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

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

May 26, 2018

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#### Explanation:

${\int}_{0}^{2} f \left(x\right) \mathrm{dx}$ 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 ${A}_{2} = 2 \cdot 1 = 2 {m}^{2}$

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