Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals.
In each case sketch the graph of the function. Keep in mind that the area bounded below the X-axis and above the graph of y=f(x)contributes as a negative number.
integral from (-1)^(3) (x-1) dx
(the -1 is suppose to be on the bottom and 3 is suppose to be on the top for the integral)
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The function is a straight line that crosses the x-axis at x=1 (y=0 when x=1).
It has gradient 1.
From -1 to 1 the line y = x - 1 is below the x-axis and from 1 to 3 it is above the x-axis.
As the function is a straight line the areas between the function and the x-axis form triangles.
Using the result that the area of a triangle is 1/2 base x height, the area below the x-axis is 1/2 times the distance along the x-axis (from -1 to 1) x the distance of the value of y at x=-1 (= -1 -1 = -2) to the x-axis (=2)
(1/2)(1- (-1)) x 2 = (1/2)(2)(2) = 2
Similarly, the area above the x-axis is
(1/2)(3-1) x 2 = 2
Therefore the total area between the function and the x-axis is 2+2 = 4
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