# What is a>0 if integral sign (0 down- a up) (2-4x+3x^2)=a?

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You need to evaluate the definite intgeral, using the property of linearity of intgeral, such that:

`int_0^a (2 - 4x + 3x^2)dx = int_0^a 2 dx - int_0^a 4x dx + int_0^a 3x^2 dx`

Using the fundamental theorem of calculus yields:

`int_0^a (2 - 4x + 3x^2)dx = 2x|_0^a - 2x^2|_0^a + x^3|_0^a`

`int_0^a (2 - 4x + 3x^2)dx = 2(a - 0) - 2(a^2 - 0) + a^3 - 0`

`int_0^a (2 - 4x + 3x^2)dx = 2a - 2a^2 + a^3`

The problem provides the information that `int_0^a (2 - 4x + 3x^2)dx = a` , hence, replacing a for `int_0^a (2 - 4x + 3x^2)dx ` yields:

`a = 2a - 2a^2 + a^3 => a^3 - 2a^2 + 2a - a = 0`

`a^3 - 2a^2 + a = 0`

Factoring out a yields:

`a(a^2 - 2a + 1) = 0`

Using zero product property and the condition provided by the problem, `a > 0` , yields:

`a^2 - 2a + 1 = 0 => (a - 1)^2 = 0 => a - 1 = 0 => a = 1`

**Hence, evaluating a, under he given conditions, yields `a = 1` .**