Understanding Integration through word problem. I am a simple guy who wants to understand the concept of integration. I have been solving problems on differentiation and integration for 8 months (Class). I know the formulas and stuff to solve college level complicated equations. But I don't understand what is the answer good for? I fail to grasp the application behind it. 
So, here is one problem i think we can use integration on.
Let's say,  A boy is travelling with speed $s=5t+10$ where $t$ is the time. 


*

*How can this question be modeled such that we need to use integration?  

*If I do $\int 5t+10\ dt$. What does this result signify?

*If i do $\int_0^{10} 5t+10\ dt$. What does this result signify and how is it different from 1 above?


It will be amazingly fruitful to me if someone gives me insight to this. Thanks.
 A: Velocity $v$ is defined as the time derivative of the position $s$:
$$
v = \frac{ds}{dt}
$$
Seen from the perspective of $s$, $s$ is an antiderivative of $v$.
$$
s = \int v(t) \, dt = \int (5 t + 10) \, dt = \frac{5}{2} t^2 + 10 t + C
$$
where $C$ is some constant.
To calculate what distance the boy traveled you need a definite integral: 
\begin{align}
s(t) - s(t_0) 
&= \int\limits_{t_0}^t ds \\  
&= \int\limits_{t_0}^t v(\hat{t}) \, d\hat{t} \\
&= \int\limits_{t_0}^t (5 \hat{t} + 10) \, d\hat{t} \\
&= \left[ 
\frac{5}{2} \hat{t}^2 + 10 \hat{t} 
\right]_{\hat{t}=t_0}^{\hat{t}=t} \\
&= \frac{5}{2}(t^2 - t_0^2)+10(t-t_0)
\end{align}
With $t_0 = 0$ and $t = 10$ we obtain
$$
s(t) = 350 + s(0)
$$
Thus after 10 time units the boy is 350 distance units further than his initial position $s_0 = s(0)$.
Interpretation:
Integration is used to solve the differential equation
$$
\dot{s}(t) = v(t)
$$
for a given velocity function $v(t)$ and initial condition $s(t_0) = s_0$
The indefinite integral just specifies the general relation between $s$ and $v$.
The indefinite integral can be used to solve the differential equation, and thus finds the trajectory $s(t)$ through $(t, s(t)) = (t_0, s_0)$ among the many possible solutions which are just different by some additive constant $C$.
