Velocity & Definite Integral I am not sure how to approach this problem.
Problem
Suppose a certain object moves in a straight line with the following velocity, where $v$ is in meters per second and $t$ is in seconds:
$v(t) = -2 + t + 3\sin(\pi t)$
Without using your calculator, but instead using properties of definite integrals and facts you know about area, determine the net change in distance of the object from $t = 0$ to time $t = 6$ and find the object's average velocity on this interval.
My impression is that the net change in distance would be:
$d = \int_{0}^{6} |v(t)| dt$
But since we cannot use our calculators I am unsure.
 A: I think net change would not have the absolute value; you would use absolute value when you want to know how much total movement there was.  So you simply evaluate
$$\int_0^6 v(t) \, dt $$
For your function, you would evaluate
$$\begin{align} \int_0^6 (-2 +t + 2 \sin{(\pi t)}\, dt &= -2 \int_0^6 dt + \int_0^6 t \, dt + 2 \int_0^6 \sin{(\pi t)} \, dt \\ &=-2 (6-0) + \frac{1}{2} (6^2 - 0^2) + \frac{2}{\pi} (1-\cos{(6 \pi)}) \\ &= -12 + 18 \\ &= 6 \end{align} $$
That is, the net distance is $6$ m.  Note that I used the fact that $\cos{(6 \pi)} = 1$.
A: We want the net change in  displacement from $t=0$ to $t=6$. This is
$$\int_0^6 v(t)\,dt,$$
no absolute values. With absolute values, you are computing the total distance travelled, which would be relevant for gasoline consumption, but is not what is asked for. 
The average velocity in our time interval  is
$$\frac{1}{6}\int_0^6 v(t)\,dt.$$
As to doing it without calculation, by properties of area, note that $\sin \pi t$ goes through $3$ full cycles. By the picture of sine, the integral of the sine part is $0$.  (There is exactly as much area below the axis as above.)
As to the integral of the $-2+t$ part, draw the line $y=-2+t$ (the usual $x$-axis is now called the $t$-axis). The integral is (sort of) the area under this curve, and "above" the $t$-axis, except that from $t=0$ to $t=2$ this area has to be viewed as negative. So it is the area of a certain triangle, $t=2$ to $t=6$, minus the area of a certain triangle, $t=0$ to $t=2$.
A picture (for the $y=-2+t$ part, forget about the sine part) is essential. 
