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Do you know how to integrate a piecewise function? If so, recall that |f(x)| = f(x) if f(x) >= 0, or -f(x) if f(x) < 0.
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just graph it and take area under curve (triangle)
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Divide it into 2 parts: when 0≤ x<4.5 and when 4.5 ≤x< ≤9
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Use substitution.

The integral of the absolute is sgn(x)[x^(2)/2]
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Two approaches:

1. Treat the absolute value as a piecewise function, and split the integral into two pieces
2. Graph the function, and use geometry, since it should look like just two triangles
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Equation the expression to 0.
U will get a value for x (called the vertice for that modulus function), lets call it 'vert'.
Now, divide you limits (0-9) into that two limits(0-vert) and (vert-9). Now work from here.

Hope u got it.

Sidenote: Can you confirm if it is modulus sign or simply (f(x)).dx?
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You need to split the integral from x=0 to x=9 into two parts. The first part is for when (2x-9) is >= 0 and the other is for when it is < 0. It has to be split at x = 9/2
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That’s a definite integral. |r| = r when r >= 0 and -r when r <= 0. Find the intervals where 2x-9 >= 0 and the intervals where 2x-9 <= 0. In the former case, the absolute value does nothing. In the latter, you will be integrating -(2x-9).
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For this problem, graph the function and then it's the area of the two triangular regions under the graph.
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