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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.
`=1/2b_2h_2 - 1/2b_1h_1`
`=1/2*3*1 - 1/2*6*2`
Therefore, `int_0^9 (1/3x-2)dx=-9/2` .
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