#2-pi/2<=int_0^2f(x)dx<=2+pi/2# ?

Given #f# continuous in #[0,2]# and differentiable in #(0,2)#. The function of #f# is inside the circle disk with center #M(1,1)# and radius #r=1#

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1 Answer
May 26, 2018

Check below

Explanation:

#int_0^2f(x)dx# expresses the area between #x'x# axis and the lines #x=0# , #x=2#.

#C_f# is inside the circle disk which means the 'minimum' area of #f# will be given when #C_f# is in the bottom semicircle and the 'maximum' when #C_f# is on the top semicircle.

Semicircle has area given by #A_1=1/2πr^2=π/2m^2#

The rectangle with base #2# and height #1# has area given by #A_2=2*1=2m^2#

The minimum area between #C_f# and #x'x# axis is #A_2-A_1=2-π/2#

and the maximum area is #A_2+A_1=2+π/2#

Therefore, #2-π/2<=int_0^2f(x)dx<=2+π/2#