# Calculate integral sign(0 to 2) minimum{x^2,3x-2} dx? tough function minimum{x^2,3x-2}

*print*Print*list*Cite

### 1 Answer

You need to convert the given function in a branched function such that:

`f(x) = {(x^2, x^2=lt3x-2),(3x-2, x^2 gt 3x-2):}`

You need to solve the inequality `x^2=< 3x-2` such that:

`x^2 - 3x + 2 =< 0`

`x^2 - 3x + 2 = 0 => x_(1,2) = (3+-sqrt(9-8))/2 => x_(1,2) = (3+-1)/2`

`x_1 = 2 ; x_2 = 1`

Hence, the expression `x^2 - 3x + 2` is negative if `x in [1,2].`

`f(x) = (x^2, x in [1,2])`

`f(x) = 3x-2, x in (-oo,1)U(2,oo)`

Hence, evaluating the definite integral over `[0,2]` yields:

`int_0^2 f(x) dx = int_0^1(3x-2)dx + int_1^2 x^2 dx` `int_0^2 f(x) dx = (3x^2/2 - 2x)|_0^1 + x^3/3|_1^2`

`int_0^2 f(x) dx = 3/2 - 2 + 8/3 - 1/3`

`int_0^2 f(x) dx = -1/2 + 7/3 =gt int_0^2 f(x) dx = 11/6`

**Hence, evaluating the definite integral under the given conditions yields `int_0^2 f(x) dx = 11/6` .**