Integration. 1 Differentiation, Integration and the Fundamental Theorem of Calculus. November 5, Introduction. 1.

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1 Integrtion November 5, Differentition, Integrtion nd the Fundmentl Theorem of Clculus 1.1 Introduction In this section the derivtive nd definite integrl re introduced in their correct setting. At Leving Cert level, integrtion is introduced merely s the inverse of differentition. This pproch simplifies things but greter understnding comes out of proper tretment. Historiclly integrtion ws developed seprtely of differentition nd the link between them ws lter discovered. The link between them; nmely tht integrtion is indeed the inverse of differentition, is known s the Fundmentl Theorem of Clculus. The topics of coordinte geometry, limits nd functions should be studied in more depth prior to thorough study clculus, but this brief section is merely intended s n exposition to id understnding. 1.2 Differentition In the figure below, the line from to b is clled secnt line. Figure 1: Secnt Line. 1

2 Integrtion 2 Introduce the ide of slope. The slope of line is something intuitive. A steep hill hs greter slope thn gentle rolling hill. The slope of the secnt line is simply the rtio of how much the line trvels verticlly s the line trvels horizontlly. Denote slope by m: m = c bc. (1) Wht bout the slope of the curve? From to b it is continuously chnging. Mybe t one point its slope is equl to tht of the secnt but tht doesn t tell much. It could be estimted, however, using ruler the slope t ny point. It would be the tngent, s shown: Figure 2: Tngent Line The bove line is the slope of the curve t t. Construct secnt line: Figure 3: Secnt line nd Tngent line Now with respect to nlytic geometry, with function f(x), the slope of this secnt is given by: m = m = f(x + h) f(x) (x + h) x f(x + h) f(x) h

3 Integrtion 3 It is pprent tht the secnt line hs slope tht is close, in vlue, to tht of the tngent line. Let h become smller nd smller: Figure 4: Secnt line pproching slope of Tngent line The slope of the secnt line is lmost identicl to tht of our tngent. Let h 0. Of course, if h = 0 there is no secnt. But if h got so close to 0 s doesn t mtter then there would be secnt nd hence slope. With respect to nlytic geometry, with function f(x), the slope of this secnt, which is indistinguishble from tht of the tngent to the point, is given by: f(x + h) f(x) m = lim. (2) h 0 h This m is the derivtive of f(x). This gives the slope of the curve t every point on the curve Nottion There re two significnt brnches of nottion used to denote the derivtive of f. The difference is just nottion. They re different wys of writing down the sme thing. This subsection is included to fight ginst simple misunderstndings. Newtonin Nottion The function is denoted f(x) nd the grph is the set of points (x, f(x)): The derivtive of f(x) is denoted f (x). Other nmes for the derivtive of f(x) include: the differentition of f(x) the derived function for f(x) the slope of the tngent t (x, f(x)) the grdient df dx

4 Integrtion 4 f x x Figure 5: Newtonin nottion for functions y x Figure 6: Leibniz nottion for functions Leibniz Nottion The function is denoted y = f(x) (e.g. y = x 2 ); nd the grph is the set of points (x, y): In this nottion, y is equivlent to f(x). However, the nottion for the derivtive of y is: dy dx. (3) It must be understood tht if y = f(x); then f (x) dy dx, (4) nd there is no notion of cnceling the ds; it is just nottion. It is n illuminting one becuse if the second grph of figure 4 is mgnified bout the secnt: If dy is ssocited with smll vrition in y f(x + h) f(x); nd dx ssocited with smll vrition in x h; then dy/dx mkes sense.

5 Integrtion 5 Figure 7: Leibniz nottion for the derivtive To reiterte if y = f(x); then dy/dx is the sme thing s: the derivtive of f(x) the differentition of f(x) the derived function for f(x) the slope of the tngent t (x, f(x)) the grdient f (x) So to emphsise f(x) nd y re freely interchngeble; s re f (x) nd dy/dx. 1.3 Integrtion Wht is the re of the shded region under the curve f(x)? Strt by subdividing the region into n strips S 1, S 2,..., S n of equl width s Figure 8.

6 Integrtion 6 Figure 8: The width of the intervl [, b] is b so the width of ech of the n strips is x = b n. Approximte the ith strip S i by rectngle with width x i nd height f(x i ), which is the vlue of f t the right endpoint. Then the re of the ith rectngle is f(x i ) x i : The re of the originl shded region is pproximted by the sum of these rectngles: A f(x 1 ) x + f(x 2 ) x + + f(x n ) x. (5) This pproximtion becomes better nd better s the number of strips increses, tht is, s n. Therefore the re of the shded region is given by the limit of the sum of the res of pproximting rectngles: Definition A = lim [f(x 1 ) x + f(x 2 ) x + f(x n ) x]. (6) If f(x) is function defined in [, b] nd x i, x re s defined bove, then the definite integrl of f from to b is b f(x) dx = lim f(x i ) x. (7)

7 Integrtion 7 So n integrl is n infinite sum. Associte lim n. Agin dx is ssocited with smll vrition in x x. 1.4 Fundmentl Theorem of Clculus Rough Version If f is function with derivtive f then b f (x) dx = f(b) f() Proof b f (x) dx = lim b f f(x + h) f(x) (x) = lim, h 0 h f (x) dx = lim f (x i ) x, [ ] f(x i + h) f(x i ) lim x. h 0 h Note tht n nd h 0 is equivlent to h x. This is becuse x let h x; b f (x) dx = lim h 0 b Note h x so x i + h x i+1 : f (x) dx = lim h 0 [ ] h lim f(x i + h) f(x i ) h 0 h, [f(x i + h) f(x i )]. 0. Hence b f (x) dx = [f(x i+1 ) f(x i )]. Note i [f(x i+1) f(x i )] forms telescoping sum: [f(x i+1 ) f(x i )] = f(x 2 ) f(x 1 ) + f(x 3 ) f(x 2 ) + f(x 4 ) f(x 3 ) +. f(x n+1 ) f(x n );

8 Integrtion 8 where ll but f(x n+1 ) f(x 1 ) cncels. But x n+1 = b nd x 1 =. Hence f(x n+1 ) = f(b) nd f(x 1 ) = f(). Hence b f (x) dx = b [f(x i+1 ) f(x i )], f (x) dx = f(b) f() The Indefinite Integrl In Leving Cert where integrls re defined s the inverse of derivtives, n indefinite integrl defines integrtion. Definition If f(x) is function nd it differentil with respect to x is f (x), then f (x) dx = f(x) + c (8) where c is clled the constnt of integrtion. Note the constnt of integrtion. It s inclusion is vitl becuse if f(x) is function with derivtive f (x) then f(x) + c lso hs derivtive f (x) s: d (f(x) + c) = df + d dx }{{} dx dx c, }{{} =f (x) =0 d dx (f(x) + c) = f (x). Geometriclly curve f(x) with slope f (x) hs the sme slope s curve tht is shifted upwrds; f(x)+ c. Note tht the constnt of integrtion cn be disregrded for the indefinite integrl. Suppose the integrnd is f (x) nd the nti-derivtive is f(x) + c. Then: The cs cncel! b f (x) dx = (f(b) + c) (f() + c), b f (x) dx = f(b) f().

9 Integrtion Conclusion Finding the derivtive of function f t x is finding the slope of the tngent to the curve t x. Integrtion menwhile mesures the re between two points x = nd x = b. The Fundmentl Theorem of Clculus sttes however tht differentition nd integrtion re intimtely relted; tht is given function f: d f(x) dx = f(x), dx d f(x) dx = f(x) + c. dx i.e. differentition nd integrtion re essentilly inverse processes. 2 Integrtion 2.1 Wht We Know From the Fundmentl Theorem of Clculus f (x) dx = f(x) + c (9) Thus: f(x) f(x) x n x (n 1) n+1 n+1 cos x sin x + c sin x cos x + c e x e x + c sec 2 x tn x + c 1 x ln x + c Also becuse d sin nx = n cos nx, nd dx d cos nx = n sin nx dx sin nx cos nx dx = + c nd n sin nx dx = cos nx n + c

10 Integrtion 10 Also, let > 0; Also d 1 dx rctn x = x 2 /.1 2 d 1 dx rctn x = 1 1 ( 2 + x 2 )/.1 2 d 1 dx rctn x = x x dx = 1 2 rctn x + c d dx rcsin x = x2 / 2 d dx rcsin x = 1 1 (2 x 2 )/ 2 d dx rcsin x = x x dx = rcsin x Properties of Integrtion Proposition (f(x) ± g(x)) dx = f(x) dx ± g(x) dx (10) k f(x) dx = k f(x) dx, where k R (11) Proof (Non-Exminble) (f(x) ± g(x)) dx = lim (f(x i ) ± g(x i )) x ( ) (f(x) ± g(x)) dx = lim f(x i ) x ± f(x i ) x (f(x) ± g(x)) dx = lim f(x i ) x ± lim g(x i ) x (f(x) ± g(x)) dx = f(x) dx ± g(x) dx

11 Integrtion 11 Let k R. k f(x) dx = lim k f(x) dx = lim k f(x) dx = k ( k lim kf(x i ) x f(x i ) x ) f(x i ) x k f(x) dx = k f(x) dx Remrks There is no direct nlogue of the product, quotient nor chin rule for integrtion. Although the substitution method below is like chin rule for integrls The Substitution Method for Evluting Integrls f(g(x))g (x) dx = f(u) du (12) where u = g(x) Proof (Non-Exminble) u = g(x) du dx = g (x) dx du = 1 g (x) dx = du g (x) So f(g(x))g (x)dx = f(u) g (x) du g (x) = f(u)du

12 Integrtion Indefinite Integrls Proposition Suppose f(x) hs n nti-derivtive F (x). Then b f(x) dx = b f(x) dx (13) Proof (Non-Exminble) b f(x) dx = F (b) F () = (F () F (b)) = b f(x) dx 2.4 Techniques of Trigonometric Integrtion Consider the integrl cos 4x cos 2x dx (14) To differentite such function is no problem; there is product rule. However, there is no such product rule for integrtion. There re functions, such s sin x dx (15) x which, becuse there is no product or quotient rule for integrtion, my not be integrted esily. In fct, (15) hs no ordinry solution in terms of elementry functions (like polynomils, exponentil functions, logrithms, trigonometric functions, inverse trigonometric functions nd their combintions). This won t hppen t LC level however so (14) my certinly be integrted. The key is to relise tht there is sum rule for integrtion so if cos 4x cos 2x cn be written s sum of integrble terms things will work out esily. Luckily such formule exist nd re in the tbles: Now (14) my be esily integrted. 2 cos A cos B = cos(a + B) + cos(a B) (16) 2 sin A cos B = sin(a + B) + sin(a B) (17) 2 sin A sin B = cos(a B) cos(a + B) (18) 2 cos A sin B = sin(a + B) sin(a B) (19)

13 Integrtion Products of Unlike Trigonometric Terms To integrte functions of the kind: cos mx cos nx (20) cos mx sin nx (21) sin mx cos nx (22) sin mx cos nx (23) convert the product to sum using (16-19) Consider now the integrl I = sin 2 θ cos 3 θ dθ (24) There is no vilble formul to express this product s sum. technique however: sin 2 θ cos 3 θ dθ = sin 2 θ cos 2 θ cos θ dθ There is substitution Now becuse cos nd sin re (up to sign) derivtives of ech other, spot tht cos θ is the derivtive of sin θ so cos θ cn ply the rôle of the g (x) in (12). Hence let u = sin θ: I = du dθ = cos θ dθ = du cos θ sin 2 θ cos 2 θ cos θ du cos θ I = u 2 cos 2 θ du Now it seems ll is lost but s cos 2 x + sin 2 x = 1 cos 2 x = 1 sin 2 x I = u 2 (1 u 2 ) du Typiclly t LC level the powers of sin nd cos in this cse will be under 3.

14 Integrtion Products of Powers of sin & cos To integrte functions of the kind: brek the one with n odd power into squre plus 1; i.e. sin n x cos m θ (25) sin 2n+1 θ = sin 2n θ sin θ nd then let u = the other one. If m nd n re both even (m = 2 = n) then use the formule: cos 2 x = 1 (1 + cos 2x) (26) 2 sin 2 x = 1 (1 cos 2x) (27) cos 3 x nd sin 3 x The integrls of cos 3 x nd sin 3 x require mix of the previous two cses. The integrnd is mnipulted s follows (similr for sin 3 x): which cn be esily integrted. 2.5 Intersecting Curves cos 3 x = cos 2 x cos x cos 3 x = 1 (1 + cos 2x) cos x 2 cos 3 x = 1 (cos x + cos 2x cos x) ( 2 ) 1 cos 3x + cos x cos x + (16) 2 2 cos 3 x = 3 cos x cos 3x Suppose the re between two curves is required: There re few fcts we must know. By definition, the re of function between x = nd x = b is the definite integrl from to b. Now consider If f(x) g(x) for x [, b] then, where A b (f, g) is the re between f(x) nd g(x) between x = nd x = b: A b (f, g) = b f(x) dx b g(x) dx = b (f(x) g(x)) dx (28)

15 Integrtion Figure 9: The re enclosed between f(x) = x 2 nd g(x) = x is bound inside the points where they intersect: x = 0, 1 Figure 10: The re between f(x) nd g(x) (shded) is the re under f(x) less the re under g(x). 2.6 Chnging Products to Sums nd Sums to Products Frequently it will be helpful to chnge sums of trigonometricl expressions into products or vice vers. For exmple to find the roots of sin 3x sin x it would very helpful to chnge this to product nd using the fct tht to find the roots. Indeed using (??):.b = 0 = 0 or b = 0 (29) sin 3x sin x = 2 cos 2x cos x

16 Integrtion 16 In contrst, some things re esier s sums. For exmple there is no product rule for integrtion; f(x)g(x) dx hs no formul in terms of f nd g. However (f(x) + g(x))dx = f(x) dx + g(x) dx (30) so it cn be hndy to write product s sum. For exmple, using (??): 2 cos 5x cos 3x dx = cos 8x dx + cos 2x dx Substitution Technique for Integrtion In generl, if we need to mke substitution, we look for the following pttern: f(g(x))g (x) dx (31) tht is we look for function nd its derivtive in the integrnd. Then let u = g(x): f(g(x))g (x) dx = du dx = g (x) dx = du g (x) f(u) g (x) du g (x) = f(u) du Exmples Spot the ptterns: 2x 2 x dx t(5 + 3t 2 ) 8 dt x 2 e x3 dx s s 3 ds dx 3x x 2 x 2 sec 2 (x 3 + 1) dx sin 2 x cos x dx

17 Integrtion 17 LIATE If we cnnot see g(x), g (x) pttern we cn use the LIATE rule. Choose u ccording to the most complicted expression in the following hierrchy: L ogrithms I nverses (inverse sine, tn) A lgebric (polynomils in x) T rigonometric E xponentil In generl this works well (lso works for Integrtion by Prts) Ares under Curves The Bsics By definition, the re of function between x = nd x = b is the definite integrl from to b. Note however tht res under the x-xis will be computed to hve negtive re so if the function looks like: Figure 11: The re under the curve from x = 1 to x = 2 will be the integrl from x = 1 to 0, minus the integrl from x = 0 to 2. Ares between Curves Now consider If f(x) g(x) for x [, b] then, where A b (f, g) is the re between f(x) nd g(x) between x = nd x = b: A b (f, g) = b f(x) dx b g(x) dx = b (f(x) g(x)) dx (32) Note if f(x) g(x) then f(x) g(x) is lwys positive so the problem of negtive re doesn t rise.

18 Integrtion 18 Figure 12: The re between f(x) nd g(x) (shded) is the re under f(x) less the re under g(x). Intersecting Curves Suppose we wnt to find the re bound by two curves, e.g.: Figure 13: The re enclosed between f(x) = x 2 nd g(x) = x is bound inside the points where they intersect: x = 0, 1 If the curves re y = f(x) nd y = g(x); there intersubsections re the set of points such tht: f(x) = g(x) (33) For exmple for f(x) = x 2 nd g(x) = x, the intersubsections occur when x 2 = x x 2 x = 0 x(x 3/2 1) = 0 x = 0, 1

19 Integrtion 19 Hence the re enclosed between the curves is given by: 1 0 ( x x 2 ) dx (34)