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Indefinite Integrals Arkansas Tech University MATH 2914: Calculus I Dr. Marcel B. Finan 5.3 Evaluating Definite Integrals: Indefinite Integrals In the previous section, we used Riemann sums to evaluate definite integrals. As you can tell, Riemann sums become hard to work with when the integrand function is complex. Therefore, a more convenient way for finding definite integrals is needed. This is provided by the following result. Theorem 5.3.1 (First Fundamental Theorem of Calculus) If f is continuous in [a; b] and F 0 = f then Z b f(x)dx = F (b) − F (a): a Recall that if s(t) is the position function of a moving object at time t then its velocity is given by v(t) = s0(t): Thus, we can apply the above theorem to find the displacement between the times a and b which is given by Z b v(t)dt = s(b) − s(a): a Recall that displacement can be positive (object moving to the right or upward) or negative (object moving to the left or downward). To find the total distance traveled, we compute Z b jv(t)jdt: a Remark 5.3.1 R b When using a f(x)dx in applications then its units is the product of the units of f(x) with the units of x: Example 5.3.1 R 2 Compute 1 2xdx: Solution. Since the derivative of x2 is 2x; F (x) = x2: Thus, we have Z 2 2xdx = F (2) − F (1) = 4 − 1 = 3 1 Indefinite Integrals Recall from Section 4.7 that a function f might have many antiderivatives. The family of all antiderivatives is given by F (x) + C where F 0(x) = f(x): 1 The notation of the general antiderivative is called an indefinite integral and is written Z f(x)dx = F (x) + C: The symbol R is the symbol of integration, f(x) is called the integrand and C is called the constant of integration. Keep in mind the relationship between f(x) and F (x) which is given by F 0(x) = f(x): Warning: The indefinite integral is a short-hand notation for a family of functions F (x) + C with the property F 0(x) = f(x) for all x: It is not to R b be confused with the definite integral a f(x)dx which is a real number. Example 5.3.2 Show that R 0dx = C. Solution. Since the derivative of a constant function is always zero, we can write Z 0dx = C Example 5.3.3 Show that R kdx = kx + C where k is a constant. Solution. Since the derivative of kx is just k; we have Z kdx = kx + C Example 5.3.4 R n xn+1 Show that x dx = n+1 + C for n 6= −1: Solution. xn+1 0 n By the power rule, if F (x) = n+1 then F (x) = x : Thus, Z xn+1 xndx = + C n + 1 x0 Note that this formula is valid only if n 6= −1 for if n = −1 we would have 0 which doesn't make sense. The case n = −1 is treated in the next example. 2 Example 5.3.5 Show that Z dx = ln jxj + C: x Solution. Suppose first that x > 0 so that ln jxj = ln x: Then (ln jxj)0 = (ln x)0 = 1 0 x : Now, if x < 0 then ln jxj = ln (−x) and by the chain rule (ln jxj) = 0 −1 1 0 1 (ln (−x)) = −x = x : Thus, in both cases (ln jxj) = x Example 5.3.6 R ax eax Show that for a 6= 0, e dx = a + C: Solution. eax 0 ax If a is a nonzero constant and F (x) = a then F (x) = e : This shows that Z eax eaxdx = + C a Example 5.3.7 R sin (ax) Show that cos (ax)dx = a + C where a 6= 0: Solution. sin (ax) 0 If a is a nonzero constant and F (x) = a then F (x) = cos (ax): This shows that Z sin (ax) cos (ax)dx = + C a Example 5.3.8 R cos (ax) Show that sin (ax)dx = − a + C where a 6= 0: Solution. cos (ax) 0 If a is a nonzero constant and F (x) = − a then F (x) = sin (ax): This shows that Z cos (ax) sin (ax)dx = − + C a Example 5.3.9 Show that R p 1 dx = arcsin x + C = − arccos x + C: 1−x2 Solution. Let F (x) = arcsin x: Then F 0(x) = p 1 : Thus, 1−x2 Z 1 p dx = arcsin x + C = − arccos x + C 1 − x2 3 Example 5.3.10 R 1 Show that 1+x2 dx = arctan x + C: Solution. 0 1 Let F (x) = arctan x: Then F (x) = 1+x2 : Thus, Z 1 dx = arctan x + C 1 + x2 Properties of Indefinite Integrals Z Z Z [f(x) ± g(x)]dx = f(x)dx ± g(x)dx: To see why this property is true, let F (x) be an antiderivative of f(x) and G(x) be an antiderivative of g(x): The result follows from the fact that d dx [F (x) ± G(x)] = f(x) ± g(x): Z Z cf(x)dx = c f(x)dx: To see this, suppose that F (x) is an antiderivative of f(x): Then R f(x)dx = d F (x)+C: But dx (cF (x)) = cf(x) so that cF (x) is an antiderivative of cf(x), that is, R cf(x)dx = cF (x) + C0: This implies Z Z Z Z cf(x)dx = cF (x)+C0 = c f(x)dx − C +C0 = c f(x)dx−cC+C0 = c f(x)dx: Note that the constant −cC + C0 is ignored since a constant of integration will result from R f(x)dx: Example 5.3.11 Find Z 3 5 sin (2x) − e−3x + − dx: x x3 Solution. Using the linearity property of indefinite integrals together with the formulas of integration obtained above we have Z 3 5 Z Z Z dx Z sin (2x) − e−3x + − dx = sin (2x)dx − e−3xdx + 3 − 5 x−3dx x x3 x cos (2x) e−3x 5 = − + + 3 ln jxj + + C 2 3 2x2 Once we have found an antiderivative of f(x), computing definite integrals is easy by the Fundamental Theorem of Calculus. 4 Example 5.3.12 R 2 2 Compute 1 3x dx: Solution. Since F (x) = x3 is an antiderivative of f(x) = 3x2; by FTC we can write Z 2 2 3 2 3 3 3x dx = x j1 = 2 − 1 = 7 1 Finally, we end this section by pointing out that not every funtion has a simple analytical antiderivative(i.e., an antiderivative that can be expressed in terms of elementary functions). For example, the function Z x sin t Si(x) = dt 0 t sin x is an antiderivative of the function x . There is no way of expressing Si(x) in terms of elementary functions. 5.
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