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Chapter 6: Techniques of Integration Winter 2016 Department of Mathematics Hong Kong Baptist University 1 / 30 §6.1 Integration by Parts In §5.6, we have introduced the method of substitution. Recall that the method of substitution can be regarded as inverse to the Chain rule for differentiation. In this section, we introduce another general method for anti-differentiation, Integration by Parts. The method for integration by parts is inverse to the Product rule for differentiation. 2 / 30 Suppose that U(x) and V (x) are two differentiable functions. According to the Product rule, d dV dU (U(x)V (x)) = U(x) + V (x) . dx dx dx Or equivalently, U(x) dV d dU = (U(x)V (x)) − V (x) . dx dx dx Integrating both sides of the above equation, we obtain Z U(x) dV dx = U(x)V (x) − dx Z V (x) dU dx. dx This leads to the so-called Integration by Parts, Z Z UdV = UV − VdU. 3 / 30 To apply the method, we first break up the given integrand into a product of two pieces, U and V 0 , then by Integration by Parts, Z Z UV 0 dx = UV − where R VU 0 dx, VU 0 dx is usually a simpler integral than R UV 0 dx. An improper choice of U and V 0 can result in making the integral more difficult rather than easier. As a common practice, we always look for a factor of the integrand that is easily integrated, and include dx with that factor to make up V 0 dx = dV . The remaining factor of the integrand is then assigned to be U. 4 / 30 Z Example 1: Find the integral xe x dx. Approach 1: Let U = e x and xdx = dV . Then dU = e x dx and V = x 2 /2. By Integration by Parts, Z Z x2 x xe dx = e x d( ) 2 Z 2 2 x x x = e − d(e x ) 2 2 Z 2 x2 x x x = e − e dx. 2 2 Z Note that the new integral to be evaluated, VdU, is becoming more complicated. 5 / 30 Approach 2: Let U = x and e x dx = dV . Then dU = dx and V = e x . Using Integration by Parts, Z Z x xe dx = x · d(e x ) Z x = xe − e x dx = xe x − e x + C . 6 / 30 The following are two useful rules of thumb for choosing U and V : (i) If the integrand involves a polynomial multiplied by an exponential function, a sine, a cosine, or some other readily integrable function, try U equals the polynomial an dV equals the rest. (ii) If the integrand involves a logarithm, an inverse trigonometric function, or some other function that is not readily integrable but whose derivative is readily calculated, try that function for U and let dV equal the rest. Of course, these “rules” come with no guarantee. 7 / 30 Z Example 2: Find the integral x · cos(x)dx. Solution: Following the rule of thumb, we let U = x and cos(x)dx = dV . Then dU = dx and V = sin(x). Using Integration by Parts, Z Z x · cos(x)dx = x · d(sin(x)) Z = x · sin(x) − sin(x)dx = x · sin(x) + cos(x) + C . 8 / 30 Z Example 3: Find the integral x 2 ln(x)dx. Solution: Following the rule of thumb, we let U = ln(x) and x 2 dx = dV . 1 Then dU = dx and V = x 3 /3. Using Integration by Parts, x Z Z x3 2 x ln(x)dx = ln(x)d( ) 3 Z 1 3 1 3 = x ln(x) − x d(ln(x)) 3 3 Z 1 3 1 2 = x ln(x) − x dx 3 3 1 3 1 = x ln(x) − x 3 + C . 3 9 9 / 30 Z Example 4: Find the integral I = e x sin(x)dx. Solution: Using Integration by Parts, Z I = e x sin(x)dx Z e x d(− cos(x)) Z x = −e cos(x) + e x cos(x)dx Z = −e x cos(x) + e x d(sin(x)) Z = −e x cos(x) + e x sin(x) − e x sin(x)dx = = −e x cos(x) + e x sin(x) − I . 1 This leads to I = (e x sin(x) − e x cos(x)) + C . 2 10 / 30 Definite Integral: If we want to evaluate a definite integral by the method of integration by parts, we must remember to include the appropriate evaluation symbol with the integrated term. 11 / 30 Z Example 5: Find the integral 2 xe −x dx. 0 Solution: Using Integration by Parts, we have Z 2 Z 2 −x xe dx = − xd(e −x ) 0 0 2 Z 2 = −xe −x + e −x dx 0 0 2 −2 = −(2e − 0) − e −x 0 = 1 − 3e −2 . 12 / 30 Z Example 6: Find the integral e x 3 ln(x)dx. 1 Solution: Using Integration by Parts, we have Z e Z e x4 3 x ln(x)dx = ln(x)d( ) 4 1 1 e Z e x 4 x4 = ln(x) − d(ln(x)) 4 1 1 4 Z e4 1 e 3 = − x dx 4 4 1 e 4 x 4 e = − 4 16 1 4 1 3e + . = 16 16 13 / 30 §6.3 Inverse Substitutions We saw in §5.6 that the method of substitution can be used to transform Z Z f (g (x))g 0 (x) dx into f (u) du by letting u = g (x). In however, the more “complicated” integral R some cases, f (g (x))g 0 (x) dx is actually simpler to evaluate. Thus, it is R sometimes useful to evaluate f (u) du by making the inverse substitution u = g (x). Here, we will only consider trigonometric substitutions to √ √ 2 2 2 2 deal with integrands of the type a − x and a + x . (The other substitutions in the textbook are optional.) 14 / 30 Z Example 7: Evaluate √ dx . 4 − x2 Solution: Letting x = 2 sin θ, Z Z dx √ = 4 − x2 Z = Z = we have dx = 2 cos θdθ, so that 2 cos θ q dθ 4(1 − sin2 θ) 2 cos θ dθ 2 cos θ 1 dθ = θ + C = arcsin x 2 + C. 15 / 30 Z Example 8: Evaluate 1/2 p 1 − 4x 2 dx. 1/4 Solution: Letting x = Z 1/2 p 1 2 sin θ, we have dx = 1 − 4x 2 dx = 1/4 Z 1 2 π/2 p 1 − sin2 θ · π/6 1 = 2 = 1 4 Z cos θdθ, so that 1 cos θ dθ 2 π/2 cos2 (θ) dθ π/6 Z π/2 (1 + cos(2θ)) dθ π/6 √ π/2 1 1 π 3 = θ + sin(2θ) = − . 4 2 12 16 π/6 16 / 30 √ Integrals involving a2 − x 2 can often be simplified by letting x = a sin(θ), or equivalently, θ = arcsin(x/a). We can rewrite any other trigonometric functions that appear by noting that cos(θ) = 1p 2 a − x 2. a √ 1 Integrals involving a2 + x 2 or a2 +x 2 can often be simplified by letting x = a tan(θ), or equivalently, θ = arctan(x/a). We can rewrite any other trigonometric functions that appear by √ 1 2 2 noting that sec(θ) = a a + x , so that tan(θ) x =√ , 2 sec(θ) a + x2 a 1 cos(θ) = =√ . 2 sec(θ) a + x2 sin(θ) = 17 / 30 Z Example 9: Evaluate x +1 dx. x2 + 9 Solution: Note that Z Z Z x +1 x dx dx = dx + . 2 2 2 x +9 x +9 x +9 For the first integral, we let u = x 2 + 9, du = 2x dx. For the second integral, we let x = 3 tan(θ), dx = 3 sec2 (θ) dθ: Z Z Z x +1 1 du 3 sec2 (θ) dx = + dθ x2 + 9 2 u 9(1 + tan2 (θ)) 1 1 = ln |u| + θ + C 2 3 x 1 1 = ln(x 2 + 9) + arctan + C. 2 3 3 18 / 30 §6.5 Improper Integrals Rb For the integral I = a f (x)dx considered previously, we have assumed that f is continuous on the closed, finite interval [a, b]. Since such a function is necessarily bounded, the integral I is also necessarily a finite number. Such integrals are called proper integrals. In this section, we generalize the definite integral to allow for two possibilities excluded in the situation described above: (i) We may have a = −∞ or b = ∞ or both (referred to as improper integrals of type I); (ii) f may be unbounded as x approaches a or b or both (referred to as improper integrals of type II). 19 / 30 Example 10: Find the area of the region A lying under the curve y = 1/x 2 and above the x-axis to the right of x = 1. Solution: The area can be expressed as Z ∞ 1 A= dx, 2 x 1 which is an improper integral of type I. To evaluate this improper integral, we interpret it as a limit of proper integrals over intervals [1, R] as R → ∞. Then Z A= 1 ∞ 1 dx x2 Z = = R lim R→∞ 1 lim (− R→∞ 1 1 R dx = lim (− ) R→∞ x2 x 1 1 + 1) = 1. R 20 / 30 21 / 30 Example 11: Find the area of the region under y = 1/x, above y = 0, and to the right of x = 1. Solution: The area is given by the improper integral of type I, A = = ∞ Z R 1 1 dx = lim dx R→∞ 1 x x 1 R lim ln(x) = lim ln(R) = ∞. Z R→∞ 1 R→∞ Note that this improper integral diverges to infinity. Comparing to y = 1/x 2 , the “spike” of y = 1/x for x > 1 is much thicker. Such a difference makes the region has infinite area. 22 / 30 23 / 30 Definition (Improper integrals of type I) If f is continuous on [a, ∞), we define the improper integral of f over [a, ∞) as a limit of proper integrals: ∞ Z Z R f (x)dx = lim f (x)dx. R→∞ a a Similarly, if f is continuous on (−∞, b], we define Z b Z f (x)dx = −∞ lim b R→−∞ R f (x)dx. In either case, if the limit exists and is a finite number, we say that the improper integral converges; if the limit does not exist, we say that the improper integral diverges. If the limit is ∞ (or −∞), we say that the improper integral diverges to infinity (or diverges to negative infinity). 24 / 30 Z ∞ Example 12: Evaluate the integral e −|x| dx. −∞ Solution: Note that this is an improper integral of type I at both endpoints. We usually break it into two separate integrals for this kind of integrals. Z ∞ Z 0 Z ∞ −|x| −|x| e dx = e dx + e −|x| dx −∞ −∞ 0 Z R e −x dx R = 2 lim (−e −x ) = 2 lim R→∞ 0 R→∞ = 2 lim (1 − e R→∞ 0 −R ) = 2. 25 / 30 Z ∞ cos(x)dx. Example 13: Evaluate the integral 0 Solution: For this improper integral, we have Z ∞ Z cos(x)dx = 0 = lim R→∞ 0 R cos(x)dx lim sin(R). R→∞ Note that the limit does not exist (and it is not ∞ or −∞). So all we can say is that the given integral diverges. 26 / 30 Definition (Improper Integrals of Type II) If f is continuous on the interval (a, b] and is possibly unbounded near a, we define the improper integral Z b Z f (x)dx = lim a c→a+ c b f (x)dx. Similarly, if f is continuous on [a, b) and is possibly unbounded near b, we define Z b Z c f (x)dx = lim f (x)dx. a c→b− a These improper integrals may converge, diverge, diverge to infinity, or diverge to negative infinity. 27 / 30 Example 14: √ Find the area of the region S lying under y = 1/ x, above the x-axis, between x = 0 and x = 1. Solution: The area A is given by Z 1 1 √ dx, A= x 0 which is an improper integral of type II since y is unbounded near x = 0. By definition, we have Z 1 1 A = lim x −1/2 dx = lim 2x 1/2 c→0+ c c→0+ c √ = lim (2 − 2 c) = 2. c→0+ This shows that the integral converges and S has a finite area of 2 square units. 28 / 30 29 / 30 Z Example 15: Evaluate the integral f (x) = √ 1/ x x −1 2 f (x)dx, where 0 if 0 < x ≤ 1 if 1 < x ≤ 2. Solution: Note that f (x) is a piecewise continuous function as in §5.4. We have Z 2 Z 1 Z 2 1 √ dx + f (x)dx = (x − 1)dx x 0 0 1 Z 1 2 1 1 √ dx + ( x 2 − x) = lim c→0+ c 2 1 x 1 = 2 + (2 − 2 − + 1) 2 5 = . 2 30 / 30