Mthemtics for Chemists 2 Lecture 1: Integrl clculus I Indefinite nd definite integrls, sic clcultion of integrls Johnnes Kepler University Summer semester 2012 Lecturer: Dvid Sevill Integrl clculus I 1/19 Integrtion: the opposite of differentition Prolem: If we hve function s(t) tht descries the distnce trveled y n oject fter time t, how do we clculte its speed t ny moment? Answer: Clculte the derivtive of s: v(t) = s (t). This function gives the instntneous speed for ny t. Prolem: If we hve function v(t) tht descries the speed of n oject t ny time t, how do we clculte its position? Answer: Clculte function s(t) such tht s (t) = v(t) nd s(0) is the initil position. Such function is clled n ntiderivtive or primitive of v(t). We will dedicte some time to lerning how to clculte primitives of functions nd wht kind of prolems we cn solve with this (prt from the one we just tried). Integrl clculus I 2/19
Definite nd indefinite integrls Consider the following in reltion to derivtives: We hve the concept of derivtive of function t point nd the concept of derivtive function. The first concept is geometric, nd it is numeric vlue ssocited to ech point in the domin. The second concept is nlytic, nd it is just function tht produces the vlues of the geometric construction. The sme hppens with integrls: There is concept of definite integrl nd concept of indefinite integrl. The definite integrl is numer with geometric mening. The indefinite integrl is function whose vlues re definite integrls. We will strt with the second concept (functions) nd then study the geometric mening of their vlues. Integrl clculus I 3/19 Severl primitives for ech function If we hve function f (x), there is one function f (x) which is its derivtive. But function cn hve mny ntiderivtives! Exmple: one primitive of the function f (x) = 1 is F(x) = x. Another one is G(x) = x + 2. Cn you think of others? Theorem: if F(x) nd G(x) re two primitives of the sme function f (x), then F(x) G(x) is constnt. (Proof: the derivtive of F(x) G(x) is zero, nd the only functions with zero derivtives re the constnts.) Another importnt difference etween differentition nd integrtion is tht the first is esy (some sic derivtives nd sic rules), ut the second one cn e very complicted! We will lern some techniques tht will help us to clculte mny integrls. Integrl clculus I 4/19
First exmples of integrls The integrl of f is written s f (x) dx or f (t) dt or f (ζ) dζ or... nd it represents ll the primitives of f. A function cn e written in ny vrile we like, so the sme is true for n integrl. Rememer tht ech function hs mny primitives! The symol ove denotes ll of them (it is esy to write ll the primitives if we know one of them). Sometimes, however, we will e interested in only one, nd we will still use the sme symol. Exmple: 2x dx = x 2, ecuse (x 2 ) = 2x. Also 2x dx = x 2 + 3. The est wy to write them ll is 2x dx = x 2 + c ecuse for every vlue of c we get primitive, nd every primitive ppers in this wy. Integrl clculus I 5/19 The first sic integrls x n dx = x n+1 n + 1 + c 1 dx x = ln x + c e x dx = e x + c if n 1 x dx = x ln + c if 1 sin x dx = cos x + c cos x dx = sin x + c Integrl clculus I 6/19
Two sic properties Two importnt properties come from the properties of derivtives: f (x) + g(x) dx = f (x) dx + g(x) dx nd c f (x) dx = c f (x) dx Exmple: we cn do now the integrl of ny polynomil. 2x 3 3x 2 + 5x 7 dx = x 4 2 x 3 + 5 2 x 2 7x + c Note tht we hve only one constnt of integrtion t the end, there is no need to put one for ech term (in the end, it will just e one constnt to e dded). Integrl clculus I 7/19 Generlizing the sic formuls We hve seen tht x n dx = x n+1 n + 1 (x + ) n dx = F(x) =? + c. How out this one: Let us guess! From ove, we could try F(x) = (x + )n+1 n + 1 : F (x) = (n + 1)(x + )n n + 1 (x + ) = (x + ) n It is not the sme, ut we hve seen tht we cn multiply y constnts. If we choose now F(x) = 1 (x + ) n+1, we relly n + 1 get tht F (x) = (x + ) n, so this F(x) must e one primitive (ll the others re F(x) + c). Integrl clculus I 8/19
More sic integrls (x + ) n (x + )n+1 dx = + c if n 1 (n + 1) 1 x + dx = 1 ln(x + ) + c e x dx = 1 ex + c sin x dx = 1 cos x + c cos x dx = 1 sin x + c 1 2 x 2 dx = rcsin x + c 1 x 2 + 2 dx = 1 rctn x + c Integrl clculus I 9/19 Difficulty in clculting integrls We hve seen tht sums nd multiplictions y constnts re ok in integrls. This is ecuse we hve nice differencition rules for sums nd product y constnts. But some other differentition rules do not give good integrtion rules. In fct there re no rules for: f (x) g(x) dx f (x)/g(x) dx (f g)(x) dx We will lern methods tht solve some integrls (ut not ll!). Integrl clculus I 10/19
The geometric mening of integrtion As we sid, the vlues of integrl functions re solutions of geometric prolem: If you hve function f (x), how cn you clculte the re etween the grph of f (x), the x-xis, nd two verticl lines x = nd x =? A Let us try to pproximte this re. We use rectngles, nd we mke the pproximtion etter nd etter (we go to the limit). Integrl clculus I 11/19 Definite integrls We divide the intervl [, ] in equl pieces of length x, = f 0 < f 1 < < f n =. Then the re is pproximtely A n f (x i ) x i=1 = x 0 x 1 x 2 = x 3 ecuse ech rectngle hs sides f (x i ) nd x. If we go to the limit, the tringles ecome infinitely thin, of width dx, nd we write the sum with different symol: A = f (x) dx We cll this definite integrl. The numers, re clled the limits of integrtion. Integrl clculus I 12/19
Exmple of definite integrl The re elow f (x) = x 2 etween 1 nd 4 is 4 1 x dx. 1 4 A primitive of f is F(x) = x 3/2 + 4 (insted of 4 it could e ny 3/2 numer c!). So the re is 4 1 ( ) ( ) x dx = F(x) 4 16 2 1 = 3 + 4 3 + 4 = 16 3 2 3 = 14 3 Integrl clculus I 13/19 From indefinite to definite integrls The min tool for clculting definite integrls is: Newton-Leiniz Axiom: if f is integrle on [, ] nd F is primitive of f, then f (x) dx = F() F() not. = F (x) It is cler tht, if we cn clculte primitive of f, we cn esily clculte the re elow f. Well it is not exctly s descried here ut we will see this in more detil. And when does function hve primitive? If f is continuous, it hs primitive. And f hs finitely jumps nd is continuous etween them, it lso hs primitive. Integrl clculus I 14/19
From definite to indefinite integrls On the other hnd, if we hve n integrle function f defined on n intervl [, ], we cn clculte the re not only etween nd, ut etween nd ny x in the intervl. With this we construct function: F(x) = x f (t) dt (Note the chnge of vrile inside the integrl.) It turns out tht this is primitive of the originl function! Fundmentl theorem of clculus: F (x) = f (x). Integrl clculus I 15/19 Properties of definite integrls From the rule properties: f (x) dx = 0 f (x) dx = f (x) dx + f (x) dx = F () F() we deduce three c f (x) dx f (x) dx = c f (x) dx Integrls over the sme intervl cn e compred: if f (x) g(x) in the intervl [, ] then f (x) dx g(x) dx Note tht sometimes definite integrls hve negtive vlues. How cn this e the re of region?? We see this next. Integrl clculus I 16/19
Appliction: clcultion of res Consider the integrl 2π 0 ( cos x) 2π 0 sin x dx. Its vlue is = cos 2π + cos 0 = 0 How cn the re elow the sine function e equl to 0? + The re tht integrls mesure is signed. If we wnt the usul re (unsigned), we hve to consider where the function is positive nd negtive: A = π 0 sin x dx 2π π sin x dx Integrl clculus I 17/19 Appliction: clcultion of lengths Integrls re good not only for res. Here we see nother ppliction: rc lengths. Prolem: given n intervl [, ] nd function f defined on it, how cn we clculte the length of the rc etween nd? length? We cn pproximte the length of very smll piece with tringle: the length is the hypothenuse. s x f ( s) 2 = ( x) 2 + ( f ) 2 s = 1 + ( f x )2 x So the length is the sum of infinitesiml pieces like these: L = ds = 1 + (f (x)) 2 dx Integrl clculus I 18/19
Another prolem with integrls There is n even worse prolem thn the difficulty of clculting integrls. If we hve function f tht is written using some elementry functions nd opertions, f cn lso e expressed in tht wy. But it is not the cse for integrls: there re functions which cn e written using elementry functions nd opertions, ut whose integrls cnnot! For exmple, e x 2 dx nd sin x 2 dx do exist s functions (ecuse of the fundmentl theorem of clculus), ut they cnnot e expressed with the functions tht we know. Integrl clculus I 19/19