Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.08 Office Hours 9:15-10:15 MTWThF Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Friday, February 15 7.5 Integrating Rational functions p410: 6,10,14,20 Main Idea: Divide and conquer. The worst there is is just c x + d ---------------- 2 n (x + ax + b) dx 2 cx+b 2(2n-1)c dx INT --------------- = ------------------ + --------- INT ----------- 2 n+1 2 2 n n 2 n (a+bx+cx ) n(4ac-b )(a+bx+cx ) (4ac-b )n (a+bx+cx ) x dx -(2a+bx) -b(2n-1) dx INT ------------- = ---------------------- + --------- INT ------------ 2 n+1 2 2 n 2 2 n (a+bx+cx ) n(4ac-b ) (a+bx+cx ) n(4ac-b ) (a+bx+cx ) / Key Words: Partial Fraction decomposition. f(x)/ g (x) Goal: Learn how to integrate any rational function. -------------------------------------------------------- Previous assignment: p403: (4),(18),(32),(33) Page 403 Problem 4 2 x + 3 x INT -------------- dx Sqrt[x+4] u = Sqrt[x+4] 2 u = x+4 2 x = u - 4 dx = 2 u du 2 2 2 (u - 4) + 3(u - 4) INT ------------------------- 2 u du u 2 2 2 2 INT (u - 4) + 3(u - 4) du 4 2 2 2 INT u -8 u + 16 + 3 u - 12 du 4 2 2 INT u -5 u + 4 du _ _ | 5 3 | | u - 5 u | 2 | ---- + ----- + 4 u | + C | 5 3 | |_ _| _ _ | 4 2 | | u - 5 u | 2 u |---- + ----- + 4 | + C | 5 3 | |_ _| _ _ | | 2 u | 4 2 | ---- | 3 u - 25 u + 60 | 15 | | |_ _| _ _ 2 Sqrt[x+4] | 2 | ----------- | 3 (x+4) - 25 (x+4) + 60 | 15 |_ _| _ _ 2 Sqrt[x+4] | 2 | ----------- | 3 (x + 8x + 16) - 25 (x+4) + 60 | 15 |_ _| _ _ 2 Sqrt[x+4] | 2 | ----------- | 3 x - x + 8 | <========= Answer 15 |_ _| / \/ -1/2 2 1/2 2/15 (1/2) (x+4) (3x - x + 8) + 2/15 (x+4) (6x-1) 2 3x - x + 8 2 (x+4)(6x-1) (1/15) --------------- + (1/15) ---------------- Sqrt[x+4] Sqrt[x+4] 2 2 3x - x + 8 + 12 x + 46 x - 8 ---------------------------------- 15 Sqrt[x+4] 2 2 15 x + 45 x x + 3 x ---------------- = ----------- It checks 15 Sqrt[x+4] Sqrt[x+4] ---------------------------------------------------------- -------------------------------------- Integrate[ (x^2 + 3x)/Sqrt[x+4],x] -------------------------------------- Page 403 Problem 18 dx INT --------------------- 2 Sqrt[ x + 4 x + 5 ] dx INT --------------------- 2 Sqrt[x + 4 x + 4 + 1] dx INT --------------------- 2 Sqrt[(x+2) + 1] u = x+2 du = dx du INT --------------------- 2 Sqrt[ u + 1] /| 2 / | Sqrt[u +1] / | / | u | / | | / | |/ t | -------------+----------------------- | 1 2 Sqrt[u +1] = Sec[t] u = Tan[t] 2 du = Sec [t] dt 2 Sec [t] dt INT --------------- = INT Sec[t] dt Sec[t] ln( Sec[t] + Tan[t] ) + C /| 2 / | Sqrt[u +1] / | / | u | / | | / | |/ t | -------------+----------------------- | 1 2 ln( Sqrt[u +1] + u ) + C 2 ln( Sqrt[x + 4 x + 5] + x+2 ) + C <=============== Answer / \/ 2 -1/2 1/2 (x + 4 x + 5) (2x + 4) + 1 -------------------------------------------------------------- 2 Sqrt[x + 4 x + 5 ] + x+2 2 -1/2 (x + 4 x + 5) ( x + 2) + 1 -------------------------------------------------------------- 2 (Sqrt[x + 4 x + 5 ] + x+2) 2 ( x + 2) + Sqrt[x + 4x + 5] -------------------------------------------------------------- _ _ | 2 | 2 |Sqrt[x + 4 x + 5 ] + x+2 | Sqrt[x + 4x + 5] |_ _| 1 --------------------- It checks. 2 Sqrt[x + 4x + 5] --------------------------------------------------------- Page 403 Problem 32 Two circles of radius b intersect as shown with their centers 2 a units apart (0 <= a < b). Find the area of the region of their overlap. ..... ..... ' / ` . ' ` . b/ '|` . . / . | . . . /_____._|_. . . . a . . . | . . .`|' . .... ' .... The problem is the same as finding the area of that portion of a circle inside a chord. We find this area and then double it. ..... 2 2 2 .' /|. x + y = b . b/ | . . / | . . /...|.... . a| . . | . ` . .| ' ''' x=b 2 2 A = INT 2 Sqrt[ b - x ] dx x=a | | /| | b/ | 2 2 | / | Sqrt[ b - x ] |/t | -----------+------------- | x | | x = b Cos[t] dx = -b Sin[t] dt 2 2 Sqrt[ b -x ] = b Sin[t] x=b A = 2 INT b Sin[t] (-b Sin[t]) dt x=a 2 x=b 2 A = -2 b INT Sin [t] dt x=a 2 x=b 1 - Cos[2 t] A = -2 b INT -------------- dt x=a 2 _ _ x=b 2 | Sin[2t] | t = 0 A = - b | t - ------- | |_ 2 _| t = ArcCos[a/b] x=a _ _ 2 | | t=0 A = -b | t - Sin[t] Cos[t] ) | |_ _| t = ArcCos[a/b] _ _ t=0 | 2 | A = |- b t + b Sin[t] b Cos[t] | | | |_ _| t = ArcCos[a/b] | /| | b/ | 2 2 | / |Sqrt[b - a ] |/t | -----------+------------- | a | | 2 2 2 A = b ArcCos[a/b] - a Sqrt[b - a ] The area from the chord to the circle is A, the area we want is 2A 2 2 2 2A = 2 b ArcCos[a/b] - 2 a Sqrt[ b - a ] <======== Answer -------------------------------------- Integrate[ 2 Sqrt[ b^2 - x^2 ],{x,a,b}] = -b^2 ArcSin[x/b] -x Sqrt[b^2-x^2] Get["font.math"]; a = 1; b = Sqrt[2]; p1 = ParametricPlot[ b{Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = ParametricPlot[ 2a {1,0} + b{Cos[t],Sin[t]},{t,0,2 Pi}]; p3 = ParametricPlot[ {a,y},{y,-Sqrt[b^2-a^2],Sqrt[b^2-a^2]}, PlotStyle->{RGBColor[1,0,0]}]; p4 = Show[p1,p2,p3,PlotLabel->"P403 p32; a=1 b=Sqrt[2]", AspectRatio->Automatic]; Display["p32.ps",p4]; ============================================= Page 403 Problem 33 Hippocrates of Chios (ca.430 B.C. showed that the two shaded regions in Figure 7 have the same area. (He squared the lune). Note that C is the center of the lower arc of the lune. Show Hippocrates' result (a) using calculus and (b) without calculus E . '''`` . ' | `. ' o D o ` ' o | o ` . o | o. A o---------+---------o B \ | / \ | / \ | / 2 Pi 2 \ | / Area quarter circle ADBE = ----- a \ | / __ 2 4 \ | /\/2 a Area triangle ABC = a \ | / 2 \/|\/ Area 1/2 circle ABE = a Pi/2 \|/ C Area lune ADBE = Area 1/2 circle - (area Quarter circle - area triangle) 2 2 2 2 = a Pi/2 - ( a Pi/2 - a ) = a . ---------------------------------------------------------------- Get["font.math"]; a = ParametricPlot[ {Cos[t],Sin[t]},{t,0,2 Pi},PlotStyle->{Thickness[0.01]}]; b = ParametricPlot[ {0,-1} + Sqrt[2] {Cos[t],Sin[t]},{t,Pi/4,3 Pi/4}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; c = ParametricPlot[ {0,-1} + t{1/Sqrt[2],1/Sqrt[2]},{t,0,Sqrt[2]}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; d = ParametricPlot[ {0,-1} + t{-1/Sqrt[2],1/Sqrt[2]},{t,0,Sqrt[2]}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; e = Show[a,b,c,d,PlotLabel->"P403 p33; Hippocrates of Chios", AspectRatio->Automatic]; Display["p33.ps",e]; ---------------------------------------------------------------- With Calculus: In the previous problem we found the area we need. We take half of the answer to have the area of the area between a chord and the circle. b = Sqrt[2] a 2 2 2 A = b ArcCos[a/b] - a Sqrt[b - a ] 2 a 2 2 A = 2 a ArcCos[----------] - a Sqrt[2 a - a ] Sqrt[2] a 2 2 A = 2 a ArcCos[1/Sqrt[2]] - a 2 2 A = 2 a Pi/4 - a 2 2 A = (1/2) a Pi - a The lune is 1/2 the circle minus the chord area or 2 2 2 2 1/2 Pi a - ( 1/2 a Pi - a ) = a . Without calculus. Top half - 1/4 of big circle + 2 triangles 2 2 2 1/2 Pi a -1/4 (Pi b ) + 2 (1/2) a 2 2 2 2 1/2 Pi a -1/2 Pi a + a = a ---------------------------------------------------------- New Material Example 3 page 405 3x-1 INT ------------ dx (x+2)(x-3) 3x-1 A B ------------ = ----------- + -------- (x+2)(x-3) x+2 x-3 3x-1 = A(x-3) + B(x+2) 3 = A+ B -1 = -3A+2B 8 = 5 B B = 8/5 -7 = -5 A A = 7/5 3x-1 7/5 8/5 ------------ = ----------- + -------- (x+2)(x-3) x+2 x-3 3x-1 INT ----------- dt = 7/5 ln(x+2) + 8/5 ln(x-3) + C (x+2)(x-3) ------------------------------------------------- Page 406 Example 4 5x+3 INT ------------ dx 3 2 x -2x -3x 5x+3 A B C ------------- = ----------- + ---------- + -------- x(x-3)(x+1) x x-3 x+1 5x+3 = A(x-3)(x+1) + Bx(x+1) + Cx(x-3) 2 x 0 = A + B + C x 5 = -2A + B -3 C 1 3 = -3A A = -1 5 = A + 4 B B = 3/2 C = -1/2 5x+3 -1 3/2 -1/2 ------------- = ----------- + ---------- + -------- x(x-3)(x+1) x x-3 x+1 5x+3 INT ----------- dx = -ln(x) + 3/2 ln(x-3) -1/2 ln(x+1) + C x(x-3)(x+1) ------------------------------------------------------------- Page 406 Example 5 x INT ------- dx 2 (x-3) x A B -------- = --------- + ------------- 2 2 (x-3) (x-3) (x-3) x = A(x-3) + B A = 1 B = 3 x 1 3 -------- dx = --------- + ------------- 2 2 (x-3) (x-3) (x-3) -1 x 3(x-3) INT ------- dx = ln(x-3) + -------- + C 2 -1 (x-3) -------------------------------------------------- Page 407 Example 7 2 6x - 3x + 1 INT -------------------- dx 2 (4x+1)(x + 1) 2 6x - 3x + 1 A Bx+C -------------------- = ---------- + -------------- 2 2 (4x+1)(x + 1) 4x+1 x + 1 2 2 6x - 3x + 1 = A(x + 1) + (Bx+C)(4x+1) 2 x 6 = A + 4 B x -3 = B + 4 C 1 1 = A + C A B C RHS 1 4 0 6 0 1 4 -3 1 0 1 1 A B C RHS 1 4 0 6 0 1 4 -3 0 -4 1 -5 A B C RHS 1 0 -16 18 0 1 4 -3 0 0 17 -17 A B C RHS 1 0 -16 18 0 1 4 -3 0 0 1 -1 A B C RHS 1 0 0 2 A = 2 0 1 0 1 B = 1 0 0 1 -1 C = -1 2 6x -3x + 1 2 x-1 -------------------- = ---------- + -------------- 2 2 (4x+1)(x + 1) 4x+1 x + 1 2 6x -3x + 1 2 x -1 ---------------- = ---------- + --------- + ------- 2 2 2 (4x+1)(x + 1) 4x+1 x + 1 x + 1 2 6x -3x + 1 2 INT ---------------- dt = 2/4 ln(4x+1) + 1/2 ln(x + 1) - ArcTan[x] + C 2 (4x+1)(x + 1) ------------------------------------------------------------------------ General Theory. (1) Divide the denominator into the numerator so that the degree of the numerator is less than the degree of the denominator. (2) Factor the denominator so that the highest degree of any factor is 2 and those factors of degree 2 have complex roots, (3) If the factor is of degree k, use denominators of 1,2, ..., k (4) If the factor is linear use "A " . If the factor i is quadratic, use "A x+B ". i i Write out the form of the decomposition for (a) 3x + 7y -------------------- 4 2 3 (x-2) (x + 3x + 5) 3 x + 2 (b) ------------ 2 (x+1)(x + 1) 2 x + 1 (c) ----------------- 2 (x + 3 x + 2) ------------------------------------------ Fortunately .... The linear factors are easy to calculate. f(c) Simply use ------ . / g (c) 3x-1 A B ------------ = ----------- + -------- (x+2)(x-3) x+2 x-3 5x+3 A B C ------------- = ----------- + ---------- + -------- x(x-3)(x+1) x x-3 x+1 x A B -------- = --------- + ------------- 2 2 (x-3) (x-3) (x-3) 6x -3x + 1 A Bx+C -------------------- = ---------- + -------------- 2 2 (4x+1)(x + 1) 4x+1 x + 1 --------------------------------------------------------