Signed Area Help
Introduction to Signed Area
Without saying so explicitly, we have implicitly assumed in our discussion of area in the last section that our function f is positive, that is its graph lies about the x-axis. But of course many functions do not share that property. We nevertheless would like to be able to calculate areas determined by such functions, and to calculate the corresponding integrals.
This turns out to be simple to do. Consider the function y = f ( x ) shown in Fig. 4.11. It is negative on the interval [ a , b ] and positive on the interval [b, c ]. Suppose that we wish to calculate the shaded area as in Fig. 4.12. We can do so by breaking the problem into pieces.
Of course, because f ≥ 0, the area between x = b and x = c is given by the integral , just as we have discussed in the last section. But our discussions do not apply directly to the area between x = a and x = b. What we can do is instead consider the function g = − f. Its graph is shown in Fig. 4.13.
Of course g is a positive function on [a, b ], except at the endpoints a and b; and the area under g —between x = a and x = b —is just the same as the shaded area between x = a and x = b in Fig. 4.14. That area is
In total, the aggregate shaded area exhibited in Fig. 4.15, over the entire interval [ a , c ], is
What we have learned is this: If f ( x ) < 0 on the interval under discussion, then the integral of f will be a negative number. If we want to calculate positive area then we must interject a minus sign.
Let us nail down our understanding of these ideas by considering some examples.
Example 1Calculate the (positive) area, between the graph of f ( x ) = x 3 − 2 x 2 − 11 x + 12 and the x -axis, between x = −3 and x = 4.
Consider Fig. 4.16.
It was drawn using the technique of Section 3.1, and it plainly shows that f is positive on [−3, 1] and negative on [1, 4]. From the discussion preceding this example, we know then that
Here we are using the standard shorthand
to stand for
F ( b ) − F ( a ).
Thus we have
Notice that, by design, each component of the area has made a positive contribution to the final answer. The total area is then
Calculate the (positive) area between f ( x ) = sin x and the x -axis for −2π ≤ x ≤ 2π.
We observe that f ( x ) = sin x ≥ 0 for −2π ≤ x ≤ −π and 0 ≤ x ≤ π. Likewise, f ( x ) = sin x ≤ 0 for −π ≤ x ≤ 0 and π ≤ x ≤ 2π . As a result
This is easily calculated to equal
2 + 2 + 2 + 2 = 8.
You Try It: Calculate the (positive) area between y = x 3 − 6 x 2 + 11 x − 6 and the x -axis.
Calculate the signed area between the graph of y = cos x + 1/2 and the x -axis, − π/2 ≤ x ≤ π.
This is easy, because the solution we seek is just the value of the integral:
Math Note: In the last example, we have counted positive area as positive and negative area as negative. Our calculation shows that the aggregate area is positive. We encourage the reader to draw a graph to make this result plausible.
You Try It: Calculate the actual positive area between the graph of y = x2 − 4, −5 ≤ x ≤ 5 and the x -axis.
You Try It: Calculate the signed area between the graph of y = x2 − 4 and the x -axis, −4 ≤ x ≤ 5.
Find practice problems and solutions for these concepts at: The Integral Practice Test.
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