyle\int_0^1 x\cos(xy)\,dy = \big[\sin(xy)\big]_{y=0}^{1} = \sin(x) - \sin(0) = \sin(x)\).

Step 3: Integrate the result with respect to \(x\).
\(\displaystyle I = \int_0^1 \sin(x)\,dx = \big[-\cos(x)\big]_0^1 = -\cos(1) + \cos(0) = 1 - \cos(1)\).

Step 4: Check the wrong options.
Options B and D add \(\cos(1)\) or \(\sin(1)\) instead of subtracting, which would come from a sign slip at the limits.
Option C mixes up \(\cos(1)\) with \(\sin(1)\) from the wrong antiderivative of \(\cos(x)\).

Final Answer:
The integral evaluates to \(1 - \cos(1)\), so option A is correct. \[ \boxed{I = 1 - \cos(1)} \]