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tut3 - √ x2 = |x| b5 x≥0 5 7 2" 0 9"& 7&...

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Unformatted text preview: √ x2 = |x| b5 x≥0 5 7! 2' " 0 9 " & 7! # &  # (! " 0! #( ) a (!  # "! % T 3 3 % 0 # $ "!  # # " & # % 0 4 3! %  # 0 " ` F (0) 1&  2' " 0 =x #)T E x2 F (x) = |x| − 1 x [−1, 1] x = −1 x < 0 Y # (! I # 0 " ( 0 9 √ "  f (x) = x>0 0=0 5 " $( 9 "& E 7! "$( E 7! " $ ( ( & 9 " P 9 E " 0! # C " )7 3 )" $!(  # ( # 0 " 9 " $( %  8 E "0 "$( " 0! # C " ) 7  # % 9! ( " 0 A 5 2 % & ( ( C " (! ( ) 0 ) "! # " 0 C (! # &  # " 0! # 4 3 ) ( ( &  # # &  # # 0 X x=1 f '' & % 07 F (x) = f (x) x ∈ [a, b] 5 9"& E [a, b] " 0 ' T & ! #" % W! 9 (! a F (z ) = z ∈ [a, b] f (t)dt, z [a, b] F ( ) 0 ) "! #" 0 C (! f 2 T 9 " P 9 E ' & % $ # "! #! " P 9 " !  # "  # E " 0 ! # C " ) 7 & 7 ! # &  # ( # & # ( Q V # % & 4H ( ) ' ) C ' & A 7 0 3 % 0   ' & # " 3 & 9 " )B   "0 UR • Q &H 1+ • Q TH 16 • Q CH 1 • Q 9H π 2 + sin 2 √ 2 − 23/4 SR 0 +∞ 0 π /2 sin3 x 3 dx = ln 3. 2 x 4 x sin x cos x π (π − 2) , 2 2 dx = tan + cot x 32 0 cos x ln(tan x)dx = − ln 2, π /2 1 1 √ − x 1 + x2 x √ dx = − ln( 2 + 1) + ln 2, F Q #! " P % & 9 " & # (! I 0 1 3  # 7 0 ' ' &H ( ' & % $ # " ! $ " ! 8 0 ' ' 0 7  # % 0 7 # &  # 80  G F 3 0 ( % & %  E $ " ' ' &  C & % 0 7 $ " ! D 0 0 ' % & 9 " & ( ) 0 ! % ) C % & 0  8 ) 0 2 7 0 ( 0  # % 0B 5 A @ " ! 3  # " ( 1&  9 ' ) 0 8 ) 0 2 7 0 # ( 0 6 5 ' 4 3! ( 2 % 1 % & ' & ! % 0 # ) # ( !  # " ! ( ' & % $ # "!    ¥     §  £    ¤  © ¨ § ¦ ¥ ¤ £ ¢¡ a π (f (x) − g (x))2dx. b [a, b] x 1% ) C f (x) − g (x) 2& ( 0 # (! # &  # E ( ! I &  # # ) 0 T & 9 # & # 0 % ( ! #! "  8 ' & 1% # "!  # " 0  # 2 T 9 # & % " $ 3 )' 0 1  # ( & 3 & (  # T "  # 9' ) 0 8 9 %! ) a % 3 )' 0 1  5 9 !' 4 4 & T " & C &' ) 3% 07  # # &  # 0 ( (! I&  # 0 # " 0 ! # ) ' 0 1 % 7 0 ( ! I &  # # & ' ( " & % # 0 # ( ! 0 9 0 # ¦ D C ! % #¦  # "  # E ' &1% #"!  # " 0 1 % ) C  # # ) 0 T & 9 # & # 0 % (! #! "  8 9"& ( 1 % ) C 0 8# " 8# T # " 0! $ %  # 2T 9 # & % " $ 3 )' 0 1  # 9 " P 0 # 9 D ( & % & 8 7! E ' 4 3 & I % 0 7 7V 5 (! I & %0 # ) 0 T & 9 1' 0 1 % % & 2  # "  8 9 # & % " $ ( 3 ) ' 0 1 % 0 7 (! 1&  8 # &  # & ' ) 3 % 0 7  # # &  # ' ' & C ¥ ¤ 5 9 I P (! " 0! # )' 0 1 % 7 0 (! I&  # E ( " 0! # ( ) a 7 0 9 "! D (!  # " V 5 ( 3 )' 0 1 7 0 " 0! # & "! 3 % # ¢ [a, b] f (x) x g (x) x g (x) y £R 0 # (! 5 (! I & C!  8 0 # # C 4 ( %  #! 8 # & % $ # " ! # 0 " 0 # $ " !  # %  # 0 " ` 5 ( " 0 ! # & ' ) C ' & C % ) 0 2 2 7! ' 4 3 ! ( 0 # E 2 % # 3 3 2 ( % 0 7 D 0 0' ( 2& 8' ` a f (x) − g (x)dx. b 2T " 1! $ (! ( 1 % ) C 0 8# (  # 2 T 9 9 " ) 0 T & % &  # E ' & 1% # "! " & " 0 2 & ( 0 # ( ! # &  #H 1% )C %  # 0 " & 7 0 4 0 # " 0 ( 2& 8' & (! C !  8 1 % ) C & % 0 7 Q # ! 1 0 % 4H # &  # % T 3 3 % 0 # ( ! ( " 0 ! # ( ) a 7 0 9 " ! D (!  # 0 9 0 # 2& 8 ' & % " $ ' 4 3 ! ( ` 5 ( 1 % ) C 0 8# 2 T 9 9 " ) 0 T ( & % & 7 0 " 0 ! # & "! 3 % # ¢ f (x) ≥ g (x) g (x) f (x) [a, b] ¡R = dy dx = 2x cos x Q CH • dy dx Q TH • Q &H • dy dx =1 √ cos x √ 2x ...
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