Question

`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`

Answer 1

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`