x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

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1 CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick properties of the definite integrl: c b b b f(x)dx = f(x)dx = b [f(x) + g(x)]dx = b f(x)dx + b f(x)dx cdx = c(b ), where c is ny constnt. c[f(x)]dx = c [f(x) g(x)]dx = b c f(x)dx = b b b b f(x)dx + f(x)dx f(x)dx f(x)dx b b g(x)dx g(x)dx 5.3 The Fundmentl Theorem of Clculus The Fundmentl Theorem of Clculus (FTC) links the two brnches of clculus: differentil clculus (this course mostly) nd integrl clculus (this chpter nd Mth 11). It does so by describing how integrtion

2 CHAPTER 5. INTEGRALS 6 nd differentition re inverse processes. Also note tht the FTC hs two prts, usully just clled Prt 1 nd Prt, but the prts re often interchnge in other textbooks. The first prt of the FTC dels with functions defined by n eqution of the form g(x) = x f(t)dt where f is continuous function on [, b] nd x vries between nd b. If f is positive, we cn think of g(x) s the re under the grph of f from to x, or the re so fr. It turns out tht g is the ntiderivtive of f. This gives: Theorem 169 (The Fundment Theorem of Clculus, Prt 1). If f is continuous on [, b], then the function g defined by g(x) = x f(t)dt for x b is continuous on [, b] nd differentible on (, b), nd g (x) = f(x). Proof. And discussion on pge s Problem 17 (Exercise 1, pge 388). Find the derivtive of g(x) = x t + 4dt. Problem 171 (Exercise 1, pge 388). Find the derivtive of G(x) = 1 x cos( t)dt. Problem 17. Find d x 4 sec(t)dt. dx 1

3 CHAPTER 5. INTEGRALS 63 In the previous section, we sw tht to work out the definite integrl s the limit of Riemnn sums is difficult. Prt of the FTC gives much simpler method: Theorem 173 (The Fundmentl Theorem of Clculus, Prt ). If f is continuous on [, b], then b f(x)dx = F(b) F(), where F is ny ntiderivtive of f, tht is, function such tht F = f. Proof. See pges Prt tell us tht if we know n ntiderivtive F of f then we cn evlute b f(x)dx simply by subtrcting the vlues of F t the endpoints of the intervl [, b]. Problem 174 (Exercise, pge 388). Evlute the integrl 1 (1 + 1 u4 5 u9 )du. Problem 175 (Exercise 6, pge 388). Evlute π π cos θdθ. Problem 176. Evlute t dt.

4 CHAPTER 5. INTEGRALS 64 Problem 177 (Exercise 5, pge 397). Evlute the integrl (3u + 1) du. 5.4 Indefinite Integrls nd the Net Chnge Theorem In Section 5.3 we used the FTC, prt, to evlute the definite integrl of function. Now we look t indefinite integrls (those without limits of integrtion). This is the equivlent to derivtives s functions s compred to derivtives t point. The nottion f(x)dx is going to be used for the ntiderivtive of f nd is clled n indefinite integrl. (fx)dx = F(x) mens F (x) = f(x). For exmple, x dx = x3 3 + C. Note. A definite integrl b f(x)dx is number, wheres n indefinite integrl f(x)dx is function (or fmily of functions).

5 CHAPTER 5. INTEGRALS 65 Tble of Indefinite Integrls c f(x)dx = [f(x) + g(x)]dx = kdx = x n dx = 1 x dx = e x dx = x dx = sin xdx = cos xdx = sec x dx = csc x dx = sec xtnxdx = csc xcot xdx = x dx = 1 dx = 1 x Note. We dopt the convention tht when formul for generl indefinite integrl is give, it is vlid only on n implied intervl. For exmple, we write 1 x dx = 1 x + C with the understnding tht it is vlid on the intervl (, ) or (,) only. Problem 178 (Exercise 14, pge 397). Find (csc t e t )dt.

6 CHAPTER 5. INTEGRALS 66 Problem 179 (Exercise 5, pge 397). Find (x + x dx. Problem 18. Find cosθ sin θ dθ Applictions Prt of the FTC sys tht if f is continuous on [, b], then b f(x)dx = F(b) F() where F is ny ntiderivtive of f. This ment tht F = f, so we could rewrite this s Consider F (x) s rte of chnge. b F (x)dx = F(b) F(). Theorem 181 (The Net Chnge Theorem). The integrl of rte of chnge is the net chnge: b F (x)dx = F(b) F(). In terms of some of the pplictions we hve lredy discussed: If the rte of growth of popultion is dn dt, then t t 1 dn dt dt = n(t ) n(t 1 ) is the net chnge in popultion during the time period from t 1 to t. If n object moves long stright line with position function s(t), then its velocity is v(t) = s (t), so t t 1 v(t)dt = s(t ) s(t 1 ) is the net chnge of position or displcement of the prticle during the time period from t q to t.

7 CHAPTER 5. INTEGRALS 67 Also, t t 1 The ccelertion of the object is (t) = v (t), so v(t) dt = totl distnce trvelled. t is the chnge in velocity from time t 1 to time t. t 1 (t)dt = v(t ) v(t 1 ) 5.5 The Substitution Rule The ntiderivtive formuls we looked t still do not tell us how to do ll integrls, here we introduce rule tht llows us to find integrls of functions such s x 1 + x dx. Here we use Theorem 18 (The Substitution Rule). If u = g(x) is differentible function whose rnge is n intervl I nd f is continuous on I, then f(g(x)g (x)dx = f(u)du. The Substitution Rule is functionlly the inverse of the Chin Rule, when using the Substitution Rule we re mking trnsformtion tht simplifies the inside function llowing us to see how to integrte the outside function. Exmple 183. Consider the integrl x 1 + x dx. Here we let u = 1 + x. Tking the derivtive of ech side with respect to x we find du dx = x. Multiplying ech side by dx gives du = xdx. We see now tht our integrnd is just 1 + x (xdx) = u du. Thus our integrl becomes udu u 3 = However, we wnt our nswer in terms of x, which is 3 3 (1 + x ) 3. + C.

8 CHAPTER 5. INTEGRALS 68 Problem 184. Clculte e 6x dx. Problem 185. Clculte x x dx. Problem 186. Clculte x (x + 1) dx. Problem 187. Clculte sec(θ) tn(θ)dθ Definite Integrls Substitution is little trickier with definite integrls becuse the limits of integrtion re in terms of prticulr vrible. There re two equivlent methods (mening both give the sme correct nswer) for evluting definite integrls with substitution:

9 CHAPTER 5. INTEGRALS evlute the indefinite integrl first, then use the FTC.. chnge limits of integrtion (by plugging them in to u(x)), then evlute the new definite integrl in terms of u. The second method bove is stted formlly in the following theorem. Theorem 188 (The Substitution Rule for Definite Integrls). If g is continuous on [, b] nd f is continuous on the rnge of u = g(x), then b f(g(x))g (x)dx = g(b) Problem 189. Evlute xdx. g() f(u)du. Problem 19. Evlute (x 1) 5 dx. Problem 191. Evlute 1 x (1 + x 3 ) 5 dx

10 CHAPTER 5. INTEGRALS 7 Problem 19. Evlute π cos x sin(sin x)dx Symmetry We cn use the following theorem to simplify clcultion of integrls of functions tht possess symmetry. Theorem 193 (Integrls of Symmetry Functions). Supposed f is continuous on [, ]. () If f is even (i.e. f( x) = f(x)), then f(x)dx = f(x)dx. (b) If f is odd (i.e. f( x) = f(x)), then f(x)dx =.

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