00:01
Hello students, so the first one is we have to find out the contour here, it is equal to integral 0 to 1, integral 0 to 1, dx into 2 minus x into y into dx into dy equal to 1 so it is already we know that, f of x y of x y dx into dy is equal to 1, so now we have to integrate this, c into 2 x square by 2 minus x cube by 3 minus x square by 2 into y, we derive this with respect to x so now we have to apply the limits 0 to 1, so here after applying the limits 0 to 1 into dy equal to 1, we get 0 to 1, here also 0 to 1, 0 to 1, c into 2 by 2 minus 1 by 3 minus 1 by 2 into y into dy equal to 1 so it is equal to c into integral 0 to 1, it is 2 by 3, so 2 by 3, this part 2 by 2 is 1, 1 minus 1 by 3 is 2 by 3 minus 1 by 2 by into dy, so now we have to integrate this term with y, so here it is c into 2 by 3 minus 1 by 2 into y square by 2 now we have to apply the limits, so it is equal to c into 2 by 3 minus 1 by 2 into y square by 2, so after applying the limits c into 2 by 3 minus 1 by 4, so it is equal to c equal to 12 by 5 now in b we have to find out fx of x, it is equal to integral minus infinity to infinity f of x comma y into dy, so it is equal to 0 to 1, 12 by 5 x into 2 minus x minus y into dy so it is equal to 0 to 1, 12 by 5 2x minus x square by 5, here 12 minus 12 x y, 12 by 5 x, here we derivate the y, y square by, so now we apply the limits 12 by 5 into 2x minus x square by 5 into 12 minus 12 x by 10, 10 into x square in place of y, we apply the limit, so that will be 24 x by 5 minus 12 x square by 5 minus 6 by 5 into x, so that is equal to minus 12 x square by 5 plus 24 minus 18 x by 5, so it is equal to minus 12 x square minus 18 x by 5, so it is a 0 less than x less than 1...