Big idea in Calculus: approximation

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1 Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion: f(x) f(x 0 )+f (x 0 )(x x 0 ) The grph of nonliner function the grph of tngent line Differentition: Clculte the rte of smll chnge Integrtion: Accumulte ll smll chnges

2 How to clculte the re? Rectngle: A = b ( = length, b = width) Tringle: A = 1 2 b Polygon: divide it to tringles, clculte ech then tke the sum

3 Wht if it is circle? Clcultion of π: Archimedes ( BC) /71 < π < 22/ Liu Hui (bout ) π , Zu Chongzhi ( ) π

4 Big ide The re of non-polygon cn be clculted from tht of polygons! (nd tke limit) Are under the curve when f(x) 0:

5 Geometric ide Use rectngle with certin height to pproximte irregulr but lmost rectngulr shpe Procedure: 1. divide the intervl [, b] into n equl subintervls with length of subintervl x = (b )/n 2. With x 0 =, x 1 = x 0 + x,, x i+1 = x i + x,, x n = b, nd choose y i so tht x i 1 y i x i. 3. Approximte the re bove [x i,x i+1 ] by f(y i ) x 4. Tke the sum (clled Riemnn sum): n f(y i ) x = f(y 1 ) x +f(y 2 ) x + +f(y n) x i=1 How to choose y i? Left-endpoint-sum (L n), right-endpoint-sum (R n), midpoint rule (M n).

6 Riemnn sum Exmple: Find re below the curve y = x 2 with 0 x 1 by using Riemnn sum pproximtion L 4, R 4 nd M 4. L 4 = [f(0)+f(1/4)+f(1/2)+f(3/4)] 0.25 = 7/32 = R 4 = [f(1/4) +f(1/2)+f(3/4)+f(1)] 0.25 = 15/31 = M 4 = [f(1/8) +f(3/8)+f(5/8)+f(7/8)] 0.25 = 21/64 = Wht if n is lrge?

7 Are s limit: Definite integrl The re under y = f(x) is the limit of Riemnn sum: n A = lim Rn = lim f(y i ) x n n i=1 is the nottion for sum (clled sigm) b f(x)dx = lim n n f(y i ) x (limit of Riemnn sum) i=1 where x = (b )/n, x 0 =, x 1 = x 0 + x,, x k+1 = x k + x,, x n = b, nd x i 1 y i x i. When n, x 0. If the limit exists, then the function f is integrble on [,b], b nd f(x)dx is the definite integrl of f from to b. (Theorem: If f is continuous, or f hs only finite number of jump discontinuities, then f is integrble.) : lower limit; b: upper limit; f(x): integrnd; : integrl sign (S=sum) definite integrl is number

8 Mening of the definite integrl When f(x) is positive, the definite integrl is the re below f(x) (nd bove y = 0) When f(x) cn be both positive nd negtive, the definite integrl is the signed re 2 Exmple: Find (2x 3)dx. 0

9 Properties of definite integrl: b ( ) b b f(x)+g(x) dx = f(x)dx + g(x)dx, b ( ) b b f(x) g(x) dx = f(x)dx g(x)dx, b b αf(x)dx = α f(x)dx, b c b f(x)dx = f(x)dx + f(x)dx, b c c f(x)dx = f(x)dx, nd f(x)dx = 0, b c b b If f(x) g(x) for x [,b], then f(x)dx g(x)dx, If m f(x) M for x [,b], then b m(b ) f(x)dx M(b ).

10 How to clculte the definite integrl? Evlution Theorem: b If f(x) is continuous on [,b], then f(x)dx = F(b) F(), where F(x) is n ntiderivtive of f(x). Recll: F(x) is n ntiderivtive of f(x) if F (x) = f(x). Exmple: 3 4 () (1+2x 4x 3 )dx; (b) xdx. 1 1 F(x) = f(x)dx is lso clled indefinite integrl of f(x). b f(x)dx (definite integrl is number), nd f(x)dx (indefinite integrl is (fmily of) function)

11 Bsic Formul Elementry integrl formuls: (see pge 277) x n dx = 1 n +1 xn+1 +C sinxdx = cosx +C 1 dx = ln x +C cosxdx = sinx +C x x dx = 1 ln x +C sec 2 xdx = tnx +C 1 dx = 1 1 x 2 sin 1 x +C 1+x 2dx = tn 1 x +C Bsic Properties of integrls: (see pge ) b b b b b [f(x)±g(x)]dx = f(x)dx ± g(x)dx c f(x)dx = c f(x)dx b c b b f(x)dx = f(x)dx + f(x)dx f(x)dx = f(x)dx c b

12 Mening nd vritions of Evlution Theorem b f(x)dx = F(b) F(), where F(x) is n nti-derivtive of f(x) b F (x)dx = F(b) F(), where F(x) is differentible function (mening: the integrl of rte of chnge is the totl chnge) t2 (t)dt = v(t 2 ) v(t 1 ) t 1 (mening: the integrl of ccelertion is the totl chnge in velocity) t2 v(t)dt = D(t 2 ) D(t 1 ) t 1 (mening: the integrl of velocity is the totl chnge in position) But the totl distnce trveled is t2 t 1 v(t) dt ( milege nd chnge of position re different)

13 Fundmentl Theorem of Clculus integrtion nd differentition re inverse process b () F (x)dx = F(b) F(), here F(x) is differentible Function F(x) (differentite) Function F (x) (integrte) b F (x)dx = F(b) F() x (b) If F(x) = f(t)dt, then F (x) = f(x). x Function f(x) (integrte) Function F(x) = f(t)dt (differentite) [ d x ] f(t)dt = f(x) dx (c) If F(x) = f(x)dx, then F (x) = f(x).

14 Smple problems Problems from Mth 111 Finl (Fll 2002) 4 1. Consider the grph of the function f(x), nd find f(x)dx x A ping-pong bll is floting down river. You record its velocity every 5 minutes for 20 minutes s shown in the tble below. Estimte the totl distnce the bll trvelled during the first 20 minutes by using 4 subintervls with left-endpoint vlues. t (min) v (m/min)

15 More problems Distnce, velocity nd ccelertion revisited: Suppose tht (t) = 2t 3, v(0) = 1. (1) Find the velocity function, nd find the totl distnce trveled during 0 t 10. (2) Drw the grph of (t) nd v(t), nd show the re on the grph representing the chnge of velocity, position, nd lso distnce trveled. t v(t) v(0) = (s)ds 0 t D(t) D(0) = v(s)ds 0 Fundmentl Theorem of Clculus: x 1. Find the derivtive of g(x) = lntdt Find the derivtive of F(x) = tn tdt x x 2 3. Find the derivtive of h(x) = 1+t 3 dt 0 4. Find the verge vlue of the function f(x) = 1/x in [1,4], nd find c [1,4] such tht f(c) = verge vlue.

16 Averge Vlue Averge vlue of function on [, b]: (think bout verge velocity) f ve = 1 b f(x)dx b Averge vlue of n numbers: n n Men-vlue Theorem for the integrl: If f(x) is continuous on [,b], then there exists c in [,b] such tht f(c) = f ve = 1 b f(x)dx, or f(c)(b ) = b b f(x)dx.

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