# 1. Applications of the Indefinite Integral

by M. Bourne

## Displacement from Velocity, and Velocity from Acceleration

High velocity train [Image source]

A very useful application of calculus is displacement, velocity and acceleration.

Recall (from Derivative as an Instantaneous Rate of Change) that we can find an expression for **velocity** by differentiating the expression for displacement:

`v=(ds)/(dt)`

Similarly, we can find the expression for the **acceleration** by differentiating the expression for velocity, and this is equivalent to finding the second derivative of the displacement:

`a=(dv)/dt=(d^2s)/(dt^2)`

It follows (since integration is the opposite process to differentiation) that to obtain the **displacement**,* *`s` of an object at time `t`* *(given the expression for velocity, `v`) we would use:

`s=int v\ dt`

Similarly, the **velocity** of an object at time `t` with acceleration `a`, is given by:

`v=inta\ dt`

### Example 1

A car starts from rest at `s=3\ "m"` from the origin and has acceleration at time `t` given by `a=2t-5\ "ms"^-2`. Find the velocity and displacement of the car at `t=4\ "s"`.

### Solution

We find the velocity using:

`v=int a\ dt`

So in this example we have:

`v=int(2t-5)\ dt`

`=t^2-5t+K`

When `t = 0`, `v = 0`, so `K =0`.

So the expression for velocity as a function of time is:

`v=t^2-5t\ text[m s]^-1`

When `t = 4`, `v = 4^2-5(4)=-4\ text[m s]^-1`

Now to find the displacement.

`s=int v\ dt`

Then

`s=int(t^2-5t)\ dt`

`=t^3/3 - (5t^2)/2 + K`

Now when `t = 0`, `s = 3`, so we substitute to obtain:

`3=0^3/3 - (5(0)t^2)/2 + K`

So `K=3` and therefore the general expression for `s` is:

`s=t^3/3 - (5t^2)/2 + 3`

When `t = 4`, `s = 4^3/3 - (5(4)^2)/2 + 3 = -15.67\ text[m]`

The graphs of the acceleration, velocity and displacement at time *t*, indicating the velocity and displacement at `t=4`.

### Example 2

A proton moves in an electric field such that its
acceleration (in cms^{-2}) is

`a = -20(1+2t)^-2`, where `t`is in seconds.

Find the velocity as
a function of time if *v* = 30 cms^{-1} when *t* = 0.

Answer

`v=int a\ dt`

So

`v=int(-20\ dt)/((1+2t)^2)`

Put `u = 1 + 2t` then `du = 2\ dt`, so `dt=(du)/2`

`v=int (-10\ du)/(u^2)`

`=int -10u^(-2)du`

`=10/u+K`

`=(10)/(1+2t)+K`

When `t = 0`, `v = 30`, so `K = 20`.

So the expression for velocity as a function of time is:

`v=((10)/(1+2t)+20)\ text[cm s]^-1`

Get the Daily Math Tweet!

IntMath on Twitter

Here are the graphs of the acceleration of the proton, and the velocity we found in Example 2.

The graphs of the acceleration and velocity at time *t*. Note `v(0)=30`.

### Example 3

A flare is ejected vertically upwards from the ground at 15 m/s. Find the height of the flare after 2.5 s.

Answer

The object has acting on it the force due to
gravity, so its acceleration is -9.8 ms^{-2}.

`v=int a dt`

`=int-9.8\ dt`

`=-9.8t+C`

Now at `t = 0`, the velocity = `15\ "ms"^-1`. So `C = 15`.

So the expression for velocity becomes:

`v=-9.8t+15`

Now, we need to find the displacement, so we integrate our expression for velocity:

`s=int v\ dt`

`=int(-9.8t + 15) dt`

`=-4.9t^2+15t+K`

Now, we know from the question that when `t = 0`, `s = 0`.

This gives `K = 0`.

So `s=-4.9t^2+15t`

At time `t = 2.5`, `s = 6.875\ "m"`.

## Displacement and Velocity Formulas

Using integration, we can obtain
the well-known expressions for displacement and velocity, given a
constant acceleration *a*,* *initial displacement zero,
and an initial velocity `v_0`:

`v=int a\ dt`

`v=at+K`

Since the velocity at `t=0` is `v_0`, we have `K=v_0`. So:

`v=v_0 + at`

Similarly, taking it another step gives:

`s=int v\ dt=int (v_0 + at)dt`

`s=v_0 t + (at^2)/2+C`

Since the displacement at `t=0` is `s=0`, we have `C=0`. So:

`s=v_0t+1/2at^2`

**Voltage across a Capacitor**

**Definition: **The current, *i *(amperes), in an electric circuit
equals the time rate of change of the **charge** *q*, (in
coulombs) that passes a given point in the circuit. We can write
this (with *t* in seconds) as:

`i=(dq)/(dt)`

By writing *i dt = dq* and **integrating**,
we have:

`q=inti\ dt`

The voltage, *V*_{C} (in
volts) across a capacitor with capacitance *C* (in farads)
is given by

`V_C=q/C`

It follows that

`V_C=1/Cinti\ dt`

You can see some more advanced applications of this at Applications of Ordinary Differential Equations.

### Example 4

The electric current (in mA) in a computer circuit as a function of time is `i = 0.3 − 0.2t`. What total charge passes a point in the circuit in `0.050` s?

Answer

The charge, *q*, is given by:

`q=inti\ dt`

`=int (0.3-0.2t) dt`

`=0.3t-0.1t^2+K`

At `t = 0`, `q = 0`, and this gives us `K = 0`.

So `q=0.3t-0.1t^2`

`[q]_[t=0.050]` `=0.3(0.050)-0.1(0.050)^2` `=0.015\ text[mC]`

(units are milli coulombs, as current `i`* *was in
`"mA"`.)

### Example 5

The voltage across an `8.50\ "nF"` capacitor in an FM receiver circuit
is initially zero. Find the voltage after `2.00\ μ"s"` if a current `i=0.042t`* *(in `"mA"`) charges the capacitor.

Answer

`V_C=1/C inti\ dt `

**Note:** `1\ "nF" = 10^-9"F"`; and `1\ μ"s" = 10^-6"s"`;

Also, `0.042t\ "mA" = 0.042 × 10^-3t\ "A"`

`V_C = (0.042 times 10^-3)/(8.5 times 10^-9)int t\ dt`

`=4.94 times 10^3 (t^2)/(2)+K`

`=2.47 times 10^3t^2 + K`

Now, we are told that when `t = 0`, `V_C= 0`.

So `K= 0`.

Thus

`V_C=2.47 times 10^3 t^2`

So when `t = 2.00\ μ"s"`, we have:

`V_C = 2.47 times 10^3(2 times 10^-6)^2`

`=9.882 times 10^-9`

`=9.88\ text[nV]`

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!