Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 36

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 .