# f (x)=2x^2-1,x>1 =x e ^(x-1),x<1 `int_(-2)^2f (x)dx`

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Since the function is represented by two different equations for `x>1` and `x<1` and since the limits of integration are in both intervals, `x>1, x<1` , you need to split the integral, such that:

`int_(-2)^2 f(x)dx = int_(-2)^1 f(x)dx + int_1^2 f(x) dx`

`int_(-2)^2 f(x)dx = int_(-2)^1 x*e^(x-1)dx + int_1^2 (2x^2 - 1)dx`

You need to use integration by parts to evaluate` int_(-2)^1 x*e^(x-1)dx` , such that:

`u = x => du = dx`

`dv = e^(x-1)dx => v = e^(x-1)`

`int_(-2)^1 x*e^(x-1)dx= x*e^(x-1)|_(-2)^1 - int_(-2)^1 e^(x-1) dx`

`int_(-2)^1 x*e^(x-1)dx = x*e^(x-1)|_(-2)^1 - e^(x-1)|_(-2)^1`

`int_(-2)^1 x*e^(x-1)dx = 1*e^0 - (-2)*e^(-3) - e^0 + e^(-3)`

`int_(-2)^1 x*e^(x-1)dx = 1 + 2/e^3 - 1 + 1/e^3`

Reducing duplicate members yields:

`int_(-2)^1 x*e^(x-1)dx = 3/e^3`

You need to evaluate the second integral `int_1^2 (2x^2 - 1)dx` using the property of linearity of integral, such that:

`int_1^2 (2x^2 - 1)dx = int_1^2 (2x^2)dx - int_1^2 dx`

`int_1^2 (2x^2 - 1)dx = (2x^3/3 - x)|_1^2`

`int_1^2 (2x^2 - 1)dx = (16/3 - 2 - 2/3 + 1)`

`int_1^2 (2x^2 - 1)dx = 14/3 - 1`

`int_1^2 (2x^2 - 1)dx = 11/3`

`int_(-2)^2 f(x)dx = 3/e^3 + 11/3`

**Hence, evaluating the definite integral, under the given conditions, yields int_(-2)^2 f(x)dx = 3/e^3 + 11/3.**

3/e^3