Indefinite Integrals

Indefinite Integrals Integration is a formalization of the process of antidifferentiation. When we differentiate a function, we lose some information about the function - although we know how it is changing we don’t know where it started from. Thus, when we go through the reverse process of differentiation, we end up with a family of functions rather than just a single function. Because of this we need an initial value to find a unique solution to a differential e quation. We define the indefinite integral in order to encompass the fact that we have a family of functions. Definition 0 .1.1 ( Indefinite In tegral). We ca ll th e se t of al l an tiderivatives of f th e indefinite integral of f , denoted by Z f (x)dx. R The symbol is an integral sign. The function f is the called the integrand and x is the variable of integration. We say that dx is a differential of x. If we know any antiderivative F to a given function f , then we simply find the indefinite integral as Z f (x)dx = F (x) + c, because the expression on the right-hand side represents all possible antiderivatives of f (x), if we let c be an arbitrary constant. When we find the indefinite integral of a function f (x) we say that we integrate the integrand f (x). Thus, the process of evaluating an integral is referred to as integration. Unfortunately, the notation we have introduced for the indefinite integral probably looks rather arcane. As we move further into the study of integration, the reason for the notation will become more clear. For the moment let us note that this notation is based on the initial developments ofRcalculus, where the notion of infinitesimal and differential were prevalent. The resemblance of and an elongated S is not accidental. This notation reflects the relationship between integration and summation. Integration is essentially the summation of the area underneath the integrand f (x) over intervals of infinitesimal length. The differential dx represents an infinitesimal change in x, which represents the intervals of infinitesimal length over which the summation involved in integration occurs. When it will not lead to confusion we will refer to the indefinite integral as simply the integral. The term indefinite integral is used to distinguish the process of indefinite and definite integration (which we will consider soon). Although both of these concepts are related to the area underneath a function, the indefinite integral is a function, whereas the definite integral is a constant, which is given by the area underneath a function over a set interval (defined by the limits of integration, which are not present in an indefinite integral). Since the process of (indefinite) integration is an inverse to differentiation, we can derive many rules for integration using rules we already know for differentiation. For instance d n+1 x = (n + 1)xn dx so we see that d  xn+1  = xn dx n + 1 This leads us to the product rule for integrals. forn 6= −1. Theorem 0.1.1 (Power Rule for Integrals). Z xn+1 xn dx = +c n+1 for every n 6= −1 (n ∈ R \ {−1}). Example 1. Evaluate the indefinite integral of x2 . Solution In this case we use the product rule, to see that Z x3 x2+1 +c= + c. x2 = 2+1 3 √ Example 2. Evaluate the indefinite integral of x. √ Solution Once we rewrite x = x1/2 we see that Z x1/2 = x1/2+1 2 + c = x3/2 + c. 1/2 + 1 3 Example 3. Evaluate the indefinite integral of x−3 . Solution We find that Z 1 x−3+1 + c = − x−2 + c x−3 = −3 + 1 2 R Example 4. Evaluate dt. Solution Rewriting the integrand we find Z Z t0+1 + c = t + c. dt = t0 dt = 0+1 Just like differentiation, integration is a linear operation. What this means is that integration satisfies the following two properties. Theorem 0.1.2 (Linearity of Integration). Let a ∈ R. Z Z 1. αf (x)dx = α f (x)dx. Z 2. Z (f + g)(x)dx = Z f (x)dx + g(x)dx. Example 5. Evaluate the indefinite integral of −2x−3 + 4x1/2 . Solution We can use the constant product rule and sum rules in conjunction to Z Z Z 1 2 −2x−3 + 4x1/2 = −2 x−3 + 4 x1/2 = −2(− x−2 ) + 4( x3/2 ) + c = x−2 + 2 3 find 8 3/2 x +c 3 In the above analysis we do not write the result as −2c + 4d, because we can easily enough choose c and d so that we have any value for the arbitrary constant. Thus, it is cleaner to just replace −2c + 4d with a single c, as both are capable of representing any arbitrary constant. Corresponding to other differentiation rules we have learned, we can evaluate a number of other integrals, such as Z ex dx = ex + c Z cos(x)dx = sin(x) + c Z sin(x)dx = − cos(x) + c Note that the negative sign corresponds to the integral of the sine function, just as the negative sign corresponds to the derivative of cosine. This list is far from exhaustive, as we have results for other trigonometric functions as well as the inverse trigonometric functions. Rather than repeating all of these results here we simply note that the corresponding rule for integration can be read directly from the result for differentiation. We’d also like to note Z 1 dx = ln |x| + c. x In the above formula we use the magnitude of x, |x|, because although x1 is defined for negative x, ln(x) is not. In fact, the natural logarithm is usually defined in terms of the above integral, and then can be shown to be the inverse of the exponential function. However, since we did not have access to integration at the time we introduced the natural logarithm, we defined it as the inverse function, and now see that it satisfies the above integral. For x > 0 d d 1 ln |x| = ln(x) = . dx dx x If we consider x < 0 then d d 1 1 ln |x| = ln(−x) = · (−1) = . dx dx −x x We cannot evaluate the above derivative at x = 0, because ln(x) is undefined for x = 0. Z   1 Example 6. Evaluate + 2ex + 3 cos(x) dx x Solution We can evaluate the above integral using the sum and constant product rules in conjunction with the integrals we have just discussed. Z  Z Z Z  1 1 x x + 2e + 3 cos(x) dx = dx + 2 e dx + 3 cos(x)dx = ln |x| + 2ex + 3 sin(x) + c. x x