# Particle's displacement after 4 seconds of motion

The velocity, v(t) $ms^{-1}$, of an object travelling along a straight line, at time $t$ seconds is given by:

$$v(t)=10e^{-\frac{1}{2}t}sin(\frac{\pi}{2}t)$$

What is the particle's displacement after 4 seconds of motion?

My solution:

I integrate $v(t)$ to get: $$s(t)=\int 10e^{-\frac{1}{2}t}sin(\frac{\pi}{2}t) \ dt = \frac{-20e^{-\frac{1}{2}t} (sin(\frac{\pi}{2}t)+\pi cos(\frac{\pi}{2}t))}{1+\pi^2}$$

Then, since the question is about finding displacement, not a distance travelled in some time, what I think I need to do is to plug 4 in the equation and find $s(4)$.

However, this yields: $$s(4) \approx -0.782$$

The answer in the textbook is $5 m$

Question

Is the textbook wrong or I made a mistake, if so, where? The integral should be fine because I checked it with Wolfram Alpha.

Taking the definite integral of an expression requires that you subtract off everything before the lower limit, $t = 0$ (which corresponds to the start of the motion).
$$s(t) = \frac{-20e^{-\frac{1}{2}t} (sin(\frac{\pi}{2}t)+\pi cos(\frac{\pi}{2}t))}{1+\pi^2} - \frac{-20\pi}{1 + \pi^2}$$