I have solved this question by myself. Thanks anyway.
K kamaruddin New member Joined Oct 30, 2005 Messages 1 Oct 30, 2005 #1 I have solved this question by myself. Thanks anyway.
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Oct 30, 2005 #2 You REALLY need to get the idea of the first one. How would you do it with a single integral? ∫022∗(2−x)dx=4\displaystyle \int_{0}^{2}{2*(2-x)} dx = 4∫022∗(2−x)dx=4 or ∫04(1/2)∗(4−y)dy=4\displaystyle \int_{0}^{4}{(1/2)*(4-y)} dy = 4∫04(1/2)∗(4−y)dy=4 Compare to the double integral version: ∫02∫02∗(2−x)1dydx=4\displaystyle \int_{0}^{2}{\int_{0}^{2*(2-x)}{1} dy} dx = 4∫02∫02∗(2−x)1dydx=4 or ∫04∫0(1/2)(4−y)1dxdy=4\displaystyle \int_{0}^{4}{\int_{0}^{(1/2)(4-y)}{1} dx} dy = 4∫04∫0(1/2)(4−y)1dxdy=4
You REALLY need to get the idea of the first one. How would you do it with a single integral? ∫022∗(2−x)dx=4\displaystyle \int_{0}^{2}{2*(2-x)} dx = 4∫022∗(2−x)dx=4 or ∫04(1/2)∗(4−y)dy=4\displaystyle \int_{0}^{4}{(1/2)*(4-y)} dy = 4∫04(1/2)∗(4−y)dy=4 Compare to the double integral version: ∫02∫02∗(2−x)1dydx=4\displaystyle \int_{0}^{2}{\int_{0}^{2*(2-x)}{1} dy} dx = 4∫02∫02∗(2−x)1dydx=4 or ∫04∫0(1/2)(4−y)1dxdy=4\displaystyle \int_{0}^{4}{\int_{0}^{(1/2)(4-y)}{1} dx} dy = 4∫04∫0(1/2)(4−y)1dxdy=4
