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

#### 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

#### General

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

#### Debsta

MHF Helper
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)

#### Splint

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

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