Integrating problem with $\sin x$

Question

I'm trying to integrate the following definite integral:

$$\begin{align}\int_{0}^{\pi}{\sin x}\,dx\end{align}$$

My try:

$$\begin{align}\int_{0}^{\pi}{\sin x}\,dx=\\\left(-\cos \pi\right)-\left(-\cos 0\right)=\\-1-(-1)=\\0\end{align}$$

But the actual result (by wolframalpha) is: 2

I'm sure that there is something I miss.

Sorry the problem was with my calculator (I forget to convert my system from Deg to Rad)

Negative cosine of pi is what? – colormegone Jan 11 '14 at 07:02 — colormegone, Jan 11 '14 at 07:02
@MohammadFakhrey That is $-\cos \pi^\circ$ – peterwhy Jan 11 '14 at 07:08 — peterwhy, Jan 11 '14 at 07:08
Please delete this post. Thank you. – Mohammad Fakhrey Jan 11 '14 at 07:09 — Mohammad Fakhrey, Jan 11 '14 at 07:09

score 4 · Accepted Answer · answered Jan 11 '14 at 07:01

4

You incorrectly calculated $-\cos\pi$:

$$-\cos\pi = -(-1) = 1.$$

answered Jan 11 '14 at 07:01

Michael Albanese

1 Answers1