Evaluate the iterated integral \int_{0}^{\pi/4} \int_{0}^{\pi/2} sin x cos 2y dx dy
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Answer and Explanation:
Thus,
=∫π40∫π20(sin(x)(−cos(2y)))dydx[Integrate with respect to y]=∫π40(−12sin(x)sin(2y))∣∣∣π20dx=∫π40((−12sin(x)sin(2(π2)))−(−12sin(x)sin(2(0))))dx=∫π400dx
Thus,
=∫π40∫π20(sin(x)(−cos(2y)))dydx[Integrate with respect to y]=∫π40(−12sin(x)sin(2y))∣∣∣π20dx=∫π40((−12sin(x)sin(2(π2)))−(−12sin(x)sin(2(0))))dx=∫π400dx
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The integrate the iterated integral over the given rectangular region R
[0,π/4]×[0×π/2][0,π/4]×[0×π/2]
you can apply Fubini's theorem integrate ∫π/40∫π/20sin(x)cos(2y)dxdy∫0π/4∫0π/2sin(x)cos(2y)dxdy with respect to x first or y.
[0,π/4]×[0×π/2][0,π/4]×[0×π/2]
you can apply Fubini's theorem integrate ∫π/40∫π/20sin(x)cos(2y)dxdy∫0π/4∫0π/2sin(x)cos(2y)dxdy with respect to x first or y.
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