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.

  1. How can this question be modeled such that we need to use integration?
  2. If I do $\int 5t+10\ dt$. What does this result signify?
  3. 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.

  • $\begingroup$ You can either understand integration through physics, or see it as the anti-derivative. $\endgroup$
    – dh16
    Apr 2 '15 at 21:41
  • $\begingroup$ I understand it from the anti-derivative point of view. What i am trying to understand for so far is from the physics point of view. I want to know why it is being used for rather than how it is used for. $\endgroup$ Apr 2 '15 at 21:43

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 $s_0 = s(0)$.


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 $C$.

  • $\begingroup$ So integrating this gives us the distance ? since we are going from t=0 to t=10, why can we not do it using the sigma operation? (submission). I still quite couldn't figure out the use of integration. $\endgroup$ Apr 2 '15 at 21:39
  • $\begingroup$ Okay. Sure!.... $\endgroup$ Apr 2 '15 at 21:40
  • 1
    $\begingroup$ Now to your comment: with Sigma operation you mean the summation of discrete values. Note that we have a smooth curve $v(t)$ where $v(t)$ might change at any arbitrary point $t$ in time. For this you need the integral. It can be interpreted as limit case of a discrete sum for discrete $v_i = v(t_i)$ with $\Delta t = t_{i+1}-t_i \to 0$. That is the historical reasoning. $\endgroup$
    – mvw
    Apr 2 '15 at 22:06
  • $\begingroup$ Wow, why i almost got this thing now. I need to work some of the problems to tame it. Thanks a lot! $\endgroup$ Apr 2 '15 at 22:09

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