2-pi/2<=int_0^2f(x)dx<=2+pi/22−π2≤∫20f(x)dx≤2+π2 ?

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

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1 Answer
Jim S
May 26, 2018

Check below

Explanation:

int_0^2f(x)dx∫20f(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

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