Is there someone to solve it?

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2 Answers
May 13, 2016

Proved in the explanation.

Explanation:

This is a definite double integral, over the square

R: #a<=x<=b and a<=y<=b#.

The order of integration can be changed, either way.

Also, #int f(y)dy#, from y=a to y=b =.#int f(x)dx#, from x=a to x=b .

The given double integral

#=int x dx int df(y) -int dx inty df(y)#, over #x in[a,b] and yin[a, b]#

#=[(b^2-a^2)/2][f(b)-f(a)]-(b-a){[yf(y)]#, between y=a and y=b}+#(b-a)int f(y) dy,# over a<=y<=b#

#= (b^2-a^2)/2(f(b)-f(a))-(b-a)(bf(b)-af(a))#

#+(b-a) intf(x) dx#, between x=a and x=b

#=(b-a) int f(x) dx#, between x=a and x=b, #-((b-a)^2/2)(f(a)+f(b))#.

May 13, 2016

#d/(dy)[(x-y)f(y)] = -f(y) +(x-y)d/(dy)f(y)# so
#(x-y)d/(dy)f(y) = d/(dy)[(x-y)f(y)]+f(y)# but
#int_a^b(int_a^bd/(dy)[(x-y)f(y)]dy)dx = -(a-b)^2(f(b)+f(a))/2#
hint. #int_a^bd/(dy)[(x-y)f(y)]dy = (x-b)f(b)-(x-a)f(a)# and concluding
#int_a^b(int_a^bf(y)dy)dx = (b-a)int_a^bf(y)dy#