# Thread: FTC Part II simple question?

1. ## FTC Part II simple question?

Hi again all,

I'm doing some assignments and one of the questions asked to evaluate the integral:

$\int_{0}^{5}\((6e^x+4cos(x)) dx$

Using FTC P. II I got that the antiderivative was $F(x)=6e^x+4sin(x)$

so I got that the integral was: $F(5) - F(0) = 884.8$

I would like to know why this is the wrong answer? Thanks

2. Originally Posted by DannyMath
the questions asked to evaluate the integral:
$\int_{0}^{5}\((6e^x+4cos(x)) dx$
Using FTC P. II I got that the antiderivative was $F(x)=6e^x+4sin(x)$
so I got that the integral was: $F(5) - F(0) = 884.8$
I would like to know why this is the wrong answer? Thanks
An arithmetic error?
I get $F(5)-F(0)=880.643$

3. HAHA I just started Physics this semester and it uses angles in degrees so I forgot that I had switched my calculator from Rad to Deg. THANKS!

4. Originally Posted by DannyMath
HAHA I just started Physics this semester and it uses angles in degrees so I forgot that I had switched my calculator from Rad to Deg. THANKS!
Ah, yes. The downfall of many, many calculations in trigonometry.

5. Geeze, I can't imagine how it would feel to have this happen on your final Trig test :S. Glad I caught it now rather than later! Cheers!