Evaluate definite integral of f(x)=minim{x^2,3x-2} for x=0 to x=2.

Before evaluating the definite integral, we'll apply the
rule of minimum to the function f(x).

minim m(x) =
min{u(x),v(x)} is:

- u(x), for
u(x)<v(x)

- v(x), for
u(x)>v(x)

We'll apply the rule of min to the
function f(x):

- x^2, for x^2<3x-2 <=>
x^2 - 3x + 2 < 0

- 3x-2, for x^2>3x-2
<=> x^2 - 3x + 2 > 0

We'll calculate
the roots of x^2 - 3x + 2 to verify where the expression is positive and where it's
negative.

 It is obvious that the roots of the expression 
x^2 - 3x + 2 are 1 and 2.

x1 + x2 = 3 =
S

x1*x2 = 2 = P

1 + 2 =
3

1 * 2 = 2

The xpression is
negative over the interval [1,2] and it is positive over the intervals (-inf.,1) U
(2,+inf.).

Now, we'll calculate the definite
integral.

Int f(x) dx = Int 1 + Int
2

Int 1 = Int f(x)x from x = 0 to x =
1

Int 2 = Int f(x) dx from x = 1 to x =
2

Int 1 = Int (3x-2)dx = Int 3xdx - 2Int
dx

Int 1 = 3x^2/2 - 2x

Int 1 =
F(1) - F(0)

F(1) - F(0) = 3/2 - 2 =
-1/2

Int 1 =
-1/2

Int 2 = F(2) -
F(1)

Int 2 = Int x^2dx

Int 2 =
x^3/3

F(2) - F(1) = 2^3/3 -
1^3/3

F(2) - F(1) = 8/3 - 1/3 =
7/3

Int 2 =
7/3

Int f(x) dx = Int 1 + Int
2

Int f(x) dx = Int 1 + Int
2

Int f(x) dx = -1/2 + 7/3

Int
f(x) dx = (-3 + 14)/6

Int f(x) dx =
11/6