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

1. ## 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 wrong?

Thanks

2. Its $sin(t)+t$ or $sin(x)+x$ .. ?!!
Big difference.

3. Originally Posted by Splint
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)

4. Originally Posted by Debsta
You are calculating cos (12 degrees). You should be calculating cos (12 radians)
Thanks Debsta, I can't believe I overlooked that.