`int_0^9((1/3)x - 2)dx` Evaluate the integral by interpreting it in terms of areas.

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Chapter 5, 5.2 - Problem 36 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lemjay | High School Teacher | (Level 3) Senior Educator

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`int_0^9 (1/3x-2)dx`

To interpret this integral in terms of area,  graph the integrand.  The integrand is the function `f(x)=1/3x-2` .

Then, shade the region bounded by `f(x)=1/3x-2` and the x-axis in the interval [0,9]. (Please refer to the attached figure.)

Notice that the bounded region forms two right triangles. The first triangle is located below the x-axis and the second triangle is located above the x-axis.

To evaluate the integral, determine the area of each triangle. Then, subtract the area of the triangle located below the x-axis from the area of the triangle located above the x-axis.

`int_0^9 (1/3x-2)dx`

`= A_(Delta_2)-A_(Delta_1)`

`=1/2b_2h_2 - 1/2b_1h_1`

`=1/2*3*1 - 1/2*6*2`

`=3/2-6`

`=3/2-12/2`

`=-9/2`

Therefore, `int_0^9 (1/3x-2)dx=-9/2` .

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