# Where have I gone wrong: Definite integrals involving trig functions

• September 4th 2011, 08:10 PM
perjac
[pls ignore] Where have I gone wrong: Definite integrals involving trig functions
EDIT: Let this be a warning to anyone relying on a calculator. I should have trusted myself. I entered the range in the opposite way to the calculator giving a negative solution.

Hi All,

Have a homework question here:
$\int^{3\pi}_{2\pi} Sin(x/4)\,dx$

I can get out:
$(-4cos(3\pi/4)) - (-4cos(2\pi/4))$

but this equals for me positive 2.8... When looking at the graph (and the solve on the calc) shouldn't it be negative? Where have I gone astray? (I think the second part of that is 0 so I assume I am doing somethign wrong in the first part).

Cheers,

J
• September 4th 2011, 08:17 PM
Educated
Re: Where have I gone wrong: Definite integrals involving trig functions
You haven't gone wrong anywhere, your answer of 2.8... is correct.

What graph are you looking at the would suggest an area under the x-axis?
Wolfram|Alpha Graph of sin(x/4)

Did you put brackets in the right place when typing it into your calculator? Did you have your calculator in radians mode?