{VERSION 5 0 "IBM INTEL NT" "5.0" }
{USTYLETAB {CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0
0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }
{CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE
"" -1 256 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0
1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0
0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times
" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }}
{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 23 "MATH 156 SAMPLE MIDTERM"
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "In this exam you will work with
the integral " }{XPPEDIT 19 1 "int(6/(1+x^2),x = 0 .. \{sqrt(3)\}/3);
" "6#-%$intG6$*&\"\"'\"\"\",&F(F(*$%\"xG\"\"#F(!\"\"/F+;\"\"!*&<#-%%sq
rtG6#\"\"$F(F6F-" }{TEXT -1 51 ". We have seen in class that the antid
erivative of " }{XPPEDIT 18 0 "6/(1+x^2);" "6#*&\"\"'\"\"\",&F%F%*$%\"
xG\"\"#F%!\"\"" }{TEXT -1 49 " is 6 arctan (x), so the value of the in
tegral is" }}{PARA 0 "" 0 "" {TEXT -1 9 "6(arctan(" }{XPPEDIT 18 0 "sq
rt(3)/3;" "6#*&-%%sqrtG6#\"\"$\"\"\"F'!\"\"" }{TEXT -1 14 ")-arctan(0)
)=6" }{XPPEDIT 18 0 "Pi/6;" "6#*&%#PiG\"\"\"\"\"'!\"\"" }{TEXT -1 1 "=
" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 56 ". So this exam has as p
urpose to find approximations of " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }
{TEXT -1 1 "." }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 262 "Answers all t
he questions. When an explanation is required, your answer should not \+
just be: \"I saw it on the graph\" or \"The numbers show that A