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Real Analysis

by: Otilia Murray I

Real Analysis MAT 125A

Otilia Murray I
GPA 3.88

Qinglan Xia

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Qinglan Xia
Class Notes
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This 11 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 125A at University of California - Davis taught by Qinglan Xia in Fall. Since its upload, it has received 111 views. For similar materials see /class/187370/mat-125a-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15
56 Substition and area between curves Last time we knew that for inde nite integral we have the substitution rule fltgltxgtgtg ltxgtdx fudu ifwe let u How about de nite integral b 91 fgxgxdx fudU a g a Why Let F be any antiderivative of f Then F is an antiderivative of fg So by the Fundamental theorem of Calculus b fgmgxdm Fltgltxgtgt 91 Fltgltbgtgt Fltgltagtgt FltugtIEZ fudU Remark There are two methods to evaluate fg xdx 0 apply the substitution rule for de nite integral 5 95 fltgltxgtgtg mac fltugtdu a 9a 0 or apply the substitution rule for inde nite integral fgxgxdx fudu some function ofx C Then substitute 05 a x b in the end Which way is better Depends on the particular problem 0 Advantage for the rst one don t need to replace u by o Advantage for the second one don t need to calculate endpoints for each substitution step Example Evaluate fo 13 dx 21 12l W 205 du 3 3 9 d 8 23u8221 0 052 1W3 m 1 1113 2 I 1 2 gt 2 Method 2We apply the substitution rule for inde nite integral by letting u x2 1 2f i du 73 23 73 2 23 x2 1 W i 2 Therefore 295 d 3 2 Jr 123z 3823 123 3 3 9 a a 0 x2 1W3 2 10 2 2 2 64 Example fag 31 Let u ln x then 4 6 dx i 4 du 62 xlnxln nx i 2 ulnu i ln4 ln2 t 111211333 1 1 4 1110112 1 m4 n n 11 ln2 ln21n2ln2 ln2 Example What is 2 sinyc5 303cm 2 Seems not that easy to evaluate But indeed the value is simply 0 Why Here the integrand 990 sinyc5 353 is an odd function on a symmetric interval 2 2 Let f be a function on a symmetric interval a a o fyc is an odd function if f 96 f96 o fyc is an even function if f ac Check g yc sin yc5 yc3 sin 305 303 g3c so 9 is odd De nite integrals of symmetric functions Theorem Let f be continuous on a symmetric interval a a foam 20afdm o If f is even then o ff is odd then fxdx 0 Example f32sinx5 mgldx 0 because sin 955 x3 is an odd function on a symmetric interval 2 2 Proof of the theorem Assume f is odd then mm fxdx0afxdx But 0 7a fltxgtdx fltxgtdx 7a 0 f udu by using substitution u x 0 fudu since f is an odd function 0 fudu 0 Therefore a a a fxdac fudu fxdx 0 7a 0 0 Example 5 3 1001 955 39 2 t 1 d st1n x7T4 an4 x lac 5 3 1001 955 5 sin 205 tan xldx0 ldx10 Ll 74 lt4 gt l 75 1001 This is because sin3 2x I tan x is an odd function on a symmetric interval 5 5 V7r4 Example qu hdx Note that f 4 is an even function on a symmetric interval So M M2Am m 2A3x 4dx 2 2i 495l3 9 2E 129 24 15 Note ffa xdx 0 as x is an odd function on a symmetric interval Area between curves 0 If f 2 0 is integrable on a b then the area between the graph of f and the XaXis is ab Suppose M 2 gm 2 o then the area between f and 990 is fab fltxgtdw bgltwgtdw o How about f 2 935 without the condition that 990 2 0 9 39 2quot 3 I 3 3 E H Note that we may always pick a large constant M so that fM29MZO Thus the area is b b me M gm Mldw W gwldw De nition If f and g are continuous function with f 2 930 on a b then the area of the region between the curves y f3c andy g30from a to b is b A W gltwgtdw What happens if we do not require f 2 930 The area between curves is b A inn gem Example Find the area of the region enclosed by the parabola y 2 902 and the line y yc Step 1 Find the intersection points of these two curves That is we solve y w y 2 302 simultaneously for ac yc2 302 302 30 20 30130 20 1 122 So the intersection points are P1 1 1 and 1322 2 On 1 2 the curve y 2 902 is above the line y 30 Therefore Step 2 Find the area Area 12 302 ycdyc 2 x2wd 1 303 302 L i2 8 1 1 4 22 3 321 6 32 1 41 2 239 Example Find the area of the region in the rst quadrant that is bounded above by y and below by the acaXis and y 902 14 Intersection points are 0 014 0 and 4 2 So the area is Method 1 fo dw 4 302 14dyc Method 2 4 4 dw 302 l4dyc 0 2 y14 y2dy 0 or method 3 integration with respect to y In general for regions like we may nd the area between two curves by integrating with respect to y instead of 90 d A my gltygtgtdy Example Find the area of the region in the rst quadrant that is bounded above by the XaXis and below by y 2352 and 30 y 3 Intersection points are 0 0 3 0 and 1 2 The area is Method 1 Integration with respect to 35 1 3 8 22d 3 acdyc 0 1 3 Method 2 Integration with respect to y


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