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

Net area

Given -

#f(x)=3x^2-x+2#

Area under the curve between

#int_1^3(3x^2-x+2)=[x^3-x^2/2+2x]_1^3#

#[3^3-3^2/2+2.3]-[1^3-1^2/2+2.1]#

#[27-9/2+6]-[1-1/2+2]#

#57/2-5/2=52-2=26#

Net area

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

The integral of ( |f(x)| ) over ([1, 3]) is:

[ \int_{1}^{3} |f(x)| , dx = \int_{1}^{3} |3x^2 - x + 2| , dx ]

To find the points where ( f(x) ) intersects the x-axis, solve ( 3x^2 - x + 2 = 0 ).

By solving this quadratic equation, you'll find two roots. If these roots fall within the interval ([1, 3]), then you need to account for the area under the x-axis. Otherwise, this area is zero.

Once you have the areas above and below the x-axis, subtract the area under the x-axis from the total area above it to get the net area.

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