REVIEW OF SIMPLE UNIVARIATE CALCULUS

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1 REVIEW OF SIMPLE UNIVARIATE CALCULUS 1. APPROXIMATING CURVES WITH LINES 1.1. The eqution for line. A liner function of rel vrible x is given by y f(x) x + b, nd b re constnts (1) The grph of liner eqution is stright line. The number is clled the slope of the function nd the number b is clled the y-intercept The point slope formul for line. Consider point P (x 1,y 1 ) nd line with slope. The line with slope pssing through the point P (x 1,y 1 ) cn be determined by picking n rbitrry point on the line denoted (x, y) nd then using the formul for the slope of line to get the liner eqution through tht point. The slope is equl to the chnge in y over the chnge in x or y y 1,x x 1 (2) x x 1 We cn then solve the eqution for either yory y 1 s follows. y y 1,x x 1 (3) x x 1 y y 1 (x x 1 ) y (x x 1 )+y 1 x +(y 1 x 1 ) 1.3. Using line to pproximte curve. If we wnt to pproximte curve with line t prticulr point, one wy to do so is pick line going through tht point tht hs the sme slope s the curve t tht point. This is clled the tngent line. In figure 1 the line pproximting the curve f t the point (x,f(x)) is shown. We know the tngent line psses through (x, f(x)) nd hence ll we need to determine is the slope. To do this choose smll number, h 0, nd on the grph of f, mrk the point (x + h, f (x + h)). The line tht psses through this point hs slope f(x+h) f(x). We cn see this h in figure 2. This line, of course, hs slope different from the tngent line. If h were lrger, the pproximtion would be worse, while if h were smller, the pproximtion would be better Intuitive definition of derivtive of f t the point [z,f(z)]. The slope of the tngent to the curve y f(x) t the point [z, f(z)] is clled the derivtive of f with respect to x t the point z. The derivtive is written s f (x) or dy or. Dte: Februry 7,

2 2 REVIEW OF SIMPLE UNIVARIATE CALCULUS FIGURE 1. Tngent to Curve Tngent to Function x, f x f x 20 tngent FIGURE 2. Tngent nd Approximte Slope Tngent to Function x h, f x h x, f x f x 20 f x h f x 10 h tngent Eqution of the tngent line to grph given the definition of derivtive. The eqution for the tngent to the grph of y f(x) t the point [, f()] is y f() f ()(x ) y f()+f ()(x ) (4)

3 REVIEW OF SIMPLE UNIVARIATE CALCULUS 3 Intuitively, this sys we pproximte the function f by its vlue t the point, then move wy from long line with slope f (x). Eqution 4 is lso clled first order Tylor series pproximtion to the function f t the point. 2. A BRIEF DISCUSSION OF LIMITS 2.1. Definition of function. A function f is set of ordered pirs (x, y), no two of which hve the sme first member. Tht is, if (x, y) f nd (x, z) f, then y z. The definition requires tht for every x in the domin of f, there is exctly one y such tht (x, y) f. It is customry to cll y the vlue of f t x nd write y f(x) (5) In R 2, function psses the verticl line test, i.e., verticl line psses through the grph of the function only once. For exmple the mpping y x 2 is function becuse for ech vlue of y there is distinct vlue of x. The unit circle x 2 + y 2 1 is not function becuse t zero y cn equl +1 nd Ide of limit. Consider number t nd function defined ner the number c, but not necessrily defined t c itself. A rough trnsltion of lim f(x) t (6) x c might be s x pproches c, f(x) pproches t or for x close to c, f(x) is close to t. Consider the exmple in figure 3 where the function is well behved t ll points. FIGURE 3. Limit of Function , f , f 0.5 t 10 f c c

4 4 REVIEW OF SIMPLE UNIVARIATE CALCULUS As x gets close to c, f(x) gets close to t. Now consider the cse where f is defined peculirly t c s in figure 4. The limit is still t. Notice tht even though f(c) is peculir point, the limit of f( )tc is t. FIGURE 4. Limit of Discontinuous Function g Now consider the function f(x) 1 x c, (7) For c5, the grph is contined in figure 5. Notice tht the limit of the function does not exist or goes to infinity s x goes to Forml definition of limit. { for ech ɛ there exists δ>0such tht lim f(x) t, if nd only if (8) x t if 0 < x c <δthen f(x) t <ɛ. For f to hve limit t c, it must be defined t ll numbers sufficiently close to c, but it need not be defined t c. Even if it is defined t c, its vlue t c is irrelevnt. The choice of δ depends on the previous choice of ɛ. We do not require tht there exist number ɛ which works for ll δ but, rther, tht for ech δ there exists ɛ which works for it Some exmples of computing limits Exmple 1. Show tht lim x 2 (2x - 1) 3. We need to find ɛ>0, δ>0pir such tht if 0 < x 2 < δ, then (2x 1) 3 < ɛ

5 REVIEW OF SIMPLE UNIVARIATE CALCULUS 5 FIGURE 5. Limit tht Does Not Exist t Point h x x 5 10 To find the pproprite level of δ, we need to find connection between the two expressions x 2 nd (2x 1) 3 First rewrite the second expression s (2x 1) 3 2x 4 2 x 2 To mke (2x 1) 3 less thn ɛ, we need only mke x 2 twice s smll. So we cn choose δ 1 2 ɛ. To see tht this works notice tht if 0 < x 2 < 1 2 ɛ, then 2 x 2 <ɛ. But this implies tht (2x 1) 3 <ɛ. This is shown grphiclly in figure Exmple 2. Show tht lim x 1 (2 3x) 5. We need to find ɛ>0, δ>0pir such tht if 0 < x ( 1) < δ then (2 3x) 5 < ɛ To find the pproprite level of δ, we need to find connection between the two expressions x ( 1) nd (2 3x) 5 First rewrite the second expression s (2 3x) 5 3x 3 3 x 1 3 x +1 nd the first expression s (x ( 1)) x +1 From these two expressions we cn conclude tht (2 3x) 5 3 x ( 1)

6 6 REVIEW OF SIMPLE UNIVARIATE CALCULUS f x FIGURE 6. Limit of 2x-1 s x 2 f x 2x 1 3 Ε 3 3 Ε x To mke (2 3x) 5 less thn ɛ, we need only mke x ( 1) three times s smll. So we cn choose δ 1 3 ɛ. To see tht this works notice tht if 0 < x ( 1) < 1 ɛ, then 3 x ( 1) <ɛ. But 3 this implies tht (2 3x) 5 <ɛ. This is shown grphiclly in figure Uniqueness of limits. If lim x c f(x) t nd lim x c f(x) m then t m Some rules for limits. If lim x c f(x) A nd lim x c g(x) B, then lim [f(x)+g(x)] A + B x lim[f(x) g(x)] A B (9b) x lim[f(x)g(x)] A.B (9c) x lim[f(x)/g(x)] A/B (provided B 0) (9d) x lim x [f(x)] p q A p q (provided A p q is defined) (9) (9e) 2.7. Equlity of functions nd equlity of limits. If the functions d nd g re equl for ll x close to (but not necessrily t x ), then

7 REVIEW OF SIMPLE UNIVARIATE CALCULUS 7 FIGURE 7. Limit of 2-3x s x -1 f x 5 Ε 5 5 Ε f x 2 3x x lim f(x) lim g(x), x x whenever either limit exists. 3. FORMAL DEFINITION OF THE DERIVATIVE AND RULES FOR DIFFERENTIATION 3.1. Definition. Let f be function defined on n open intervl (, b), nd ssume tht c (, b). Then f is sid to be differentible t c whenever the limit f(x) f(c) lim (10) x c x c exists. The limit, denoted by f (c), is clled the derivtive of f t c. Another wy to express the definition is (x 0)f f(x 0 + h) f(x 0 ) (x 0 ) lim (11) h 0 h Where the derivtive is defined, it is equl to the slope of the tngent to the curve f(x) t the point x 0 s shown in figure Using the limit concept to compute derivtive.

8 8 REVIEW OF SIMPLE UNIVARIATE CALCULUS FIGURE 8. The Derivtive of f t x 0 is the Slope of the Tngent Line t x 0 f x 10 tngent line 7.5 x 0,f x x Procedure. : Add h to (h 0) nd compute f(+h). b: Compute f() c: Compute the chnge in the vlue function: f(+h) - f(). f ( + h) f () d: For h 0, form the quotient h. e: Simplify the frction in d s much s possible. Whenever possible, cncel h from both the numertor nd denomintor. f(+h) f () f: f () is the number tht h pproches s h tends to zero Exmple. Let the function be f(x) x 3. : f(+h) (+h) h+3h 2 +h 3 b: f() x 3 c: f(+h) - f() 3 2 h+3h 2 +h 3 f ( + h) f () d: h 32 h+3h 2 +h 3 h 3 e: 2 h+3h 2 +h h+h 2 h f: As h 0, the expression goes to Some useful rules of differentition.

9 REVIEW OF SIMPLE UNIVARIATE CALCULUS Constnt functions. The derivtive of constnt is zero. f(x) f (x) 0 f(x) 3 0 (12) Additive nd multiplictive constnts. Consider the exmple y b + f(x) dy f (x) y f(x) dy f (x) If y 5+4f(x) then dy 4f (x) (13) (14) Power rule. Consider s n exmple If f(x) x n then f (x) nx n 1 (15) Iff(x) x 4 (16) f (x) 4x Sum nd difference rule. The derivtive of sum is the sum of the derivtives. The sme holds true for differences. So let H(x) f(x) + g(x). Then H (x) f (x) + g (x). Specificlly d[f(x)+g(x)] + dg(x) f (x)+g (x) (17) Consider s n exmple y 3x 2 +5x +3 f (x) 6x +5 (18)

10 10 REVIEW OF SIMPLE UNIVARIATE CALCULUS Product Rule. Let H(x) f(x)g(x). Then H (x) d[f(x).g(x)] H (x) d[f(x).g(x)] f(x). dg(x) (x) + g(x).df f(x).g (x)+g(x).f (x) We sometimes sy tht derivtive of the product of two functions is equl to the derivtive of the first times the second, plus the second times the derivtive of the first. Consider the following exmple where f(x) (3x+2) nd g(x) x 2 nd H(x) x 2 (3x + 2) 3x 3 + 2x 2. (19) H (x) f (x) dg(x) + g (x) (3x + 2)(2x) +(x 2 )(3) 6x 2 +4x +3x 2 (20) Tking the derivtive directly we obtin 9x 2 +4x H(x) 3x 3 +2x 2 H (x) 9x 2 +4x (21) Quotient Rule. If f nd g re differentible t x nd g(x) 0, then H( ) f( ) is differentible g( ) t x, nd H (x) d [ ] f(x) g(x) g(x) f (x) dg(x) [g(x)] 2 g(x) f (x) f(x) g (x) [g(x)] 2 Consider the following exmple where f(x) (3x - 5) nd g(x) (x - 2). (22) H (x) d H(x) 3x 5 x 2 [ ] f(x) g (x) g(x) f (x) dg(x) (g(x)) 2 (x 2) (3) (3x 5) (1) (x 2) 2 3x 6 3x +5 (x 2) 2 1 (x 2) 2 (23)

11 REVIEW OF SIMPLE UNIVARIATE CALCULUS Second nd higher order derivtives. The derivtive of function f is clled the first derivtive of f. If f is lso differentible, then we cn differentite f in turn. We cll f the second derivtive of f. We use the following nottion for the second derivtive ( ) d df(x) Consider the following exmple (f ) f (x) d2 f(x) 2 d2 y 2 (24) f(x) (3x +1)(x 2 +5) (3)(x 2 +5)+(3x + 1) (2x) 3x x 2 +2x 9x 2 +2x +15x ( ) d 18x +2 (25) The derivtive of d2 f(x) (f ) is clled the third derivtive of f. We use the nottion f or f (3) or d 3 f(x) 3 2 for this derivtive. For the nth derivtive we use the nottion f (n) or dn f(x) n The chin rule for composite functions Definition. Let f be defined on n open intervl S, nd let g be defined on f(s), nd consider the composite function g f defined on S by the eqution (g f )(x) g[f(x)] (26) Assume tht there is point c in S such tht f(c) is n interior point of f(s). If f is differentible t c nd if g is differentible t f(c), then g f is differentible t c nd we hve This cn lso be written (g f ) (c) g [f(c)] f (c) (27) dg[f(x)] dg[f(x)] df(x) g (f (x)) f (x) (28) Exmples. 1: Exmple 1

12 12 REVIEW OF SIMPLE UNIVARIATE CALCULUS f(x) x 2 +2x g(x) 3x 2 g(f(x)) 3 (x 2 +2x) 2 3(x 4 +4x 3 +4x 2 ) 3x 4 +12x 3 +12x 2 (29) dg[f(x)] dg[f(x)] 6(x 2 +2x)(2x +2) 6(2x 3 +2x 2 +4x 2 +4x) 12x 3 +36x 2 +24x If we simply tke the derivtive of (g f) we obtin 2: Exmple 2 g(f(x)) 3x 4 +12x 3 +12x 2 dg[f(x)] 12x 3 +36x 2 +24x (30) As nother exmple let y u 3 nd let u [1 - (2 + 3x) 2 ]. We cn then find dy/ s dy d [ 1 (2+3x) 2 ] 3 3[1 (2 + 3x) 2 ] 2 d [1 (2+3x)2 ] 3[1 (2 + 3x) 2 ] 2 [( 2)(2 + 3x) d (3x)] 3[1 (2 + 3x) 2 ] 2 [( 6)(2 + 3x)] (31) 3[1 (2 + 3x) 2 ] 2 [( 12 18x)] [1 (2+3x) 2 ] 2 ( 36 54x)

13 REVIEW OF SIMPLE UNIVARIATE CALCULUS Exponentil nd logrithmic functions Exponentil functions - e x. f (x) e x d(ex ) e x (32) Logrithmic functions - log (x). f (x) log (x) d(log(x)) 1 x (33) p x nd p f(x). For exmple d(p x ) d(p f(x) ) p x log p p f(x) log p (34) d(4 x ) 4 x log 4 d(5 3x2 ) 5 3x2 log 5(6x) Exponentil functions nd the chin rule. d[e g(x) ] e g(x) d[g(x)] (35) For exmple d[e 2x3 ] e 2x3 (6x 2 ) Logrithms nd the chin rule. d[log(g(x))] d[g(x)] g (x) g (x) g(x) (36) For exmple, d[log(2x 2 +3x)] 4x +3 2x 2 +3x 3.7. Implicit Differentition.

14 14 REVIEW OF SIMPLE UNIVARIATE CALCULUS The Ide of Implicit Differentition. Sometimes we cnnot solve n eqution for y s function of x, tht is we cnnot find n explicit formul for x. We my be ble to compute dy/ in some cses, however. Consider first n exmple where x 3 3xy 2 + y 3 0. While n explicit formul for y is not vilble, we cn differentite both sides of the eqution using the chin rule to obtin 3x 2 3x ( 2y dy ) x 3 3xy 2 + y y 2 Then we cn collect terms in dy nd rerrnge to obtin 3x 2 3 y 2 (6xy 3 y 2 ) dy dy 3x2 3 y 2 6 xy 3 y 2 At the point (2, -1) the derivtive is (-3/5). x2 y 2 2 xy y 2 +3y 2 dy 0 (37) Procedure to find dy/ for n implicit eqution in y nd x. 1: Differentite ech side of the eqution with respect to x, considering y s function of x. 2: Solve the resulting eqution for dy/ Liner pproximtions nd differentils Liner pproximtions. The liner pproximtion to f bout the point is given by f (x) f () +f ()(x ) (x close to ) (39) This is lso the eqution for the tngent t the point Definition of differentil. Consider differentible function y f(x) nd let denote n rbitrry chnge in the vrible x. The expression f (x) is clled the differentil of y f(x), nd it is denoted by dy (or df), so tht dy f (x). (40) Note tht dy is proportionl to with f (x) the fctor of proportionlity. Note tht if x chnges by, then the corresponding chnge in y f(x) is y f (x + ) f (x). (41) Now consider the liner pproximtion to function in eqution 39. Replce x by x + nd by x s follows f(x + ) f(x) +f (x)(x + x)(x + close to x) f(x + ) f(x) f (x)()(x + close to x) Now replce f(x + ) f(x) with y from eqution 41 nd f (x)() with dy from 40 to obtin (38) (42)

15 REVIEW OF SIMPLE UNIVARIATE CALCULUS 15 f(x + ) f(x) f (x)()(x + close to x) y f (x)() dy (x + close to x) y dy f (x)()(x + close to x) The differentil dy is not the ctul increment in y s x is chnged to x +, but rther the chnge in y tht would occur if y continued to chnge t the fixed rte f (x) s x chnges to x +. figure 9 illustrtes the difference between y nd dy. (43) FIGURE 9. Difference Between y nd dy y x 0, f x 0 f x x 0,f x 0 dy y x 0 x 0 x 3.9. Rules for differentils. 1: d(f + bg) df + bdg 2: d(f ( ) g) gdf + fdg 3: d f gdf fdg g g, g Exmple problems. For ech of the following find dy. : y 3x 2 +7x 5 b: y x (x 2 +3) c: y (x 8) (7x +4) d: y x x 2 +1

16 16 REVIEW OF SIMPLE UNIVARIATE CALCULUS Economic Exmple. Consider the following simple mcro model Y C + I C f(y ) (44) N g(y ) where Y is income, C is consumption, I is investment nd N is employment. Using differentils we cn find how Y nd N chnge s we chnge I. First find the differentil for ech eqution to obtin dy dc + di dc f (Y )dy (45) dn g (Y ) dy Now substitute for dc in the differentil of the ntionl income identity nd solve for dy to obtin dy dc + di f (Y ) dy + di dy (1 f (Y )) di 1 dy 1 f (Y ) di Similrly we cn obtin dn s (46) dn g (Y ) dy g (Y ) 1 f (Y ) di (47) Now if g (Y ) > 0 nd 0 f (Y ) 1, then s investment increses so does employment.

17 REVIEW OF SIMPLE UNIVARIATE CALCULUS ANTIDERIVATIVES 4.1. Definition of n Antiderivtive. A function F is clled the ntiderivtive of f on n intervl I if F (x) df/ f(x) for ll x in I. If F is n ntiderivtive of f on n intervl I, then the most generl ntiderivtive of F on I is F(x) + c, where c is n rbitrry constnt Exmples. 1: Let F(x) 4x nd f(x) F (x) 4. Then prticulr ntiderivtive of f(x) 4 is F(x) 4x nd generl ntiderivtive of f(x) 4x +c. For exmple G(x) 4x+5 hs derivtive G (x) 4 s does H(x) 4x+22. 2: Let F(x) ln x nd f(x) 1/x. Then prticulr ntiderivtive of f(x) 1/x is F(x) ln x for x > 0. f(x) is not defined for x Antidifferentition Formuls. TABLE 1. Antidifferentition Formuls Function f(x) c f(x) f(x) + g(x) x n x,x -1 n + 1 n +1 1/x ln x e x e x Antiderivtive F(x) c F(x) F(x) + G(x)

18 18 REVIEW OF SIMPLE UNIVARIATE CALCULUS 5. INTEGRATION 5.1. Intuition. The re under curve f(x) between x nd x b is clled the integrl of f between nd b. In figure10 the re beneth the curve between nd b is the vlue of the integrl of f(x) evulted from to b. f x FIGURE 10. Integrtion nd Are b x 5.2. Riemnn integrl (Rudin [1, p. 120]). Let P {x 0,x 1,x 2,..., x n } be prtition of the closed intervl [, b] where x 0 x 1... x n 1 x n b nd let x i x i -x i 1. Suppose f is rel bounded function defined on [,b]. Corresponding to ech prtition P of [, b] define the following M i sup f(x) (x i 1 x x i ), m i inf f(x) (x i 1 x x i ), n U(P, f) M i x i n M i (x i x i 1 ) L(P, f) i1 n m i x i i1 i1 i1 n m i (x i x i 1 ). (48)

19 REVIEW OF SIMPLE UNIVARIATE CALCULUS 19 Consider the function defined in figure 11 over the intervl [,b]. The lower sum (L(P, f) n i1 m i x i n i1 m i(x i x i 1 ))uses the lowest vlue of f(x) in ny intervl. In figure 12, the intervl [,b] is divided into ten prts. The pproximtion to the re is better thn when the prtition is six. FIGURE 11. Lower Sum with Six Divisions FIGURE 12. Lower Sum with Ten Divisions The upper sum (L(P, f) n i1 M i x i n i1 M i(x i x i 1 ))uses the highest vlue of f(x) in ny intervl. In figure 13, the intervl [,b] is divided into six prts while in figure 14, it is divided into ten prts.

20 20 REVIEW OF SIMPLE UNIVARIATE CALCULUS FIGURE 13. Upper Sum with Six Divisions FIGURE 14. Upper Sum with Ten Divisions Now we cn define the upper nd lower Riemnn integrls of f over [, b] s b b f(x) inf U(p, f), f(x) sup L(P, f), where the inf nd sup re tken over ll prtitions P of [,b]. If the upper nd lower integrls re equl we sy tht f is Riemnn integrble on [,b] nd write the common vlue of eqution 49 s b (49) f(x) (50)

21 REVIEW OF SIMPLE UNIVARIATE CALCULUS Riemnn-Stieltjes integrl (Rudin [1, p. 122]). Let α be monotoniclly incresing function on [, b]. Corresponding to ech prtition P of [, b] define α i α(x i ) α(x i 1 ) (51) where it is cler tht α i 0. Now define U(P, f, α) L(P, f, α) n M i α i, i1 (52) n m i α i. where M i nd m i re defined s in 48. Now define the following upper nd lower integrls s b b i1 fdα inf U(p, f, α), fdα sup L(P, f, α), where the inf nd sup re tken over ll prtitions P of [,b]. If the upper nd lower integrls re equl we write the common vlue of eqution 53 s (53) or sometimes s b fdα (54) b f(x) dα(x). (55) The ide is tht we consider the upper sum for ll possible prtitions nd tke its lowest vlue. Then we consider the lower sum for ll possible prtitions nd tke its highest vlue. These two vlues should be the sme nd re the integrl of f(x) over the intervl [, b] Fundmentl theorem of clculus. If f is Riemnn integrble on [,b] nd if there is differentible function F on [,b] such tht F df/ f then b f(x) F (b) F (). (56) Consider for exmple the function F(x) 2x where 3 nd b 5. In this cse F f 2 nd b f(x) F (b) F () F (5) F (3) 2(5) 2(3) Or consider for exmple the function F(x) 3x 2 + 2x where 1 nd b 4. In this cse F f 6x + 2 nd (57)

22 22 REVIEW OF SIMPLE UNIVARIATE CALCULUS 4 1 b f(x) F (b) F () (6x +2) F (4) F (1) [3 (4 2 ) + 2(4)] [3 (1 2 ) + 2(1)] [48 + 8] [3 + 2] (58) 5.5. Properties of the definite integrl. 1: c c (b ), where c is ny constnt b 2: 3: 4: b [ f (x) +g(x)] b f (x) + b g(x) b cf(x) c b f (x) b [ f (x) g(x)] b f (x) b g(x) 5.6. Nottion used in evluting definite integrls. So for exmple G (x) b G(b) G() (59) 4 1 b f(x) F (x) b (6x +2) ( 3x 2 +2x ) 4 1 F (b) F () [ 3(4 2 ) + 2(4) ] [ 3(1 2 ) + 2(1) ] [48 + 8] [3 + 2] (60) 5.7. Some useful integrtion formuls. : powers b: reciprocls b x n xn+1 n +1 b bn+1 n +1 n+1, n 1 (61) n +1 c: exponentils b 1 x ln(x) b ln(b) ln() ln(b/) (62) b e x e x b e b e (63)

23 REVIEW OF SIMPLE UNIVARIATE CALCULUS Indefinite Integrls. Recll tht function F is clled n ntiderivtive of f on n intervl I if F (x) f(x) for ll x in I. An lterntive nottion for the ntiderivtive of f is clled the indefinite integrl nd is written ntiderivtive of f(x) f(x) (64) Opertionlly we then hve f(x) F (x) mens F (x) f(x) (65) Note tht the definite integrl b f(x) is number while the indefinite integrl f(x) is function. Note tht we lso hve the following b f(x) 5.9. Differentition of Integrl Expressions. f(x) b (66) Some fundmentl rules. The derivtive of n indefinite integrl with respect to the vrible of integrtion is just the integrnd d f(x) f(x) (67) Similrly the integrl of derivtive of function with respect to vrible is just tht function plus constnt F (x) F (x) +c (68) For definite integrl where F (x) f(x), this gives t f(x) F (x) t F (t) F () (69) Differentition of n integrl expression. First consider the cse where the lower limit of integrtion () is fixed d dt t f(x) d [ F (t) F ()] dt F (t ) f (t ) Similrly when the upper limit of integrtion is fixed d dt t f(x) F (x) t F () F (t) f(x) d [F () F (t)] t dt F (t) f(t) And in generl if (t) nd b(t) re differentible nd f(x) is continuous we obtin (70) (71)

24 24 REVIEW OF SIMPLE UNIVARIATE CALCULUS b(t) d f(x) f[b(t)] b (t) f [(t)] (t) (72) dt (t) We cn prove this by letting F (x) f(x) nd writing v f(x) F (v) F (u). Now let u u b(t) nd v b(t), write out the definition of the integrl nd then differentite d b(t) (t) [ b(t) (t) f(x) F [b(t)] F [(t)] ] f(x) d [F [b(t)] F [(t)]] F [b(t)] b (t) F [(t) (t)] f[b(t)] b (t) f[(t)] (t) (73) Integrtion Tble. Here is tble listing integrls for vriety of functions. [f(x) +g(x)] f(x) + g(x) cf(x) c f(x) b f(x) c f(x) + b f(x) c sin (x) cos ( x )+c x n +1 xn n +1 + c (n 1) cos (x) sin (x) +c 1 1 x ln x + c x 2 +1 tn 1 (x) +c e x e x + c sec 2 (x) tn (x) +c ln x x ln x x + c csc 2 ( x ) cot (x) +c e x 1 ex + c sec (x) tn (x) sec (x) +c p x px ln p + c csc (x) cot (x) csc (x) +c 1 1 x 2 sin 1 (x) +c Integrtion using substitution Forml Definiion of Integrtion by Substitution. Consider the function y F(g(x)). Using the chin rule we cn compute the derivtive dy/ s follows. dy d [F (g(x))] F (g(x)) g (x) (74) Now consider the indefinite integrl of the rightmost expression in eqution 74. F (g(x)) g (x) F (g(x)) + c (75) Now ssume tht F is n ntiderivtive for function f. We then hve f(g(x)) g (x) F (g(x)) g (x) F (g(x)) + c (76) By mking the substitution u g(x) we cn often clculte the integrl f(g(x)) g (x) by computing f(u) du F (u) +c nd then mking the substitution u g(x). Specificlly f(g(x)) g (x) f(u) du (77) u g(x),du g (x)

25 REVIEW OF SIMPLE UNIVARIATE CALCULUS 25 The ide is to write the originl integrl in n lterntive form mking the substitution u g(x) nd du g (x), computing the integrl f(u) du F (u) + c, nd then substituting bck in for u Exmple 1. Consider the following exmple. Find 1 (78) (3+2x) 2 Let u 3 + 2x. Then du 2. We cn then write the following 1 (3 + 2x) (3+2x) u du ( ) u + c ( ) (79) (3+2x) + c ( ) 1 + c 2(3 + 2x) ( ) 1 (3+2x) 1 + c 2 We cn check the nswer by differentiting ( ) ( ) d 1 1 (3 + 2 x) 1 + c ( 1) (3 + 2 x) 2 (2) 2 2 ( ) 2 (3 + 2 x) 2 (80) 2 1 (3+2x) Exmple 2. Consider nother exmple. Find x 2 + b 2 x 2 (81) Let u 2 +b 2 x 2. Then du 2b 2 x. We cn then write the following x 2 + b 2 x 2 1 (2 b 2 x) 2 b b 2 x 2 1 udu 2 b b 2 ( ) 2 u 3/2 + c b 2 u3/2 + c 1 3 b 2 (2 + b 2 x 2 ) 3/2 + c (82)

26 26 REVIEW OF SIMPLE UNIVARIATE CALCULUS We cn check the nswer by differentiting d [ ] 1 3 b 2 (2 + b 2 x 2 ) 3/2 ]+c ( )( ) 3 1 ( 2 + b 2 x 2 ) 1/2 2 b 2 x 2 3 b 2 ( )( 3 2 b 2 ) x ( 2 + b 2 x 2 ) 1/2 2 3 b 2 x ( 2 + b 2 x 2 ) 1/2 (83) x 2 + b 2 x Integrtion by prts Procedure for Integrtion by Prts. Consider the function h(x) f(x)g(x). Using the product rule nd differentiting we obtin d h(x) Now integrte both sides of the expression s follows d h(x) d f(x) g(x) f(x) g (x) +f (x) g(x) (84) d f(x) g(x) f(x) g (x) + f (x) g(x) f(x) g(x) + c f(x) g (x) + f (x) g(x) f(x) g (x) f(x) g(x) f (x) g(x) + c f(x) g (x) f(x) g(x) f (x) g(x) We cn use this informtion to compute f(x) g (x) by computing f (x) g(x) when the second integrl is esier to compute. In prctice we typiclly mke the following substitutions (85) u f(x), du f (x), dv g (x) v g(x) (86) Exmple 1 of Integrtion by Prts. Consider the following exmple. Find xe x (87) Set u x, dv e x du, v e x (88) Then

27 REVIEW OF SIMPLE UNIVARIATE CALCULUS 27 xe x udv uv vdu xe x e x xe x e x + c (89) Exmple 2 of Integrtion by Prts. Find xe x (90) (x +1) 2 Set Then u xe x, dv 1 (x +1) 2 du (xe x + e x ), v 1 xe x (x +1) 2 udv uv xex x +1 xex x +1 xex x +1 + vdu x +1 1 x +1 (xex + e x ) 1 x +1 ex (x +1) e x xex x +1 + ex + c xex x +1 + ex (x +1) + c x +1 ex x +1 + c We cn check the nswer by differentiting [ d e x ] x +1 + c (x +1)ex e x (x +1) 2 xe x (x +1) 2 (91) (92) (93)

28 28 REVIEW OF SIMPLE UNIVARIATE CALCULUS REFERENCES [1] Rudin, Wlter. Principles of Mthemticl Anlysis, 3 rd Edition. New York: McGrw Hill, 1976.