Calculus for Physics II: Integrals

The question this answers

A car's speedometer reads 20 m/s for ten seconds. You know it went 200 metres, because distance is speed times time. Now suppose the needle is climbing the whole time, from 0 up to 20 m/s. The product $v\,t$ is meaningless: which $v$ would you use?

The last lesson went the other way. Given position, the derivative $v = dx/dt$ told you the speed at an instant, by zooming in until the curve looked straight. Now you have the speed at every instant and you want the distance back. Differentiation chopped a curve into slopes; integration is the machine that glues slopes back into a curve. Almost every physics quantity you will meet (work, impulse, charge, moment of inertia, the centre of mass of a lopsided object) is a product where one factor refuses to hold still, and every one of them is solved the same way.

Slice it, add it, shrink it

Here is the trick, and it never changes. Chop the ten seconds into slivers so short that over one sliver the velocity has no time to change. Call a sliver $\Delta t$. Over that sliver the car really does move $v\,\Delta t$, because for that instant it may as well be moving at constant speed. Add the slivers:

$\Delta x \approx \sum_i v(t_i)\,\Delta t$

That is an approximation, and you can see the error: inside each sliver you froze $v$ at its starting value, so if $v$ is rising you undercounted. Make the slivers thinner and the error shrinks. Take the limit as $\Delta t \to 0$ and the sum becomes exact. That limit is written with a stretched-out S, which is what the integral sign is:

$x(t) = x_0 + \int_0^{t} v(t')\,dt'$

Read it aloud as what it is: "add up $v$ times $dt$, from 0 to $t$." The $dt$ is not decoration. It is the width of the sliver, and it carries units. If $v$ is in m/s and $dt$ is in s, every term in the sum is in metres, so the answer is in metres. Checking that the integrand times $dx$ has the units of your answer is the fastest error check you own.

Because every term is a height times a width, the sum is an area. The integral of $v(t)$ is the area under the velocity-time graph. Area below the axis counts as negative, which is exactly right: driving backwards should subtract from your displacement.

When the graph is made of straight lines, just find the area

You do not need any calculus to integrate a triangle. A great many problems hand you a graph built from straight segments and expect you to break it into rectangles and triangles and add.

Worked example 1. A train starts from rest, rises linearly to 12 m/s over 4.0 s, holds 12 m/s until $t = 10$ s, then falls linearly to zero at $t = 14$ s. How far does it go?

Three pieces. The first is a triangle of base 4.0 s and height 12 m/s: $\tfrac{1}{2}(4.0)(12) = 24$ m. The middle is a rectangle, 6.0 s wide and 12 m/s tall: $(6.0)(12) = 72$ m. The last is a triangle of base 4.0 s and height 12 m/s again: 24 m. Total $24 + 72 + 24 = 120$ m.

The same picture works one level up. The area under an acceleration-time graph is a change in velocity, because $\Delta v = \sum a\,\Delta t$. And the area under a force-position graph is work, because $W = \sum F\,\Delta x$. Same machine, different labels on the axes.

Integration as undoing a derivative

Curves need more than geometry. The shortcut is that the derivative rules run backwards. If differentiating $t^{n+1}$ gives $(n+1)t^n$, then to integrate $t^n$ you raise the power and divide by the new power:

$\int t^n\,dt = \frac{t^{n+1}}{n+1}\ (n \neq -1), \qquad \int \frac{dt}{t} = \ln|t|, \qquad \int e^{kt}\,dt = \frac{e^{kt}}{k}$

The $n = -1$ case is excluded because raising the power would give $t^0/0$. That gap is exactly where the logarithm lives, and it is why logs show up in rockets and in drag problems and nowhere else.

Each time you integrate, you gain one unknown constant, because every function that differs by a constant has the same derivative. Physics fixes it with an initial condition. Two integrations, two conditions: you need both the starting velocity and the starting position before a trajectory is determined. That is why $F = ma$ alone predicts nothing until you say where the thing started and how fast it was going.

Worked example 2. A cart starts from rest at the origin with $a(t) = 12 - 4t$ in SI units. Where is it at $t = 3.0$ s?

Integrate once for velocity:

$v(t) = \int_0^{t}(12 - 4t')\,dt' = 12t - 2t^2$

Check it at $t=0$: $v = 0$, which matches "starts from rest", so the constant is zero. Integrate again:

$x(t) = \int_0^{t}(12t' - 2t'^2)\,dt' = 6t^2 - \frac{2t^3}{3}$

At $t = 3.0$ s: $6(9) = 54$ and $\tfrac{2}{3}(27) = 18$, so $x = 54 - 18 = 36$ m.

Notice what you did not do: you did not reach for $x = \tfrac12 at^2$. That formula is itself the result of integrating a constant $a$, and here $a$ is not constant. Using it would have been a guess.

Any product with a varying factor

Once you see the pattern, it is everywhere. Take the defining relationship, write it for an infinitesimal piece, and add the pieces up.

Impulse is force times time, so for a changing force $J = \int F\,dt$. Charge is current times time, so $q = \int i\,dt$. The mass of a rod with varying thickness is $m = \int \lambda\,dx$, where $\lambda$ is the mass per unit length. And work is force times displacement, so for a varying force

$W = \int_{x_1}^{x_2} F(x)\,dx$

A spring pulls back with $F = kx$ when stretched a distance $x$, so the work you must do to stretch it from its natural length out to $x$ is $\int_0^{x} kx'\,dx' = \tfrac12 kx^2$. That is not a formula to memorise; it is the area of a triangle of base $x$ and height $kx$. With $k = 250$ N/m and $x = 0.12$ m: $\tfrac12(250)(0.12)^2 = \tfrac12(250)(0.0144) = 1.8$ J.

A capacitor works the same way. Pushing charge $dq$ onto a capacitor already holding $q$ costs $dW = (q/C)\,dq$, because $q/C$ is the voltage it is already fighting. So the stored energy is $\int_0^{Q} q\,dq/C = Q^2/2C = \tfrac12 CV^2$. With $C = 4.7\ \mu$F and $V = 200$ V: $\tfrac12(4.7\times10^{-6})(200)^2 = \tfrac12(4.7\times10^{-6})(4\times10^{4}) = 0.094$ J.

Slicing up an object

Problems about rods, disks, chains and tanks all use one recipe, and it is worth writing down as steps because the setup is where people lose the marks.

1. Draw one representative slice and give its position a symbol.
2. Write the slice's mass (or charge, or volume) in terms of that symbol and its thickness: this is your $dm$.
3. Write what you want for that one slice.
4. Integrate over the range the slices actually cover.

Worked example 3. A disk of radius $R = 0.50$ m and total mass $M = 2.0$ kg has areal density growing as $\sigma(r) = \sigma_0 r$. Find its moment of inertia about the central axis.

The right slice is a thin ring of radius $r$ and thickness $dr$, because every point on it is the same distance from the axis. Its area is circumference times thickness, $2\pi r\,dr$, so its mass is $dm = \sigma_0 r \cdot 2\pi r\,dr = 2\pi\sigma_0 r^2\,dr$.

You do not know $\sigma_0$, so get it from the total mass:

$M = \int_0^{R} 2\pi\sigma_0 r^2\,dr = \frac{2\pi\sigma_0 R^3}{3}$

Now the moment of inertia, which weights each bit of mass by $r^2$:

$I = \int_0^{R} r^2\,dm = \int_0^{R} 2\pi\sigma_0 r^4\,dr = \frac{2\pi\sigma_0 R^5}{5}$

Divide one by the other and the unknown $\sigma_0$ cancels: $I/M = \tfrac{3}{5}R^2$, so $I = 0.6(2.0)(0.50)^2 = 0.6(2.0)(0.25) = 0.30$ kg·m². Forming the ratio to kill an unknown constant is a move you will use constantly.

The same recipe handles a chain being lifted off the floor. When a length $y$ already hangs vertically, its mass is $\lambda y$ and its centre of mass is at height $y/2$, so raising the next bit costs $dW = \lambda g y\,dy$, and $W = \tfrac12 \lambda g L^2$. With $\lambda = 2.0$ kg/m, $L = 8.0$ m: $\tfrac12(2.0)(9.8)(64) = 627$ J.

When the integral is not over time

Sometimes you know force as a function of position or of speed, not of time, and integrating $dt$ is hopeless. The escape is a single identity from the chain rule:

$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx}$

Multiply both sides by $dx$ and integrate, and you get $\int v\,dv = \int a\,dx$, that is $\tfrac12 v^2 - \tfrac12 v_0^2 = \int a\,dx$. That is the work-energy theorem, derived in one line.

Worked example 4. An 80 kg skydiver glides horizontally and feels only quadratic drag, $F = -cv^2$ with $c = 0.25$ kg/m. Over what distance does the speed fall to half?

Time is not asked for, and distance is, so use the position form:

$m\,v\frac{dv}{dx} = -cv^2 \quad\Longrightarrow\quad m\frac{dv}{v} = -c\,dx$

One factor of $v$ cancelled on each side, which is why this problem is tractable at all. Integrate from $v_0$ to $v_0/2$ on the left and 0 to $x$ on the right. The left side is the $n = -1$ case, so it gives a logarithm:

$m\ln\frac{v_0/2}{v_0} = -cx \quad\Longrightarrow\quad x = \frac{m}{c}\ln 2$

The starting speed drops out entirely. $x = (80/0.25)(0.693) = 320 \times 0.693 = 222$ m.

The same "integrate over position" move gives gravitational energy far from Earth, where $g$ is not constant. The force is $GM_Em/r^2$, so the work against it from $R$ to $R+h$ is

$W = \int_{R}^{R+h}\frac{GM_Em}{r^2}\,dr = GM_Em\left(\frac{1}{R} - \frac{1}{R+h}\right)$

For a 500 kg payload lifted 500 km from the surface, with $GM_E = 3.986\times10^{14}$: $1/R = 1.5699\times10^{-7}$ and $1/(R+h) = 1.4556\times10^{-7}$, a difference of $1.143\times10^{-8}$. Multiply by $GM_Em = 1.993\times10^{17}$ to get $2.28\times10^{9}$ J. Using $mgh$ instead would have given $2.45\times10^{9}$ J, about 8% high, because gravity weakens on the way up.

How to recognise that you need an integral

Reach for this machinery when you see any of these:

• A quantity that is normally a product, with one factor varying. Force times distance, current times time, density times length. If the varying factor were frozen you would just multiply; since it is not, you integrate.
• A graph you are asked for a total from. "How far", "what impulse", "how much work" off a plot means area, and signed area at that.
• A rate given as a function of time. $a(t)$, $i(t)$, a burn rate. You want the accumulated amount, so integrate.
• An extended object that is not uniform. Rods, disks, cones, chains: slice it, write $dm$, integrate.
• A force that depends on position or speed. Then no kinematic equation applies. Either integrate $F\,dx$ for energy, or use $a = v\,dv/dx$.

And the negative signal: if the acceleration is genuinely constant and nothing is extended or non-uniform, do not integrate. The kinematic equations already did it for you.

Derivatives and integrals are now both in hand. The next tool is the one that makes hard integrals and hard equations survivable at all: knowing when a term is small enough to throw away. That is approximations and small-angle expansions.