Distance covered by integrating the Velocity of a Body For a National Board Exam Review

The velocity of a body is given by v(t) = sin(pi*t) where the velocity
  is given in meters per second and t is given in seconds. The distance
  covered in meters between  t = 0.25 and t = 0.5.

Answer is 0.2251m
This is supposed to be simple and straightfoward; so I put in my calculator:
$${ \int_{0.25}^{0.5} sin(\pi t) dt = 5.14 \times 10^{-3}}$$ 
I am getting: 
What am I doing wrong?
 A: It looks like your integration went a little wrong. It's important to write out the steps since syntax mistakes can be frustrating with calculators. 
$\int_{0.25}^{0.5}\sin(\pi t) dt = -\frac{1}{\pi}[cos(\pi t)]_{0.25}^{0.5} = -\frac{1}{\pi}(\cos(0.5\pi) - \cos(0.25\pi)) = 0.225079... \approx 0.2251$
A: I think there is some error in the integration. 
Notice, the velocity $v(t)=\sin (\pi t)$ is the function of time $t$ which is varying with time 
Now, consider  a very small time interval $dt$ then the elementary distance $ds$ covered in elementary time $dt$ is given as $$ds=v(t)dt$$ Now, the total distance covered by body between $t=0.25$ & $t=0.5$ $$=\int_{0.25}^{0.5}v(t)dt$$ $$=\int_{0.25}^{0.5}sin (\pi t)dt$$ $$=\frac{-1}{\pi}\left[\cos (\pi t)\right]_{0.25}^{0.5}$$ $$=\frac{-1}{\pi}\left[\cos \frac{\pi}{2}-\cos \frac{\pi}{4}\right]$$ $$=\frac{-1}{\pi}\left[0-\frac{1}{\sqrt 2}\right]$$ $$=\frac{1}{\pi\sqrt 2}$$ Hence, we have $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{distance covered}=\frac{1}{\pi \sqrt{2}}\approx 0.225079079\ m}}$$
