Intergral at a point sint+t t=12&0.

Sep 2009
37
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
 
Jan 2010
564
242
Kuwait
Its \(\displaystyle sin(t)+t\) or \(\displaystyle sin(x)+x\) .. ?!!
Big difference.
 

Debsta

MHF Helper
Oct 2009
1,284
585
Brisbane
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
You are calculating cos (12 degrees). You should be calculating cos (12 radians)