Let $v(t)$ be the velocity. You are told that $v(0)=50$ mi/h; you are also told that when brakes are applied, acceleration, $v'(t)$, is $-26\ \mathrm{ft}/\mathrm{sec}^2$. It's better to put everything in the same units, so let's change velocity to feet per second:
$$50\frac{\mathrm{mi}}{\mathrm{hr}} = 50\left(\frac{5280\text{ feet}}{1\text{ mi}}\right)\left(\frac{1\text{ hr}}{3600\text{ sec}}\right) = \frac{220}{3}\text{ft}/\text{sec}.$$
So $v(0) = \frac{220}{3}$ ft/sec. $v'(t) = -26$ ft/sec${}^2$.
By the First Fundamental Theorem of Calculus, we have:
$$v(t) -v(0) = \int_0^t v'(x)\,dx = \int_0^t(-26)\,dx = -26x\Bigm|_0^t = -26t.$$
Therefore,
$v(t) = v(0) - 26t = \frac{220}{3}-26t$.
The car comes to a stop when $v(t)=0$. Solving $v(t)=0$ we get
$$t = \frac{220}{78} = \frac{110}{39}\text{ seconds.}$$
How far did it travel? The distance traveled is the integral of the velocity, so
$$\begin{align*}
\text{distance traveled} &= \int_0^{110/39}v(t)\,dt\\
&= \int_0^{110/39}\left(\frac{220}{3} - 26t\right)\,dt \\
&= \frac{220}{3}t - 13t^2\Bigm|_0^{110/39}\\
&= \frac{24200}{117} - \frac{157300}{1521}\\
&\approx 103.4188\text{ feet.}
\end{align*}$$
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.
Best Answer
Without using calculus:
Separate $v(t)$ into two components, its constant component, and its cosine component.
From the constant component, it's easy to start by observing that a constant speed of 20 ft/sec for 10 sec yields 200ft. Hang onto this number.
From the cosine component, we know that cosine has a period of $2\pi$. At $t=2pi$, effect of the term has completely nullified itself: the amount of velocity above the $x$-axis has exactly the same profile as that below the $x$ axis. So let's look at $(2\pi,10)$.
Now, we know that $3\pi \approx 9.4 < 10$. From $2\pi$ to $3\pi$, we have the same effect: the velocity above the $x$ axis "neutralizes" the velocity below.
So now we're left with just a little tail of negative velocity between $3\pi$ and $10$. It's only about a half-second worth of time, and in that time the velocity goes from -7 to -5.87. Let's split the difference, and consider it to have a constant velocity in that half-second window of -6.43 ft/s. Multiplying this quantity by $(10-3\pi)$, and adding to the previous result, and we get about 196 feet.
Let's compare this to the integral:
$$\begin{align*}\int_0^10 \left(20+7\cos t\right)\ dt &= \left.20t\right|_0^10 + \left.7\sin t \right|_0^10 \\ &= 200(10-0) + 7\sin(10)-7\sin(0) \\ &= 200+7\sin(10) \\ &= 200-3.808 \\ &\approx 196.19. \end{align*}$$