Calculus II. George Voutsadakis 1. LSSU Math 152. Lake Superior State University. 1 Mathematics and Computer Science

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1 Calculus II George Voutsadakis Mathematics and Computer Science Lake Superior State University LSSU Math 52 George Voutsadakis (LSSU) Calculus II February 205 / 88

2 Outline Techniques of Integration Integration by Parts Trigonometric Integrals Trigonometric Substitution Hyperbolic and Inverse Hyperbolic Functions The Method of Partial Fractions Improper Integrals Numerical Integration George Voutsadakis (LSSU) Calculus II February / 88

3 Integration by Parts Subsection Integration by Parts George Voutsadakis (LSSU) Calculus II February / 88

4 Integration by Parts Integration By Parts Recall the Product Rule for Derivatives: (f(x)g(x)) = f (x)g(x)+f(x)g (x); Integrateboth sides with respect to x: (f(x)g(x)) dx = [f (x)g(x)+f(x)g (x)]dx; Since integration is the reverse operation of differentiation, we have (f(x)g(x)) dx = f(x)g(x); Moreover, because of the sum rule for integrals: [f (x)g(x)+f(x)g (x)]dx = Putting all these together: f(x)g(x) = f (x)g(x)dx + f (x)g(x)dx + f(x)g (x)dx; f(x)g (x)dx; Finally, subtract to get the Integration By Parts Formula: f (x)g(x)dx = f(x)g(x) f(x)g (x)dx; George Voutsadakis (LSSU) Calculus II February / 88

5 Integration by Parts Alternative Form We came up with the formula f (x)g(x)dx = f(x)g(x) Use two new variables u and v as follows: Set f(x)g (x)dx; u = g(x) v = f(x) du = g (x)dx dv = f (x)dx Now substitute into the formula above to get the uv-form of the By Parts Rule: udv = uv vdu; George Voutsadakis (LSSU) Calculus II February / 88

6 Integration by Parts Example I (fg-form) f (x)g(x)dx = f(x)g(x) f(x)g (x)dx; x cosxdx = x(sinx) dx = x sinx (x) sinxdx = x sinx sin xdx = x sinx ( cosx)+c; = x sinx +cosx +C; George Voutsadakis (LSSU) Calculus II February / 88

7 Integration by Parts Example I (uv-form) udv = uv vdu; We want to compute x cosxdx; Set u = x and dv = cosxdx; Then du dx = du = dx; Moreover dv dx = cosx v = sinx; x cosxdx = udv = uv vdu = x sinx sin xdx = x sinx +cosx +C; George Voutsadakis (LSSU) Calculus II February / 88

8 Integration by Parts Example II (fg-form) f (x)g(x)dx = f(x)g(x) f(x)g (x)dx; xe x dx = x(e x ) dx = xe x (x) e x dx = xe x e x dx = xe x e x +C; George Voutsadakis (LSSU) Calculus II February / 88

9 Integration by Parts Example II (uv-form) udv = uv vdu; We want to compute xe x dx; Set u = x and dv = e x dx; Then du dx = du = dx; Moreover dv dx = ex v = e x ; xe x dx = udv = uv vdu = xe x e x dx = xe x e x +C; George Voutsadakis (LSSU) Calculus II February / 88

10 Integration by Parts Example III (fg-form) f (x)g(x)dx = f(x)g(x) f(x)g (x)dx; x 7 lnxdx = ( 8 x8 ) lnxdx = 8 x8 lnx 8 x8 (lnx) dx = 8 x8 lnx x 8 8 x dx = 8 x8 lnx x 7 dx 8 = 8 x8 lnx 8 8 x8 +C = 8 x8 lnx 64 x8 +C; George Voutsadakis (LSSU) Calculus II February / 88

11 Integration by Parts Example III (uv-form) udv = uv vdu; We want to compute x 7 lnxdx; Set u = lnx and dv = x 7 dx; Then du dx = x du = dx; Moreover x dv dx = x7 v = 8 x8 ; x 7 lnxdx = udv = uv vdu = 8 x8 lnx = 8 x8 lnx 64 x8 +C; 8 x8 x dx = 8 x8 lnx 8 x 7 dx George Voutsadakis (LSSU) Calculus II February 205 / 88

12 Integration by Parts Example IV: Applying By Parts Twice x 2 cosxdx = x 2 (sinx) dx = x 2 sinx (x 2 ) sinxdx = x 2 sinx 2x sinxdx = x 2 sinx 2x( cosx) dx = x 2 sinx [ 2x cosx (2x) ( cosx)dx] = x 2 sinx +2x cosx 2cosxdx = x 2 sinx +2x cosx 2sinx +C; George Voutsadakis (LSSU) Calculus II February / 88

13 Integration by Parts Example V: Applying By Parts Twice x 2 e 3x dx = x 2 ( 3 e3x ) dx = x 2 3 e3x (x 2 ) 3 e3x dx = x 2 3 e3x 2x 3 e3x dx = x 2 3 e3x 2x( 9 e3x ) dx = x 2 3 e3x [2x 9 e3x (2x) 9 e3x dx] = x 2 3 e3x 2x 9 e3x e3x dx = 3 x2 e 3x 2 9 xe3x e3x +C; George Voutsadakis (LSSU) Calculus II February / 88

14 Integration by Parts Example VI: Integral of lnx f (x)g(x)dx = f(x)g(x) f(x)g (x)dx; ln xdx = (x) lnxdx = x lnx x(lnx) dx = x lnx x x dx = x lnx dx = x lnx x +C; George Voutsadakis (LSSU) Calculus II February / 88

15 Integration by Parts Example VII: Returning to the Original Form e x cosxdx = (e x ) cosxdx = e x cosx e x (cosx) dx = e x cosx + e x sinxdx = e x cosx + (e x ) sinxdx = e x cosx +e x sinx e x (sinx) dx = e x cosx +e x sinx e x cosxdx Thus, e x cosxdx = e x (cosx +sinx) e x cosxdx, and, hence, 2 e x cosxdx = e x (cosx+sinx) e x cosxdx = 2 ex (cosx+sinx)+c; George Voutsadakis (LSSU) Calculus II February / 88

16 Integration by Parts Example VIII f (x)g(x)dx = f(x)g(x) f(x)g (x)dx; (x 2)(x +4) 8 dx = (x 2)[ 9 (x +4)9 ] dx = 9 (x 2)(x +4)9 (x 2) 9 (x +4)9 dx = 9 (x 2)(x +4)9 (x +4) 9 dx 9 = 9 (x 2)(x +4)9 9 0 (x +4)0 +C = 9 (x 2)(x +4)9 90 (x +4)0 +C; George Voutsadakis (LSSU) Calculus II February / 88

17 Integration by Parts Example IX: A Definite Integral By Parts π/4 0 x sin2xdx = π/4 0 x( 2 cos2x) dx = 2 x cos2x π/4 π/4 0 0 π/4 = 2 x cos2x π/ (x) ( 2 cos2x)dx 2 cos2xdx = 2 x cos2x π/ sin2x π/4 0 = (0 0)+( 4 0) = 4 ; George Voutsadakis (LSSU) Calculus II February / 88

18 Trigonometric Integrals Subsection 2 Trigonometric Integrals George Voutsadakis (LSSU) Calculus II February / 88

19 Trigonometric Integrals Odd Powers of sinx sin 3 xdx = = u=cos x = = = = sin 2 x sinxdx ( cos 2 x)sinxdx ( u 2 )( du) (u 2 )du 3 u3 u +C 3 cos3 x cosx +C; George Voutsadakis (LSSU) Calculus II February / 88

20 Trigonometric Integrals Odd Power of sinx or cosx sin 4 x cos 5 xdx = = = u=sin x = = = = = sin 4 x cos 4 x cosxdx sin 4 x(cos 2 x) 2 cosxdx sin 4 x( sin 2 x) 2 cosxdx u 4 ( u 2 ) 2 du u 4 (u 4 2u 2 +)du (u 8 2u 6 +u 4 )du 9 u9 2 7 u7 + 5 u5 +C 9 sin9 x 2 7 sin7 x + 5 sin5 x +C George Voutsadakis (LSSU) Calculus II February / 88

21 Trigonometric Integrals Double-Angle Identities cos 2 x = +cos2x, sin 2 x = cos2x ; 2 2 sin 4 xdx = (sin 2 x) 2 dx = ( cos2x ) 2 dx 2 = ( 4 2 cos2x + 4 cos2 2x)dx = ( 4 2 cos2x + 4 (+cos4x ))dx 2 = ( 4 2 cos2x cos4x)dx = ( cos2x + 8 cos4x)dx = 3 8 x 4 sin2x + sin4x +C; 32 George Voutsadakis (LSSU) Calculus II February / 88

22 Trigonometric Integrals Integrals of Tangent and Secant tan xdx sec xdx sinx = cosx dx u ( du) = u du = ln u +c u=cos x = = ln cosx +C = ln cosx +C = ln secx +C; secx(tanx +secx) = dx tanx +secx sec 2 x +tanx secx = dx tanx +secx u du = ln u +C u=tan x+sec x = = ln tanx +secx +C; George Voutsadakis (LSSU) Calculus II February / 88

23 Trigonometric Integrals Tangent and Secant I Recall the identity +tan 2 x = sec 2 x; tan 3 x sec 4 xdx = tan 3 x sec 2 x sec 2 xdx = tan 3 x(+tan 2 x)sec 2 xdx u=tan x = u 3 (+u 2 )du = (u 5 +u 3 )du = = 6 u6 + 4 u4 +C 6 tan6 x + 4 tan4 x +C; George Voutsadakis (LSSU) Calculus II February / 88

24 Trigonometric Integrals Tangent and Secant II Again, we will use +tan 2 x = sec 2 x; tan 3 x sec 3 xdx = tan 2 x sec 2 x tanx secxdx = (sec 2 x )sec 2 x tanx secxdx u=sec x = (u 2 )u 2 du = (u 4 u 2 )du = = 5 u5 3 u3 +C 5 sec5 x 3 sec3 x +C; George Voutsadakis (LSSU) Calculus II February / 88

25 Trigonometric Substitution Subsection 3 Trigonometric Substitution George Voutsadakis (LSSU) Calculus II February / 88

26 Trigonometric Substitution Integrals Involving a 2 x 2 Integrals Involving a 2 x 2 x = asinθ, dx = acosθdθ, a 2 x 2 = acosθ; x Example: Evaluate 2 dx; Set x = sinθ, π 2 θ π 2 ; Then dx = cos θdθ; Moreover, x 2 = sin 2 θ = cos 2 θ = cosθ; Note, also, that θ = sin x and cosθ = x 2 ; x +cos2θ 2 dx = cosθcosθdθ = dθ = 2 2 θ + 4 sin2θ +C = 2 θ + 2 sinθcosθ+c = 2 sin x + 2 x x 2 +C; George Voutsadakis (LSSU) Calculus II February / 88

27 Trigonometric Substitution A Trigonometric Identity Show that sinθtanθ+cosθ = cosθ ; sinθtanθ+cosθ = sinθ sinθ cosθ +cosθ = sin2 θ cosθ +cosθ = sin2 θ cosθ + cos2 θ cosθ = sin2 θ +cos 2 θ cosθ = cosθ ; George Voutsadakis (LSSU) Calculus II February / 88

28 Trigonometric Substitution Integral of tan 2 θ Evaluate tan 2 θdθ; sin tan 2 2 θ θdθ = cos 2 θ dθ = sin 2 θsec 2 θdθ = sin 2 θ(tanθ) dθ = sin 2 θtanθ (sin 2 θ) tanθdθ = sin 2 θtanθ 2sinθcosθ sinθ cosθ dθ = sin 2 θtanθ 2sin 2 θdθ = sin 2 θtanθ ( cos2θ)dθ = sin 2 θtanθ (θ 2 sin2θ)+c = sin 2 θtanθ θ+sinθcosθ +C = sin 2 θtanθ+sinθcosθ θ +C = sinθ[sinθtanθ+cosθ] θ +C sinθ θ +C = tanθ θ+c; cosθ preceding slide = George Voutsadakis (LSSU) Calculus II February / 88

29 Trigonometric Substitution Integrals Involving (a 2 x 2 ) n/2 Example: Evaluate x 2 (4 x 2 ) 3/2dx; Set x = 2sinθ, π 2 θ π 2 ; Then dx = 2cosθdθ; Moreover, (4 x 2 ) 3/2 = (4 4sin 2 θ) 3/2 = (4cos 2 θ) 3/2 = 8cos 3 θ; Note, also, that θ = sin ( x 2) and tanθ = x 4 x 2 ; x 2 4sin 2 (4 x 2 ) 3/2dx = θ sin 2 8cos 3 θ 2cosθdθ = θ cos 2 θ dθ = tan 2 θdθ preceding slide = tanθ θ+c = x sin ( x 4 x 2 2 ) +C; George Voutsadakis (LSSU) Calculus II February / 88

30 Trigonometric Substitution Integral of sec 3 θ Recall that secθdθ = ln secθ +tanθ +C; Evaluate sec 3 θdθ; sec 3 θdθ = secθsec 2 θdθ = secθ(tanθ) dθ = secθtanθ (secθ) tanθdθ = secθtanθ secθtan 2 θdθ = secθtanθ secθ(sec 2 θ )dθ = secθtanθ (sec 3 θ secθ)dθ = secθtanθ sec 3 θdθ+ sec θdθ; Therefore, 2 sec 3 θdθ = secθtanθ+ secθdθ 2 sec 3 θdθ = secθtanθ+ln secθ+tanθ +C, i.e., sec 3 θdθ = 2 secθtanθ+ ln secθ +tanθ +C; 2 George Voutsadakis (LSSU) Calculus II February / 88

31 Trigonometric Substitution Integrals Involving x 2 +a 2 Integrals Involving x 2 +a 2 x = atanθ, dx = asec 2 θdθ, 4x Example: Evaluate 2 +20dx; 4x Note 2 +20dx = x 2 +a 2 = asecθ; 4(x x 2 +5)dx = dx; Set x = 5tanθ, π 2 < θ < π 2 ; Then dx = 5sec 2 θdθ; Moreover, x 2 +5 = 5tan 2 θ +5 = 5sec 2 θ = ( 5secθ; ) x 5 Note, also, that θ = tan and secθ = cosθ = x ; George Voutsadakis (LSSU) Calculus II February / 88

32 Trigonometric Substitution Integrals Involving x 2 +a 2 (Cont d) Recall x 2 +5 = 5secθ, dx = 5sec 2 θdθ ( x 5 x θ = tan ), secθ = 2 +5 ; 5 x dx = 2 [ 0 5 x x ln 5 5 ( 5secθ) 5sec 2 θdθ = 0 ] 2 secθtanθ+ 2 ln(secθ+tanθ) x x ln +C = ( x x 5 )+C = ( x 2 +5+x 5 )+C; sec 3 θdθ preceding problem = George Voutsadakis (LSSU) Calculus II February / 88

33 Trigonometric Substitution Integrals Involving x 2 a 2 Integrals Involving x 2 a 2 x = asecθ, dx = asecθtanθdθ, x 2 a 2 = atanθ; Example: Evaluate x 2 x 2 9 dx; Set x = 3secθ; Then dx = 3secθtanθdθ; Moreover, x 2 9 = 9sec 2 θ 9 = 9tan 2 θ = 3tanθ; Note, also, that θ = ( x ) x sec and sinθ = 2 9 ; 3 x x 2 x 2 9 dx = 9sec 2 θ3tanθ 3secθtanθdθ = 9secθ dθ = cosθdθ = x 9 9 sinθ+c = 2 9 +C; 9x George Voutsadakis (LSSU) Calculus II February / 88

34 Trigonometric Substitution Completing the Square Example: Evaluate (x 2 6x +) 2dx; Note (x 2 6x +) 2dx = [(x 2 6x +9)+2] 2dx = u=x 3 [(x 3) 2 +2] 2dx = (u 2 +2) 2du; Set u = 2tanθ; Then du = 2sec 2 θdθ; Moreover, u 2 +2 = 2tan 2 θ+2 = 2sec 2 θ; Note, also, that θ = tan ( u 2 ) and sinθ = u u 2 +2, cosθ = 2 u 2 +2 ; George Voutsadakis (LSSU) Calculus II February / 88

35 Trigonometric Substitution Completing the Square (Cont d) Recall u = x 3 u 2 +2 = 2sec 2 θ, du = 2sec 2 θdθ sinθ = u, cosθ = 2 u 2 +2 u 2 +2 ; 2sec 2 (u 2 +2) 2du = θ 2sec 2 (2sec 2 θ) 2dθ = θ 4sec 4 θ dθ = 2 2 cos 2 θdθ = (+cos2θ)dθ = 2 8 (θ ( sin2θ)+c ) = 8 θ + 8 sinθcosθ+c = 2 u 2 u 2 8 tan u 2 +2 u C = ( 2 x 3 x 3 8 tan )+ 2 4(x 2 6x +) +C; George Voutsadakis (LSSU) Calculus II February / 88

36 Hyperbolic and Inverse Hyperbolic Functions Subsection 4 Hyperbolic and Inverse Hyperbolic Functions George Voutsadakis (LSSU) Calculus II February / 88

37 Hyperbolic and Inverse Hyperbolic Functions Hyperbolic Functions and Derivatives Definition of Hyperbolics sinhx = ex e x 2 tanhx = sinhx coshx sechx = coshx Basic Derivatives coshx = ex +e x 2 cothx = coshx sinhx cschx = sinhx (sinhx) = coshx (tanhx) = sech 2 x (sechx) = sechx tanhx (coshx) = sinhx (cothx) = csch 2 x (cschx) = cschx cothx George Voutsadakis (LSSU) Calculus II February / 88

38 Hyperbolic and Inverse Hyperbolic Functions Basic Integral Formulas Basic Integrals sinhxdx = coshx +C coshxdx = sinhx +C sech 2 xdx = tanhx +C csch 2 xdx = cothx +C sechx tanhxdx = sechx +C Example: Calculate x cosh(x 2 )dx; x cosh(x 2 )dx u=x2 = cschx cothxdx = cschx +C 2 coshudu = 2 sinhu +C = 2 sinh(x2 )+C; George Voutsadakis (LSSU) Calculus II February / 88

39 Hyperbolic and Inverse Hyperbolic Functions Powers of sinhx and coshx Calculate sinh 4 x cosh 5 xdx; sinh 4 x cosh 5 xdx = sinh 4 x(cosh 2 x) 2 coshxdx = sinh 4 x(+sinh 2 x) 2 coshxdx u=sinhx = u 4 (+u 2 ) 2 du = u 4 (u 4 +2u 2 +)du = (u 8 +2u 6 +u 4 )du = 9 u u7 + 5 u5 +C = 9 sinh9 x sinh7 x + 5 sinh5 x +C; Calculate cosh 2 xdx; cosh 2 xdx = 2 (+cosh2x)dx = 2 (x + 2 sinh2x)+c = 2 x + sinh2x +C; 4 George Voutsadakis (LSSU) Calculus II February / 88

40 Hyperbolic and Inverse Hyperbolic Functions Hyperbolic Substitutions (instead of Trig Substitutions) Instead of trigonometric substitutions, one may sometimes perform hyperbolic substitutions to calculate an integral: The Method For expressions of the form x 2 +a 2, instead of x = atanθ, we may use x = asinhu; In that case dx = acoshudu; x2 +a 2 = acoshu; For expressions of the form x 2 a 2, instead of x = asecθ, we may use x = acoshu; In that case dx = asinhudu; x2 a 2 = asinhu; George Voutsadakis (LSSU) Calculus II February / 88

41 Hyperbolic and Inverse Hyperbolic Functions Example of Hyperbolic Substitution x Example: Calculate 2 +6dx; We set x = 4sinhu; Then dx = 4coshudu, x 2 +6 = 6sinh 2 u +6 = 4coshu; Moreover, u = sinh x 4, sinhu = x x sinh 4 and coshu = 2 2 u + = +; Therefore, 6 x 2 +6dx = 4coshu4coshudu = 6cosh 2 udu = 8(+cosh2u)du = 8u +4sinh2u +C = 8u +8sinhucoshu +C = 8sinh x x x C; George Voutsadakis (LSSU) Calculus II February / 88

42 Hyperbolic and Inverse Hyperbolic Functions Integrals of Inverse Hyperbolic Functions Integrals Involving Inverse Trigonometric Functions dx x 2 + = sinh x +C; dx x 2 = cosh x +C; (x > ) dx x 2 = tanh x +C; ( x < ) dx x 2 = coth x +C; ( x > ) dx x x = 2 sech x +C; (0 < x < ) dx x +x = 2 csch x +C; (x 0) George Voutsadakis (LSSU) Calculus II February / 88

43 Hyperbolic and Inverse Hyperbolic Functions Examples of Inverse Hyperbolic Integrals Evaluate the following integrals: 4 dx x2 ; xdx x 4; 0.6 dx x2 = cosh x 4 2 = cosh 4 cosh 2; 0.2 xdx u=x x 4 = du u 2 = 2 tanh u = 2 (tanh 0.36 tanh 0.04); George Voutsadakis (LSSU) Calculus II February / 88

44 The Method of Partial Fractions Subsection 5 The Method of Partial Fractions George Voutsadakis (LSSU) Calculus II February / 88

45 The Method of Partial Fractions Outline of Partial Fractions Method To integrate a rational function f(x) = P(x), we write it as a sum of Q(x) simpler rational functions that can be integrated directly; For example, to integrate x 2 dx: We decompose the fraction into partial fractions: 2 Then, work as follows x 2 dx = x 2 = 2 x 2 x + ; = 2 [ ] 2 x 2 dx x + x dx 2 x + dx = 2 ln x ln x + +C; 2 George Voutsadakis (LSSU) Calculus II February / 88

46 The Method of Partial Fractions Distinct Linear Factors I Evaluate x 2 7x +0 dx; Factor the denominator: x 2 7x +0 = (x 2)(x 5); Decompose into partial fractions: x 2 7x +0 = A x 2 + B (x 2)(x 5) x 5 x 2 7x +0 = A(x 2)(x 5) B(x 2)(x 5) + = x 2 x 5 { A(x 5)+B(x 2) } = { (A+B)x +( 5A 2B) } A+B = 0 A = B 5A 2B = 5B 2B = { } A = 3 B = ; So we get 3 x 2 7x +0 = 3 x x 5 ; George Voutsadakis (LSSU) Calculus II February / 88

47 The Method of Partial Fractions Distinct Linear Factors I (Cont d) We obtained So, we have x 2 7x +0 dx = x 2 7x +0 = 3 x x 5 ; [ ] 3 x dx x 5 x 2 dx + 3 x 5 dx = 3 = 3 ln x 2 + ln x 5 +C; 3 George Voutsadakis (LSSU) Calculus II February / 88

48 The Method of Partial Fractions Distinct Linear Factors II Evaluate x 2 +2 (x )(2x 8)(x +2) dx; x 2 +2 (x )(2x 8)(x +2) = A x + B 2x 8 + C x +2 (x )(2x 8)(x +2)(x 2 +2) A(x )(2x 8)(x +2) = + (x )(2x 8)(x +2) x B(x )(2x 8)(x +2) C(x )(2x 8)(x +2) + 2x 8 x +2 x 2 +2 = A(2x 8)(x +2)+B(x )(x +2)+C(x )(2x 8); Now, we get: x = 3 = A ( 6) 3 A = 6 ; x = 4 8 = B 3 6 B = ; x = 2 6 = C ( 3) ( 2) C = 6 ; Therefore, we obtain x 2 +2 (x )(2x 8)(x +2) = 6 x + 2x x +2 ; George Voutsadakis (LSSU) Calculus II February / 88

49 The Method of Partial Fractions Distinct Linear Factors II (Cont d) We obtained x 2 +2 (x )(2x 8)(x +2) = 6 x + 2x x +2 ; Now, we integrate x 2 +2 (x )(2x 8)(x +2) dx [ ] 6 = x + 2x dx x +2 = 6 x dx + 2x 8 dx ++ 6 x +2 dx = 6 ln x + 2 ln 2x 8 + ln x +2 +C; 6 George Voutsadakis (LSSU) Calculus II February / 88

50 The Method of Partial Fractions Long Division First... x 3 + Evaluate x 2 4 dx; Numerator has higher degree than denominator! Start by performing the long division (x 3 +) (x 2 4); x x 2 4 x 3 + x 3 4x 4x + It has quotient x and remainder 4x +; Thus, x 3 + 4x + x 2 = x + 4 x 2 4 = x + 4x + (x 2)(x +2) ; George Voutsadakis (LSSU) Calculus II February / 88

51 The Method of Partial Fractions...Breaking Into Partial Fractions Next... We found x3 + x 2 4 = x + 4x + (x 2)(x +2). Decompose the second fraction: 4x + (x 2)(x +2) = A x 2 + B (4x +)(x 2)(x +2) x +2 (x 2)(x +2) A(x 2)(x +2) B(x 2)(x +2) + 4x + = x 2 x +2 A(x +2)+B(x 2); x = 2 9 = 4A A = 9 4 ; x = 2 7 = 4B B = 7 4 ; This gives x x 2 4 = x + 4 x x +2 ; = George Voutsadakis (LSSU) Calculus II February / 88

52 The Method of Partial Fractions...and Integrating We got Hence, we have x 3 + x 2 4 dx = x x 2 4 = x + 4 x x +2 ; = [ ] x + x dx x +2 xdx x 2 dx x +2 dx = 2 x ln x ln x +2 +C; 4 George Voutsadakis (LSSU) Calculus II February / 88

53 The Method of Partial Fractions Repeated Linear Factors 3x 9 Evaluate (x )(x +2) 2dx; Decompose into partial fractions: 3x 9 (x )(x +2) 2 = A x + B x +2 + C (x +2) 2 (3x 9)(x )(x +2) 2 (x )(x +2) 2 = A(x )(x +2) 2 B(x )(x +2)2 + x x +2 3x 9 = A(x +2) 2 +B(x )(x +2)+C(x ); x = 6 = 9A A = 2 3 ; x = 2 5 = 3C C = 5; x = 0 9 = 4A 2B C B = 4A C+9 2 = 2 3 ; So we get + C(x )(x +2)2 (x +2) 2 3x 9 (x )(x +2) 2 = x + 3 x (x +2) 2; George Voutsadakis (LSSU) Calculus II February / 88

54 The Method of Partial Fractions Repeated Linear Factors (Cont d) We just got 3x 9 (x )(x +2) 2 = x + 3 x (x +2) 2; 3x 9 Now, we integrate (x )(x +2) 2dx = [ ] x + 3 x (x +2) 2 dx x dx + 2 dx +5 3 x +2 = 2 3 = 2 3 ln x ln x +2 5 x +2 +C; (x +2) 2dx George Voutsadakis (LSSU) Calculus II February / 88

55 The Method of Partial Fractions Irreducible Quadratic Factors 8 Evaluate (x +3)(x 2 +9) dx; Decompose the fraction 8 (x +3)(x 2 +9) = A Bx +C + x +3 x 2 8(x +3)(x2 +9) +9 (x +3)(x 2 +9) A(x +3)(x 2 +9) + (Bx +C)(x +3)(x2 +9) x +3 x 2 8 = +9 A(x 2 +9)+(Bx +C)(x +3); x = 3 8 = 8A A = ; x = 0 8 = 9+3C 3C = 9 C = 3; x = 8 = 0+(B +3) 4 8 = 4B +2 B = ; 8 Therefore (x +3)(x 2 +9) = x +3 + x +3 x 2 +9 ; = George Voutsadakis (LSSU) Calculus II February / 88

56 The Method of Partial Fractions Irreducible Quadratic Factors (Cont d) We found that 8 (x +3)(x 2 +9) = x +3 8 Now we integrate: (x +3)(x 2 +9) dx [ ] x +3 = + x +3 x 2 dx +9 x +3 = x +3 dx + x 2 +9 dx = x +3 dx x x 2 dx x +3 x 2 +9 ; x 2 +9 dx = ln x +3 2 ln(x2 +9)+3 3 tan x 3 +C = ln x +3 2 ln(x2 +9)+tan x 3 +C; George Voutsadakis (LSSU) Calculus II February / 88

57 The Method of Partial Fractions Repeated Quadratic Factors Evaluate 4 x x(x 2 +2) 2dx; 4 x Decompose the fraction x(x 2 +2) 2 = A Bx +C + x x 2 +2 A =,B =,C = 0,D = 2,E = ; So, we get 4 x x(x 2 +2) 2 = x + x 2x x (x 2 +2) 2; Integrating, we get 4 x x(x 2 +2) 2dx [ = x + x ] 2x x (x 2 +2) 2 dx = x dx x x 2 +2 dx 2x (x 2 +2) 2dx = ln x 2 ln(x2 +2) x 2 +2 (x 2 +2) 2dx; + Dx +E (x 2 +2) 2 (x 2 +2) 2dx George Voutsadakis (LSSU) Calculus II February / 88

58 The Method of Partial Fractions The Integral (x 2 +2) 2 dx Set x = 2tanθ; Then dx = 2sec 2 θdθ, x 2 +2 = 2tan 2 θ+2 = 2sec 2 θ, θ = tan x x 2, sinθ = 2 x 2 +2, cosθ = x 2 +2 ; (x 2 +2) 2dx = 2sec 2 4sec 4 θdθ = θ 2 2 cos 2 θdθ = (+cos2θ)dθ = (θ sin2θ)+c = 8 θ+ 8 sinθcosθ+c = 2 x 2 x 2 8 tan x 2 +2 x C = 2 x x 8 tan + 2 4(x 2 +2) +C; George Voutsadakis (LSSU) Calculus II February / 88

59 Improper Integrals Subsection 6 Improper Integrals George Voutsadakis (LSSU) Calculus II February / 88

60 Improper Integrals Overview of Improper Integrals Suppose, we wanted to determine the amount of area under the graph of f(x) = e x over the unbounded interval [0, ); This is given by the improper integral 0 e x dx; It is called improper because it represents the area of an unbounded region; To compute such an integral, we first introduce an artificial bound R > 0 and we compute instead the proper integral R 0 e x dx = e x R 0 = ( e R ( )) = e R ; Finally, we push R towards + : 0 e x dx = lim R R 0 e x dx = lim R ( e R ) = 0 = ; George Voutsadakis (LSSU) Calculus II February / 88

61 Improper Integrals Formal Definitions Definitions of Improper Integrals If, for some fixed a, the function f(x) is integrable on [a,b] for all b > a, then define the improper integral of f(x) over [a, ) by a f(x)dx = lim R R a f(x)dx; The integral converges if the limits exists and is finite, and it diverges, otherwise; If, for some fixed a, the function f(x) is integrable on [b,a] for all b < a, then define the improper integral of f(x) over (,a] by a f(x)dx = lim R a R f(x)dx; The integral converges if the limits exists and is finite, and it diverges, otherwise; George Voutsadakis (LSSU) Calculus II February / 88

62 Improper Integrals Formal Definitions (Cont d) Definition of Third Type of Improper Integral If, for all a < 0,b > 0, the function f(x) is integrable on [a,0],[0,b], then we define the improper integral of f(x) over (, ) by f(x)dx = 0 f(x)dx + 0 f(x)dx; The integral converges if both integrals on the right converge and it diverges, otherwise; George Voutsadakis (LSSU) Calculus II February / 88

63 Improper Integrals Example of Improper Integral I Show that 2 x3dx converges and compute its value; We first calculate R x3dx = 2x 2 2 R 2 = 8 2R 2; Therefore, we obtain: 2 x3dx = lim R R 2 x3dx = lim ( R 8 2R 2) = 8 0 = 8 ; George Voutsadakis (LSSU) Calculus II February / 88

64 Improper Integrals Example of Improper Integral II Determine whether If so, compute its value; We first calculate R x dx converges; x dx = ln x R = ln R ; Therefore, we obtain: dx = lim x R R x dx = lim [ ln R ] = lim [ lnr] = ; R R George Voutsadakis (LSSU) Calculus II February / 88

65 Improper Integrals The p-integral Show that, for a > 0, a lim R a xpdx = lim ( R p x pdx = R R a p a p p a p p, if p > diverges, if p x p dx = lim ) ; R x p p If p >, then p < 0, so lim R R p = 0 and, therefore, a p a xpdx = p ; If p <, then p > 0, so lim R R p = ; Therefore, the integral diverges; R If p =, then dx = lim dx = lim (lnr lna) = ; x R x R a George Voutsadakis (LSSU) Calculus II February / 88 a R a = ;

66 Improper Integrals Using L Hôpital s Rule Recall how L Hôpital s Rule is applied: Calculate x + lim x 0 e x (= xe x dx; xe x dx = x( e x ) dx xe x e x dx = xe x e x = x + e x ; 0 xe x dx = lim R R 0 ) L Hôpital (x +) = lim x (e x ) = lim x e x = 0; By Parts = xe x R + dx = lim ( ) = 0 = ; R er George Voutsadakis (LSSU) Calculus II February / 88

67 Improper Integrals Application: Escape Velocity The earth exerts a gravitational force of magnitude F(r) = G Mem on r an object of mass m at distance r from its center; 2 Find the work required to move the object infinitely far from the earth; W = F(r)dr = G M R em r e r e r 2 dr = lim G M em R r e r 2 dr = R ( GM e m lim R r e r 2dr = GM em lim ) R = R [ r r e GM e m lim ] = G M em J; R r e R r e Calculate the escape velocity v esc on the earth s surface; The escape velocity must provide kinetic energy at least as big as the work required to move the object infinitely far from the earth; 2 mv2 esc GM em vesc 2 r 2GM e v esc e r e 2GMe r e ; George Voutsadakis (LSSU) Calculus II February / 88

68 Improper Integrals Application: Perpetual Annuity An investment pays a dividend continuously at a rate of $6,000 per year; Compute the present value of the income stream if the interest rate is 4% and the dividends continue forever. PV = 0 lim 6000 T Pe rt dt = lim T 0.04 e 0.04t 5000 ( ) = $50,000; T 0 T e 0.04t dt = = lim T (e 0.04T ) = George Voutsadakis (LSSU) Calculus II February / 88

69 Improper Integrals Improper Integrals for Infinite Discontinuities at Endpoints We determine the amount of area under the graph of f(x) = x over the interval [0,]; This is given by the improper integral 0 x dx; It is improper because it represents the area of an unbounded region; To compute such an integral, we first introduce an artificial bound 0 < R < and we compute instead the proper integral R x /2 dx = 2 x R = 2 2 R = 2 2 R; Finally, we push R towards 0 from the right: dx = lim dx = lim x R 0 + x R 0 +(2 2 R) = 2 0 = 2; 0 R George Voutsadakis (LSSU) Calculus II February / 88

70 Improper Integrals Definitions of Integrals with Infinite Discontinuities Integrants with Infinite Discontinuities If f(x) is continuous on [a,b) but discontinuous at x = b, we define b a f(x)dx = lim R b R a f(x)dx; If f(x) is continuous on (a,b] but discontinuous at x = a, we define b a f(x)dx = lim R a + b R f(x)dx; In both cases the improper integral converges if the limit exists and it diverges otherwise; George Voutsadakis (LSSU) Calculus II February / 88

71 Improper Integrals Examples of Improper Integrals Calculate 9 0 dx; x Calculate 9 0 dx = lim x R R x /2 dx = lim R 0 +[2 x 9 R ] = lim R 0 +(6 2 R) = 6 0 = 6; /2 0 x dx; /2 0 x dx = lim R 0 + /2 R x dx = lim R 0 +[lnx /2 R ] = lim R 0 +(ln lnr) = ; 2 George Voutsadakis (LSSU) Calculus II February / 88

72 Improper Integrals p-integral Revisited Show that, for a > 0, a 0 lim R 0 + a 0 x pdx = a xpdx = lim ( R 0 + R a p p R p p 0 a p p, if p < diverges, if p x p dx = lim ) ; x p a R 0 + p R If p <, then p > 0, so lim = 0; Therefore, R 0 +R p a a p 0 xpdx = p ; If p >, then p < 0, so lim = and, therefore, the R 0 +R p integral diverges; a a If p =, then dx = lim dx = lim x R 0 + x R 0 +(lna lnr) = ; R George Voutsadakis (LSSU) Calculus II February / 88 ; =

73 Improper Integrals An Additional Example Evaluate dx; 0 x 2 First, recall the formula x 2 dx = sin x +C; We compute R 0 dx = x 2 sin x R 0 = sin R sin 0 = sin R; Therefore, we get 0 dx = lim x 2 R R 0 dx = lim R = π x 2 R sin 2 ; George Voutsadakis (LSSU) Calculus II February / 88

74 Improper Integrals The Comparison Test for Improper Integrals Comparison Test for Improper Integrals Assume that f(x) g(x) 0 for x a; If If a a f(x)dx converges, then g(x)dx diverges, then a a g(x)dx also converges; f(x)dx also diverges; The Comparison Test may also be applied for improper integrals with infinite discontinuities at the endpoints; George Voutsadakis (LSSU) Calculus II February / 88

75 Improper Integrals Applying the Comparison Test I Show that e x x dx converges; Note that for x, we get 0 x 0 e x e x ; x Therefore, by the comparison test, to show that converges, it suffices to show that computation: e x dx = lim R lim R [( e x ) R ] = R lim R (e e R ) = e ; e x dx = e x x dx e x dx converges; Here is the George Voutsadakis (LSSU) Calculus II February / 88

76 Improper Integrals Applying the Comparison Test II Show that dx converges; x 3 + Note that for x, we get x 3 x 3 + x 3 x comparison test, to show that to show that dx converges; x 3 This is, however, true, since this is a p-integral, with p = 3 2 > ; x 3 + ; By the x 3 dx converges, it suffices x 3 + George Voutsadakis (LSSU) Calculus II February / 88

77 Improper Integrals Applying the Comparison Test III Does dx converge? x +e 3x Note that for x, we get e 3x x +e 3x 0 x +e 3x e 3x; By the comparison test, to show dx converges, it x +e 3x suffices to show e3xdx converges; lim R e3xdx = lim R [ 3 e 3x R ] = [ 3e 3 ] 3e 3R = 3e 3; George Voutsadakis (LSSU) Calculus II February / 88

78 Improper Integrals Applying the Comparison Test IV /2 Does 0 x 8 +x2dx converge? Note that for 0 < x 2, we get x 8 x 2 x 8 +x 2 x 2 +x 2 = 2x 2 0 2x 2 comparison test, to show 2 /2 0 x2dx diverges; /2 This is, however, true, since this is a p-integral, with p = 2 > ; 0 x 8 +x2; By the x 8 +x2dx diverges, it suffices to show George Voutsadakis (LSSU) Calculus II February / 88

79 Numerical Integration Subsection 7 Numerical Integration George Voutsadakis (LSSU) Calculus II February / 88

80 Numerical Integration Trapezoidal Rule Trapezoidal Rule The N-th trapezoidal approximation to b a f(x)dx is T N = 2 x(y 0 +2y + +2y N +y N ), where x = b a N and y j = f(a+j x). Error Bound for T N If f (x) exists and is continuous, and, for all x in [a,b], f (x) K 2, then b Error(T N ) = f(x)dx T N K 2(b a) 3 2N 2 ; a George Voutsadakis (LSSU) Calculus II February / 88

81 Numerical Integration Applying the Trapezoidal Rule Consider the integral 3 sin(x 2 )dx; Calculate T 4 ; We divide [, 3] into four subintervals of length x = 3 4 = 2 ; Then, we apply the formula for T 4 : T 4 = 2 x(y 0+2y +2y 2 +2y 3 +y 4 ) = 2 2 (f()+2f(.5)+2f(2)+2f(2.5)+f(3)) = 4 (sin+2sin(.52 )+2sin4+2sin(2.5 2 )+sin9) ; George Voutsadakis (LSSU) Calculus II February / 88

82 Numerical Integration Midpoint Rule Midpoint Rule The N-th midpoint approximation to b a f(x)dx is M N = x(f(c )+f(c 2 )+ +f(c N )), where x = b a N andc j = a+(j 2 ) x is the midpoint of [x j,x j ]. Error Bound for M N If f (x) exists and is continuous, and, for all x in [a,b], f (x) K 2, then b Error(M N ) = f(x)dx M N K 2(b a) 3 24N 2 ; a George Voutsadakis (LSSU) Calculus II February / 88

83 Numerical Integration Applying the Midpoint Rule Consider the integral 4 Calculate its exact value; xdx; 4 Find M 6 ; x /2 dx = 2 3 x 3 4 = 2 4 (8 ) = ; M 6 = x[f(c )+f(c 2 )+f(c 3 )+f(c 4 )+f(c 5 )+f(c 6 )] = 2 [f(.25)+f(.75)+f(2.25)+f(2.75)+f(3.25)+f(3.75)] = 2 ( ) 4.669; George Voutsadakis (LSSU) Calculus II February / 88

84 Numerical Integration Error and Bound of the Midpoint Rule Consider again the integral 4 4 xdx; We found that x /2 dx M ; Find the exact value of Error(M 6 ); 4 Error(M 6 ) = xdx M = 0.002; Find the theoretical error bound and verify that the actual error is less than that bound; Note that f (x) = 2 x /2 and f (x) = 4 x 3/2 = 4 x 3; Therefore, for x 4, f (x) 4 = K 2; This shows that K 2 (b a) N 2 = = 0.009; Therefore, 28 Error(M 6 ) K 2(b a) 3 24N 2 ; George Voutsadakis (LSSU) Calculus II February / 88

85 Numerical Integration Example (Accuracy of Trapezoidal) Find how large N should be so that T N approximates an error of at most 0 4 ; We are asked to determine N, so that Error(T N ) 0 4 ; Since Error(T N ) K 2 (b a) 3 2N 2, it suffices to find N, such that K 2 (b a) ; Note that f (x) = 2N 2 2xe x2 and f (x) = 2e x2 + 4x 2 e x2 = (4x 2 2)e x2 ; Its graph over the interval [0,3] is given in the figure; Observe that f (x) 2 = K 2 ; K 2 (b a) e x2 dx with 2N (3 0)3 2N N N N ; 2 George Voutsadakis (LSSU) Calculus II February / 88

86 Numerical Integration Simpson s Rule Simpson s Rule For N even, the N-th midpoint approximation to b a f(x)dx is S N = 3 x[y 0 +4y +2y y N 3 +2y N 2 +4y N +y N ], where x = b a N and y j = f(a+j x); Error Bound for M N If f (4) (x) exists and is continuous, and, for all x in [a,b], f (4) (x) K 4, then b Error(S N ) = f(x)dx S N K 4(b a) 5 80N 4 ; a George Voutsadakis (LSSU) Calculus II February / 88

87 Numerical Integration Applying Simpson s Rule Consider the integral 4 We divide[2, 4] into eight subintervals of length x = = 4 ; Then, we apply the formula for S 8 : S 8 = 3 x(y 0+4y +2y y 6 + 4y 7 +y 8 ) = 2 +x 3 dx; Calculate S 8 ; 3 4 (f(2)+4f(2.25)+2f(2.5)+ +2f(3.5)+4f(3.75)+f(4)) = 2 ( ) 0.74; George Voutsadakis (LSSU) Calculus II February / 88

88 Numerical Integration Applying Simpson s Rule 3 Consider the integral x dx; Calculate S 8 ; S 8 = 34 (f()+4f(.25)+2f(.5)+ + 2f(2.5) + 4f(2.75)+ f(3)) = 2 ( ).099; Find a bound for Error(S 8 ); Compute carefully: f (x) = x 2, f (x) = 2 x 3, f (x) = 6 x 4, f (4) (x) = 24 x 5; Thus, on [,3], we have f (4) (x) 24 = K 4 ; Error(S 8 ) K 4(b a) 5 80N 4 = ; George Voutsadakis (LSSU) Calculus II February / 88