Question

solve the problem given ​

Answer 1

Answer:

Find the value of the integral

∫20∫20f(x+y) dx dy

where f(t) denotes the greatest integer ≤t

Solution Attempt: Let [y] represent the greater integer ≤y and {y} the fractional part of y.

Let x+y=t⟹dx=dt. Hence, ∫20∫20f(x+y) dx dy=∫20∫y+2yf(t) dt dy.

∫y+2yf(t) dt=∫y+2y[t]dt.

Case 1 If y is an integer

∫y+2y[t]dt=∫y+1y[t]dt+∫y+2y+1[t]dt=[y]+[y]+1=2[y]+1.

Case 2 : If y is not an integer

∫y+2y[t]dt=∫[y]+1y[t]dt+∫[y]+2[y]+1[t]dt+∫y+2[y]+2[t]dt.

=[y](1+{y})+([y]+1)+([y]+2)({y})

=[y]+[y]{y}+[y]+1+[y]{y}+2{y}

=2[y]+1+2[y]{y}+2{y}

=2y+1+2[y]{y}

Answer 2

here is your solution...

1.used than(u+v)=tanU+tanV1-tanU-tanV identify for decomposing arctan(2x-11+x-x2)

2.i used x=1-u transform in I integral

3.after summing 2 integrals I found result.

1=\inint10arcta2x-11+x-x2]dx

=\int 10 arctan(2x-11-x(x+1))dx

=\int1o arctan x DC +\int 10 arctan(x-1)dx

after using x=1-u an dx=-dutransforms

1=int 01arctan(1-u)(-du)+\int01arctan(u)(-du)

=\int(u-1)du+\int 01arctanu du

=-\int10 arctanu du-\int 10 arctan(u-1) du

=-\int 10 arctanx dx-\int 10 arctan(x-1) dx

after summing 2 integrals

2i=0 hence 1=0

I think it helps u