Properties of Definite Integrals

Properties of Definite Integrals We have seen that a definite integral represents the area underneath a function over a given interval. There are numerous useful properties of definite i ntegrals w orth s tudying, s o t hat w e c an become adept at using and manipulating them. Suppose f and g are both Riemann integrable functions. In light of the fundamental theorem of calculus, and that b Z f (x)dx = F (b) − F (a), a for F and antiderivative of f , we must have that Z b Z a f (x)dx, f (x)dx = F (a) − F (b) = − a b which in a sense states that by integrating in the opposite direction, or backwards, we pick up a negative sign. Let’s also notice that if we integrate over an interval of 0 length, then we must find 0 area, so Z a f (x)dx = 0. a Just as we constant multiple and sum rules for indefinite integrals, they must hold for definite integrals, so Z b Z b kf (x)dx = k f (x)dx, a and Z a b Z (f (x) + g(x))dx = a b Z f (x)dx + a b g(x)dx. a To illustrate the first property, suppose you are finding the area of a square. If you double the length of one of the sides, then the original square will fit twice inside the enlarged one, so the area of the enlarged square is twice that of the original square. The second property says that if we split an object into two pieces, the sum of the area should be the area of the sum. In a similar vein, it follows that Z Z Z c b f (x)dx = a c f (x)dx + a f (x)dx, b where here we are simply splitting the area in a different way (using a vertical rather than horizontal line). Finally, if f (x) ≥ g(x) on (a,b), then it must follow that there is more area under f (x), so Z b Z f (x)dx ≥ a b g(x)dx. a A special case of this property is looking at the rectangles formed by the minimum and maximum values of f (or least upper bounds and greatest lower bounds). The rectangle formed by the max of f must have a greater area than under f , which must have a greater area than under the minimum value of f . We summarize all of these results in the following theorem. Theorem 0.1.1 (Properties of the Definite Integral). Let f and g be Riemann integrable functions, k ∈ R. It follows that Z a Z b 1. f (x)dx = − f (x)dx b a 1 Z a f (x)dx = 0 2. a Z b Z b f (x)dx kf (x)dx = k 3. a a Z b b Z c Z b Z g(x)dx b Z f (x)dx ≥ c f (x)dx b a a 6. Z f (x)dx + f (x)dx = 5. b a a a Z Z f (x)dx + (f (x) + g(x))dx = 4. b if f (x) ≥ g(x), x ∈ (a, b) g(x)dx, a a Using the above rules, we can now integrate a wider variety of functions. For instance, we are now capable of dealing with piecewise-defined functions. Z 4 Example 1. Evaluate f (x)dx for 0  0≤x<2  x 4 2≤x<3 f (x) =  3x2 3 ≤ x Solution To solve this problem we need to split the integral into multiple pieces, according to where the function is defined differently. Z 4 Z 2 Z 3 Z 4 3 4 x2 2 + 4x + x3 = (2 − 0) + (12 − 8) + (64 − 9) = 61. f (x)dx = xdx + 4dx + 3x2 dx = 2 0 2 3 0 0 2 3 Z 5 Example 2. Evaluate |x|dx. −5 Solution We do not know how to directly integrate |x|, but we can write it in terms of two functions we do now how to integrate. Using the fact that we can split the integral over the limits of integration, we find that Z 5 Z 0 Z 5 Z 0 Z 5 |x|dx = |x|dx + |x|dx = (−x)dx + xdx −5 −5 x2 0 = − 2 −5 + 0 2 x 5 2 0 −5 =− 02 2 − −52 2 + 52 2 − 0 02 2 =2· 52 = 52 = 25. 2 This is really part of a more general phenomenon. When we have a function which is symmetric about the y-axis, and we integrate over an interval [−a, a], we find a value that is twice the integral from [0, a]. For this reason we can see that Z π/2 Z π/2 π/2 = 2(sin(π/2) − sin(0)) = 2. cos(x)dx = 2 cos(x)dx = 2 sin(x) −π/2 0 0 We have previously said that the definite integral finds the ‘signed’ area underneath a function. We know that functions like the sine and cosine function have certain symmetries about the x-axis that cause their definite integrals to be 0 over certain limits. If we want to find the amount of conventional area between such a function and the x-axis, we need to integrate the magnitude of the function. 2 Example 3. Find the conventional area between sin(x) and the x-axis over [0, 2π]. Solution In order to solve this problem we integrate | sin(x)|. However, we do not know how to integrate such a function, so we need to break it up into intervals where we know sine is positive and negative, so we can remove the absolute value sign. Z 2π Z π Z 2π 2π π | sin(x)|dx = sin(x)dx + (− sin(x))dx = − cos(x) + cos(x) 0 0 0 π π = − cos(π) + cos(0) + cos(2π) − cos(π) = 1 + 1 + 1 + 1 = 4. 3