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t12and0
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Intergral at a point sint+t t=12&0.
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...
Splint
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May 18, 2010
Replies: 3
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Calculus
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