Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5. - 5.3) Remrks on the course. Slide Review: Sec. 5.-5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description of motion motivted the cretion of Clculus Problem of Motion: Slide 2 Given x(t) find v(t) Differentil Clculus. Given v(t) find x(t) Integrl Clculus. Derivtives nd integrls re opertions on functions. One is the inverse of the other. This is the content of the Fundmentl theorem of Clculus.
Mth 20B Integrl Clculus Lecture 2 An integrl is sum of infinite mny terms Definition (Integrl of function) Let f(x) be function defined on intervl x [, b]. The integrl of f(x) in [, b] is the number given by Slide 3 n f(x)dx = lim f(x n i ) x, i=0 if the limit exists. Given nturl number n we hve introduced prtition on [, b] given by x = (b )/n. We denoted x i = (x i + x i )/2, where x i = + i x, i = 0,,, n. This choice of the smple point x i is clled midpoint rule. An integrl is sum of infinite mny terms Continuous functions re integrble. The sum of infinite mny terms is finite. Slide 4 Theorem If f(x) is continuous in [, b], then f(x) dx = lim R n, exists. n Nottion: f(x) dx is clled the definite integrl of f(x) from to b. Notice: f(x) dx is number.
Mth 20B Integrl Clculus Lecture 3 Properties deduced from the definition Slide 5 f(x) dx = f(x) dx; b f(x) dx = 0; c dx = c(b ); (f ± g) dx = f dx ± g dx; More properties deduced from the definition Slide 6 c f(x) dx = c f(x) dx = f 0 f g c f(x) dx; f(x) dx + f dx 0; f dx m f M m(b ) c g x; f(x) dx; f dx M(b ).
Mth 20B Integrl Clculus Lecture 2 4 Integrtion cn be used to define new functions from old ones Slide 7 Theorem 2 If f(x) is continuous in [, b], then F (x) = f(s) ds, x [, b], is continuous functions nd F () = 0. Exmples: ln(x) = x s ds, x2 = 2s ds. 0 Fundmentl Theorem of Clculus Slide 8 Review: New function using integrtion. Fundmentl Theorem of Clculus. (Sec. 5.3) Integrtion tbles. (Sec. 5.4)
Mth 20B Integrl Clculus Lecture 2 5 Integrtion cn be used to define new functions from old ones Slide 9 Theorem 3 If f(x) is continuous in [, b], then F (x) = f(s) ds, x [, b], is continuous functions nd F () = 0. Exmples: ln(x) = x s ds, x2 = 2s ds. 0 Derivtion nd integrtion re opertions on functions, nd they re inverse of ech other Slide 0 Theorem 4 (Fundmentl Theorem of Clculus) If f(x) is continuous in [, b] nd c is ny constnt, then F (x) = is differentible nd f(s) ds + c, x [, b], F (x) = f(x).
Mth 20B Integrl Clculus Lecture 2 6 The FTC justifies the nme ntiderivtion for integrtion Slide Definition 2 The function F (x) given by F (x) = f(s) ds + c for ny constnt c is clled the ntiderivtive of f(x). The ntiderivtive of given function is not unique. Two ntiderivtives of the sme function cn differ by constnt. Nottion: f dx lso denotes n ntiderivtive of f. Here re two simple reformultions of the FTC Slide 2 Corollry If f(x) is continuous in [, b], then f(s) ds = F (b) F (), where F (x) is ny ntiderivtive of f(x). Corollry 2 If F (x) is differentible in [, b], then F (s) ds = F (b) F ().
Mth 20B Integrl Clculus Lecture 2 7 The insight from the FTC cn be used to construct integrtion tbles Slide 3 dx = x + c, x n dx = n + xn+ + c, n, dx = ln(x) + c, x e x dx = e x + c. The insight from the FTC cn be used to construct integrtion tbles Slide 4 sin(x) dx = cos(x) + c, cos(x) dx = sin(x) + c, dx + x2 = rctn(x) + c, dx x 2 = rcsin(s) + c.
Mth 20B Integrl Clculus Lecture 3 8 FTC nd the chin rule F (x) = f(t) dt + c, F () = c, F (x) = f(x). Slide 5 IF (x) = F (g(x)), IF (x) = F (g(x)) g (x). IF (x) = Exmple: f(x) = g(x) sin(x) f(t)dt + c, IF (x) = f(g(x)) g (x). ln(t) dt, f (x) = ln(sin(x)) cos(x). Substitution rule Slide 6 Review: Derivtion Integrtion re inverse opertions. Chin rule Substitution rule. Exmples.
Mth 20B Integrl Clculus Lecture 3 9 Derivtion nd integrtion re inverse opertions Slide 7 F (x) = f(s) ds + c F () = c, F (x) = f(x). In other words, F (x) = f(x) dx F (x) = f(x). In still other words, F (x) dx = F (x). Derivtion tbles Integrtion tbles (x n ) = nx n nx n dx = x n + c, Slide 8 [sin(x)] = cos(x) cos(x) dx = sin(x) + c. Derivtion rules Integrtion rules Chin rule Substitution rule. Product rule Integrtion by prts.
Mth 20B Integrl Clculus Lecture 3 0 Chin rule provides wy to compute some integrls Slide 9 Find the primitive (ntiderivtive) of 2x cos(x 2 ). cos(x 2 )2x dx = [sin(x 2 )] dx, = sin(x 2 ) + c. Chin rule provides wy to compute some integrls Slide 20 Find the primitive (ntiderivtive) of cos(x)/ sin(x). sin(x) cos(x) dx = [ln(sin(x))] dx, = ln(sin(x)) + c.
Mth 20B Integrl Clculus Lecture 3 Here is the generl cse of this inverse form of chin rule Slide 2 Find the primitive of f(g(x)) g (x) knowing tht the primitive of f(x) is F (x), tht is, F (x) = f(x). f(g(x)) g (x) dx = F (g(x)) g (x) dx, = [F (g(x))] dx, = F (g(x)) + c. The substitution rule is technique to use the inverse form of the chin rule efficiently Slide 22 Recll, F (x) = f(x), then f(g(x)) g (x) dx = F (g(x)) g (x) dx. Introduce u = g(x), then denote du = g (x) dx. Substitute these expressions in the right hnd side bove: f(g(x)) g (x) dx = F (u) du, = F (u) + c, = F (g(x)) + c.
Mth 20B Integrl Clculus Lecture 3 2 The substitution rule on definite integrls chnges the limits of integrtion Slide 23 Theorem 5 (Chnge of vrible) Let g(x) be differentible in [, b] with g (x) continuous in [, b]. Let f(u) be continuous for u = g(x) nd x [, b]. Then, f(g(x)) g (x) dx = g(b) g() f(u) du. Proof of the chnge of vrible theorem Slide 24 Let F (x) be primitive of f(x), so F (x) = f(x), nd F (d) F (c) = d c f(u) du. f(g(x)) g (x) dx = = F (g(x)) g (x) dx, [F (g(x))] dx, = F (g(b)) F (g()), = g(b) g() f(u) du.