# What is the net area between #f(x) = 3x^2-4x+2# and the x-axis over #x in [1, 3 ]#?

To work this out, we need to evaluate the definite integral between 3 and 1.

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To find the net area between ( f(x) = 3x^2 - 4x + 2 ) and the x-axis over ( x ) in the interval ([1, 3]), you need to integrate the absolute value of the function from ( x = 1 ) to ( x = 3 ).

The function intersects the x-axis at points where ( f(x) = 0 ). To find these points, solve the equation ( 3x^2 - 4x + 2 = 0 ) for ( x ). The solutions are ( x = 1 ) and ( x = 2/3 ). However, since the interval is ([1, 3]), we only need to consider ( x = 1 ) as the lower bound.

Integrating the absolute value of ( f(x) ) from ( x = 1 ) to ( x = 3 ) gives:

[ \int_{1}^{3} |3x^2 - 4x + 2| dx ]

Split the interval ([1, 3]) into two intervals: ([1, 2/3]) and ([2/3, 3]), and integrate separately.

For the interval ([1, 2/3]), ( |3x^2 - 4x + 2| = 3x^2 - 4x + 2 ).

For the interval ([2/3, 3]), ( |3x^2 - 4x + 2| = -(3x^2 - 4x + 2) ).

Calculate both integrals:

[ \int_{1}^{2/3} (3x^2 - 4x + 2) dx ] [ \int_{2/3}^{3} -(3x^2 - 4x + 2) dx ]

Then, sum the absolute values of both results to find the net area between the function and the x-axis over the given interval.

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