The area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O


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1 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the definite integrl of f etween nd. Exmple 1 The grph of y = sin x is given elow: y π A 3π 5π 4 π O B π π π x Descrie ech shded region using the integrl nottion. A = B = Exmple Find 4 x dx.
2 Exmple 3 Find 3 1 (x + 3) dx. Exmple 4 Find 4 t dt. Exmple 5 Find x dx.
3 3 So fr we hve defined the definite integrl of function only when f is nonnegtive. If f is positive for some x vlues nd negtive for others, then f(x) dx is the sum of res ove the xxis, counted positively, nd res elow the xxis, counted negtively. x y = f(x) For exmple, in the ove picture, suppose the shded re ove the xxis is nd the shded re elow the xxis is 3. Then Exmple 6 Find (x + 1) dx. f(x) dx = + ( 3) = 1. Exmple 7 It is known tht π π sin x dx = 1. Using this, compute. 3π sin x dx. 1 rcsin x dx
4 4 Section 5.3 The Fundmentl Theorem nd Interprettions In previous section, we studied how to compute integrls when the given functions re simple ones, such s lines or semicircles. However, we still do not know how to compute generl integrls. In this section, we will study the Fundmentl Theorem of Clculus, one of the most importnt theorems in Clculus, which mkes it possile to compute the integrl of huge collection of functions. Exmple 1 Consider (x + 1) dx. We need definition to stte the Fundmentl Theorem. Definition Let f e function. If F (x) = f(x) for ll x, then we sy tht F is n ntiderivtive of f. Exmple F (x) = 1 3 x3 is n ntiderivtive of f(x) = x. Note tht G(x) = 1 3 x3 + is lso n ntiderivtive of f. In generl, if F (x) is n ntiderivtive of f(x), then so is F (x) + C for ny constnt C. Moreover, if F nd G re oth ntiderivtives of f, then F nd G differ y only constnt. The Fundmentl Theorem of Clculus Let f e continuous function on the intervl [, ] nd f(x) = F (x) (so F is n ntiderivtive of f). Then Proof f(x) dx = [F (x)] = F () F ().
5 5 Exmple 3 Compute 3 nswer to the result from the previous section. 1 (x + 3) dx using the Fundmentl Theorem of Clculus. Compre your Exmple 4 Compute (x + 1) dx exctly. Exmple 5 Find π π sin x dx. Exmple 6 Compute π cos x dx. Is the result wht you expected? Exmple 7 Compute 1 dx x. Here dx 1 is nother nottion for 1 + x x dx.
6 6 Section 5.4 Theorems out Definite Integrls Definition Let f e continuous function on n intervl [, ]. The verge of f from to is defined to e 1 f(x) dx. Remrk Why verge? Look t the following picture: y h y = f(x) x Let s determine h so tht the re of the rectngle equls tht of the shded region. Note tht h must stisfy Therefore, h( ) = h = 1 f(x) dx. f(x) dx. In words, the verge of f from to is the height of rectngle whose se is nd whose re is the sme s the re under the grph of f from to. Exmple 1 Find the verge of f(x) = x on the intervl [1, 3].
7 7 Definition Let <. Let f e continuous function on [, ]... f(x) dx =. f(x) dx = f(x) dx. Theorem 1 Let f e continuous function. Then no mtter wht,, nd c re, c Proof We consider only two cses: Cse I: < < c f(x) dx = f(x) dx + c f(x) dx. Cse II: < c < Theorem Let f nd g e continuous functions nd let c e constnt.. (f(x) ± g(x)) dx = f(x) dx ± g(x) dx.. Proof cf(x) dx = c f(x) dx.
8 8 Exmple Compute π sin x dx + π cos x dx. Theorem 3 Let f e continuous function.. If f is odd, then. If f is even, then Proof f(x) dx =. f(x) dx = f(x) dx. Exmple 3 Compute ( ) 1 x 11 1x x 5 + 3x + 4 dx. Theorem 4 Let f nd g e continuous functions.. If f(x) g(x) for x, then f(x) dx. If m f(x) M for x, then m( ) Proof g(x) dx. f(x) dx M( ).
9 9 We close this section with the re etween two curves. f(x) f(x) f(x) g(x) g(x) g(x) x x x When oth f nd g re positive over the intervl [, ], then it is cler tht the re etween f nd g is given y (f(x) g(x))dx. Remrk Even when f nd g re not necessrily ove the xxis, still the re etween f nd g is given y (f(x) g(x))dx. Why? Exmple 4 Find the re etween the curves y = f(x) nd y = g(x), where f(x) = x 4x + 5 nd g(x) = x + 4x 1. Exmple 5 Find the shded re elow: cos x O sin x
10 1 Exmple 6 Mesuring Income Inequlity (dpted with permission from Michel Olinick, Middleury College) The Lorenz Curve L(x) for country gives the proportion of the totl income erned y the lowest proportion x of the popultion. Think of the individuls rrnged from poorest (lowest) to richest (highest). Exmple: L(.) =.15 sys tht the poorest. of the popultion erns.15 of the totl income. Restting in percents, this sys tht the poorest % of the popultion erns 15% of the totl income.. Wht does L(.9) =.7 sy in percents? For this function, wht percentge of the totl income does the richest 1% ern?. If you know L(.) =.13 nd L(.4) =.3, wht percent of the totl income does the second % of the popultion ern? c. Explin why the Lorenz Curve must hve the following properties: L() =, L(1) = 1, L is n incresing function on [, 1]. ( ) d. Suppose tht L(x) = x for ll x, x 1 for country. Wht does this tell you out the country s income inequlity? e. Suppose tht L(x) = for x < 1 nd L(1) = 1 for country. Wht does this tell you out the country s income inequlity? f. The Gini Index G for country with Lorentz Curve L(x) is defined to e G = 1 (x L(x)) dx. i. Check tht L(x) = x x3 meets the requirements (( ) ove) for Lorentz Curve nd clculte the corresponding Gini Index. ii. Determine the numer p so tht country with Lorenz Curve L(x) = x p hs Gini Index of.5. g. Wht is the smllest vlue G cn hve? the lrgest? h. Explin why Gini Index for country mesures the income inequlity of the country.
11 11 We close this chpter with the forml definition of the definite integrl using the concept of Riemnn Sum. Recll tht f(x) dx represents (in generlized sense) the re of the region enclosed y the xxis, lines x = nd x =, nd the grph y = f(x). Mensurtion y Prts Let s consider circle with rdius r. We ll know tht the re of the circle is equl to πr. But why? π n r
12 1 Definition A sequence is n infinite list of numers s 1, s, s 3,..., s n,... We cll s 1 the first term, s the second term, nd so on. s n is the generl term, sy n th term. Exmple 7 1, 4, 9, 16, is sequence. To e more precise, s 1 = 1, s = 4, s 3 = 9, nd so on. In this exmple, the generl n th term is given y s n =. Exmple 8 Give the first six terms of the following sequences:. s n = n(n+1). t n = n+( 1)n n Exmple 9 Consider the sequence 1, 1,, 3, 5, 8, 13,... Cn you red off the pttern? Wht is the next term? Exmple 1 Find the generl n th term s n when the first few terms of the sequence re. 4, 7, 1, 13, 16, 19,...., 4, 6, 8, 1, 1,... c. 1, 3, 5, 7, 9, 11,... d. 1,, 4, 8, 16, 3,... e. 1, 1, 1, 1, 1, 1,... f.,,,,,,... g. 1, 3, 3 4, 4 5, 5 6, 6 7,... Sequences cn lso e defined recursively, y giving n eqution relting the n th term to the previous terms (clled recursion formul) nd s mny of the first terms s re needed to get strted. Exmple 11 Give the first few terms of the following recursively defined sequences.. s n = s n for n > 1 nd s 1 = 4.. s n = s n 1 for n > 1 nd s 1 = 1. c. s n = s n 1 + s n for n > nd s 1 = s = 1. d. s n = s n 1 n for n > 1 nd s 1 = 1.
13 13 Exmple 1 Consider recursive formul given y { sn 1 s n = 3s n if s n 1 is even if s n 1 is odd. With s 1 = 4, write the first few terms of the sequence.. With s 1 = 5, write the first few terms of the sequence. c. With s 1 = 7, write the first few terms of the sequence. d. Try other initil terms. Does your nswer lwys end up with 4,, 1, 4,, 1,...? We now consider the limit of sequence. Definition If s n pproches fixed numer L s n, then we sy tht s n converges to L nd L is the limit of s n. When there is no such L, we sy tht the sequence diverges. Exmple 13 Do the following sequences converge or diverge? If sequence converges, find its limit.. s n = 1 n d. s n = ( 1)n n. s n = 5n 1 n 3n+5 e. s n = sin n n c. s n = n f. s n = 1 + ( 1) n
14 14 Definition A finite series is the sum of finitely mny terms of sequence. For given sequence n, the sum of the first k terms, clled the k th prtil sum nd denoted y S k, is defined y S k = k i = k 1 + k. i=1 Exmple 14 Let n = n + 1. Then 5 i = = = 35. i=1 Remrk. i is dummy vrile in the expression ove, tht is, we could hve used express the sme finite series.. The summtion nottion cn e used in different wy. For exmple, k j or j=1 k n to n=1 5 i = = 5 i=3 nd p:prime p 1 p = = 17. Exmple 15 Simplify n. k = n. k=1. n k = n. k=1 c. n k 3 = n 3. k=1
15 15 Definition Lower nd upper Riemnn sums of function y = f(x) on the intervl [, ]. Exmple 16 Compute 1 x dx using Riemnn sums.
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