Evaluate the above double integral. Round up your answer to the nearest integer.
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Consider the inner integral first.
The 4 cos y contributes zero because it is integrated over an integer multiple of its period.
The 2 x y contributes zero because y is an odd function, and it is integrated over an interval centred on the origin.
Integrating the remaining term is easy because as far as the inner integral is concerned it is a constant.
Making use of these observations the double integral simplifies to
∫ − 1 7 2 9 1 7 2 9 2 8 π sin 2 x d x
Transform this into an integrable form with a trig. identity to get
1 4 π ∫ − 1 7 2 9 1 7 2 9 1 − cos 2 x d x
Now everyday integration gives, to the nearest integer, the answer 1 5 2 0 5 7