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Study Guides > Business Calculus

Reading: The Definite Integral Applied to Area

We have already used integrals to find the area between the graph of a function and the horizontal axis. Integrals can also be used to find the area between two graphs. If f(x) ≥ g(x) for all x in [a,b], then we can approximate the area between f and g by partitioning the interval [a,b] and forming a Riemann sum (figure 1). The height of each rectangle is top – bottom, f(ci) – g(ci) so the area of the ith rectangle is (height) × (base) = {f(ci) – g(ci)} × ∆x . This approximation of the total area is a Riemann sum.
Figure 1 Figure 1
The limit of this Riemann sum, as the number of rectangles gets larger and their width gets smaller, is the definite integral [latex] \int_{a}^{b} [f(x) - g(x)]dx [/latex]. The area between two curves f(x) and g(x), where f(x) ≥ g(x), between x = a and x = b is [latex] \int_{a}^{b}(f(x) - g(x)) dx [/latex] The integrand is “top – bottom.” Make a graph to be sure which curve is which.


Find the area bounded between the graphs of f(x) = x and g(x) = 3 for 1 ≤ x ≤ 4. (figure 2)
Figure 2 Figure 2


Always start with a graph so you can see which graph is the top and which is the bottom. In this example, the two curves cross, and they change positions; we’ll need to split the area into two pieces. Geometrically, we can see that the area is 2 + ½ = 2.5. Writing the area as a sum of definite integrals, we get: Area = [latex] \int_{1}^{3}(3 -x)dx + \int_{3}^{4}(x-3)dx [/latex] These integrals are easy to evaluate using antiderivatives: [latex-display] \int_{1}^{3}(3 -x)dx = (3x - \frac{x^2}{2})_{0}^{3} = ((9 -\frac{9}{2}) - (15 - \frac{25}{2})) = 2 [/latex-display] [latex-display] \int_{3}^{4}(x-3)dx = (\frac{x^2}{2} - 3x)_{3}^{4} = ((\frac{16}{2} - 12) - (\frac{9}{2} - 9)) = \frac{1}{2} [/latex-display] The two integrals also tell us that the total area between f and g is 2.5 square units, which we already knew. Note that the single integral [latex] \int_{1}^{4}(3 - x)dx = 1.5 [/latex] is not the area we want in this problem. The value of the integral is 1.5, and the value of the area is 2.5. That’s because for the triangle on the right, the graph of y = x is above the graph of y = 3, so the integrand 3 – x is negative; in the definite integral, the area of that triangle comes in with a negative sign. In this example, it was easy to see exactly where the two curves crossed so we could break the region into the two pieces to figure separately. In other examples, you might need to solve an equation to find where the curves cross.


Two objects start from the same location and travel along the same path with velocities [latex] v_A(t) = t+3 [/latex] and [latex] v_B(t) = t^2 - 4t + 3 [/latex] meters per second (figure 3). How far ahead is A after 3 seconds?
Figure 3 Figure 3


Since , the "area" between the graphs of  and represents the distance between the objects. After 3 seconds, the distance apart [latex-display] =\int_{0}^{3}(v_A(t) - v_B(t))dt = \int_{0}^{3}((t+3)-(t^2 - 4t + 3))dt = \int_{0}^{3}(5t - t^2)dt [/latex-display] [latex] =(\frac{5}{02}t^2 - \frac{t^3}{3})_{0}^{3} = (\frac{5}{2} \times 9 - \frac{27}{3}) - (0) = 13.5 [/latex] meters.

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  • Business Calculus. Provided by: Washington State Colleges Authored by: Dale Hoffman and Shana Calaway. Located at: https://docs.google.com/file/d/0B1lkHWwO61QEM0gwOFhES2N5Tlk/edit. License: CC BY: Attribution.