S Splint Sep 2009 37 1 May 18, 2010 #1 Hi, Trying to work this one out, \int_0^{12} (\sin{t} + {t}), dx The answer shows = (72 - cos12) - (0-1) =(73-.84) =72.16 I calculate (72-.98) - (0-1) which equals 70.02. Anyone willing to have a quick look and try and see where I went wrong? Thanks

Hi, Trying to work this one out, \int_0^{12} (\sin{t} + {t}), dx The answer shows = (72 - cos12) - (0-1) =(73-.84) =72.16 I calculate (72-.98) - (0-1) which equals 70.02. Anyone willing to have a quick look and try and see where I went wrong? Thanks

General Jan 2010 564 242 Kuwait May 18, 2010 #2 Its \(\displaystyle sin(t)+t\) or \(\displaystyle sin(x)+x\) .. ?!! Big difference.

D Debsta MHF Helper Oct 2009 1,284 585 Brisbane May 18, 2010 #3 Splint said: Hi, Trying to work this one out, \int_0^{12} (\sin{t} + {t}), dx The answer shows = (72 - cos12) - (0-1) =(73-.84) =72.16 I calculate (72-.98) - (0-1) which equals 70.02. Anyone willing to have a quick look and try and see where I went wrong? Thanks Click to expand... You are calculating cos (12 degrees). You should be calculating cos (12 radians) Reactions: Splint and HallsofIvy

Splint said: Hi, Trying to work this one out, \int_0^{12} (\sin{t} + {t}), dx The answer shows = (72 - cos12) - (0-1) =(73-.84) =72.16 I calculate (72-.98) - (0-1) which equals 70.02. Anyone willing to have a quick look and try and see where I went wrong? Thanks Click to expand... You are calculating cos (12 degrees). You should be calculating cos (12 radians)

S Splint Sep 2009 37 1 May 18, 2010 #4 Debsta said: You are calculating cos (12 degrees). You should be calculating cos (12 radians) Click to expand... Thanks Debsta, I can't believe I overlooked that.

Debsta said: You are calculating cos (12 degrees). You should be calculating cos (12 radians) Click to expand... Thanks Debsta, I can't believe I overlooked that.