Calculus in Physics — Integration

In short

Integration is the reverse of differentiation — it reconstructs a quantity from its rate of change. If you know how velocity changes with time, integration gives you the total displacement. If you know how force varies with position, integration gives you the total work done. Geometrically, the integral is the area under a curve. Formally, \int_a^b f(t)\,dt adds up infinitely many infinitely thin slices of f(t) between t = a and t = b.

An autorickshaw pulls away from a traffic signal in Bengaluru. For the first 3 seconds, its speedometer reads different values at different times — 0 m/s at the start, 2 m/s after half a second, 6 m/s after 1.5 seconds, 14 m/s at 3 seconds. The speed keeps changing. How far did it travel in those 3 seconds?

If the speed were constant — say, a steady 10 m/s — the answer would be simple: distance = speed × time = 10 × 3 = 30 metres. But the speed is not constant. It is different at every instant. You cannot just multiply a single speed by 3 seconds because there is no single speed to multiply.

This is the problem integration solves. It takes a quantity that varies continuously — velocity, force, pressure, charge density — and adds up its total effect over an interval. Differentiation broke things apart (rate of change at an instant); integration puts them back together (total accumulation over time).

Adding up thin slices — the area under a velocity-time graph

Start with something concrete. Suppose the autorickshaw's velocity at time t is v(t) = 4t + 2 m/s. At t = 0, it is already moving at 2 m/s. At t = 1 s, it is at 6 m/s. At t = 3 s, it reaches 14 m/s.

If the velocity were constant at, say, 6 m/s for 1 second, the displacement in that 1-second window would be 6 \times 1 = 6 m. On a velocity-time graph, this is the area of a rectangle: height = 6, width = 1.

That suggests a strategy: chop the time interval into small pieces, treat the velocity as roughly constant in each piece, and add up the rectangular areas. The sum of all those rectangles approximates the total displacement.

Six rectangles approximate the area under the velocity-time line $v = 4t + 2$. Each rectangle uses the velocity at its left edge as its height. The total area of the rectangles approximates the displacement — but notice the gaps between the rectangle tops and the line. More rectangles means smaller gaps.

With 6 rectangles (each 0.5 s wide), you add up the areas:

\text{Sum} = 0.5 \times [v(0) + v(0.5) + v(1) + v(1.5) + v(2) + v(2.5)]

= 0.5 \times [2 + 4 + 6 + 8 + 10 + 12] = 0.5 \times 42 = 21 \text{ m}

This is an approximation. The rectangles miss the triangular bits between their tops and the actual line. If you used 100 rectangles, the gaps would be tiny. If you used 1000, tinier still. If you used infinitely many infinitely thin rectangles — each of width dt, where dt is so small it might as well be zero — the gaps vanish entirely. The sum becomes exact.

That infinite sum of infinitely thin slices is the integral:

\text{Displacement} = \int_0^3 v(t)\,dt = \int_0^3 (4t + 2)\,dt

The integral sign \int is a stretched-out "S" — it stands for summa, the Latin word for sum. The dt at the end tells you what variable you are summing over and reminds you of the infinitesimal width of each slice.

The indefinite integral — undoing differentiation

Before computing the definite integral (with limits), step back and ask: what function, when differentiated, gives you 4t + 2?

You already know from differentiation that:

\frac{d}{dt}(2t^2 + 2t) = 4t + 2

Why: the power rule says \frac{d}{dt}(t^n) = nt^{n-1}. Differentiating 2t^2 gives 4t. Differentiating 2t gives 2. Together: 4t + 2.

So 2t^2 + 2t is a function whose derivative is 4t + 2. This function is called an antiderivative of 4t + 2.

But wait — the function 2t^2 + 2t + 7 also has derivative 4t + 2, because the derivative of the constant 7 is zero. So does 2t^2 + 2t - 100. In fact, 2t^2 + 2t + C for any constant C is an antiderivative.

The indefinite integral is the family of all antiderivatives:

\int (4t + 2)\,dt = 2t^2 + 2t + C

Why: differentiating the right side gives back 4t + 2, confirming the integration is correct. The constant C is called the constant of integration — it accounts for the fact that differentiation destroys information about the starting value.

Physically, the constant C has a clear meaning. If v(t) = 4t + 2 is velocity, then \int v\,dt = 2t^2 + 2t + C is position. The constant C is the position at time t = 0 — the starting point. Differentiation tells you the rate of change but forgets where you started. Integration reconstructs the whole journey, but you need to supply the starting point separately.

The definite integral — with limits

When you want the total displacement between t = 0 and t = 3, you compute the definite integral:

\int_0^3 (4t + 2)\,dt

The numbers 0 and 3 are the limits of integration — the lower limit (where you start summing) and the upper limit (where you stop). To evaluate this, find the antiderivative and subtract its value at the lower limit from its value at the upper limit:

Step 1. Find the antiderivative.

F(t) = 2t^2 + 2t

Why: you can drop the constant C for a definite integral because it cancels in the subtraction F(3) - F(0).

Step 2. Evaluate at the upper limit.

F(3) = 2(3)^2 + 2(3) = 18 + 6 = 24

Step 3. Evaluate at the lower limit.

F(0) = 2(0)^2 + 2(0) = 0

Step 4. Subtract.

\int_0^3 (4t + 2)\,dt = F(3) - F(0) = 24 - 0 = 24 \text{ m}

Why: this subtraction computes the net area under the curve between t = 0 and t = 3. The C that appears in both F(3) and F(0) cancels, so the constant of integration is irrelevant for definite integrals.

The autorickshaw travels 24 m in the first 3 seconds. Compare this with the rectangle approximation of 21 m — the rectangles underestimated because they used the left-edge velocity, which is always lower than the actual velocity on a rising line.

The fundamental theorem of calculus — differentiation and integration are inverses

Everything above rests on one deep fact. Differentiation breaks a quantity into its instantaneous rate of change. Integration reassembles those rates into the original quantity. They undo each other:

\frac{d}{dt}\left[\int_a^t f(\tau)\,d\tau\right] = f(t)

Why: the integral from a to t accumulates the area under f as t increases. The rate at which that area grows, at any instant t, is just the height of the function at that instant — which is f(t). Geometrically: adding a thin vertical strip of width dt and height f(t) changes the area by f(t)\,dt, so the rate of change is f(t).

And the evaluation formula:

\int_a^b f(t)\,dt = F(b) - F(a) \quad \text{where } F'(t) = f(t)

This is the fundamental theorem of calculus. In physics notation, the most important instance is:

\int_a^b \frac{dx}{dt}\,dt = x(b) - x(a)

If you integrate velocity (the rate of change of position) over a time interval, you get the change in position. If you integrate acceleration over a time interval, you get the change in velocity. If you integrate force over a distance, you get the work done. The pattern is always the same: integrating a rate of change gives the total change.

Standard integration rules — the toolkit

Just as differentiation has a set of rules (power rule, chain rule, product rule), integration has corresponding rules. Here are the ones you will use most in physics.

Power rule for integration

\int t^n\,dt = \frac{t^{n+1}}{n+1} + C \qquad (n \neq -1)

Why: this is the reverse of the power rule for differentiation. Differentiating \frac{t^{n+1}}{n+1} gives \frac{(n+1)\,t^n}{n+1} = t^n, confirming the formula.

The exception n = -1 matters because \frac{t^0}{0} is undefined. The integral of t^{-1} = 1/t is the natural logarithm:

\int \frac{1}{t}\,dt = \ln|t| + C

Trigonometric integrals

\int \sin t\,dt = -\cos t + C

Why: differentiating -\cos t gives \sin t. The negative sign comes from the derivative of cosine being -\sin t — reversing that flips the sign.

\int \cos t\,dt = \sin t + C

Why: differentiating \sin t gives \cos t directly.

Exponential integral

\int e^{at}\,dt = \frac{1}{a}\,e^{at} + C

Why: differentiating \frac{1}{a}e^{at} brings down the factor a from the chain rule, giving \frac{a}{a}e^{at} = e^{at}.

Summary table

Function f(t)	Integral \int f(t)\,dt	Note
t^n (for n \neq -1)	\dfrac{t^{n+1}}{n+1} + C	Power rule
1/t	\ln\lvert t \rvert + C	Special case n = -1
\sin t	-\cos t + C	Sign flip
\cos t	\sin t + C	Direct reverse
e^{at}	\frac{1}{a}e^{at} + C	Chain rule in reverse
\text{constant } k	kt + C	Area of a rectangle

Two properties make computing integrals easier. A constant factor pulls out:

\int k\,f(t)\,dt = k\int f(t)\,dt

And the integral of a sum is the sum of the integrals:

\int [f(t) + g(t)]\,dt = \int f(t)\,dt + \int g(t)\,dt

Why: both properties follow from the fact that differentiation has the same properties (the derivative of a sum is the sum of derivatives, and constants pull out of derivatives). Since integration reverses differentiation, it inherits these rules.

When physics requires integration

Differentiation appears in physics wherever a rate shows up — velocity is the derivative of position, acceleration is the derivative of velocity. Integration appears whenever you need to add up a continuously varying quantity.

Here are the situations where integration is unavoidable:

Non-uniform velocity. If a body's velocity changes with time (non-uniform motion), you cannot compute displacement using s = vt because there is no single v. You must integrate: s = \int_{t_1}^{t_2} v(t)\,dt.

Variable force and work. If a force varies with position — a spring force F = -kx, gravitational force F = -GMm/r^2, or any force that is not constant — the work done is not simply F \times d. You must integrate: W = \int_{x_1}^{x_2} F(x)\,dx.

Continuous mass distributions. Finding the centre of mass of a rod whose density varies along its length requires integration: x_{\text{cm}} = \frac{\int x\,dm}{\int dm}. You cannot just average two endpoints because the mass is spread continuously.

Charge distributions. The electric field due to a charged rod, ring, or disk comes from adding up contributions from each infinitesimal charge element — that sum is an integral.

Moment of inertia. For a continuous body, I = \int r^2\,dm replaces the discrete sum I = \sum m_i r_i^2.

In each case, the pattern is the same. You have a quantity that varies continuously, and you need its total effect. You chop it into infinitesimal pieces, compute the effect of each piece, and integrate.

Worked examples

Example 1: Displacement from a velocity function

An autorickshaw accelerates from a traffic signal. Its velocity is v(t) = 4t + 2 m/s, where t is in seconds. Find the displacement in the interval t = 0 to t = 3 s.

Setup. The velocity is not constant — it increases linearly with time. The displacement is the area under the v-t graph from t = 0 to t = 3.

s = \int_0^3 v(t)\,dt = \int_0^3 (4t + 2)\,dt

Step 1. Find the antiderivative of 4t + 2.

\int (4t + 2)\,dt = 4 \cdot \frac{t^2}{2} + 2t = 2t^2 + 2t

Why: use the power rule on 4t = 4t^1, giving 4 \cdot \frac{t^2}{2} = 2t^2. The integral of the constant 2 is 2t.

Step 2. Evaluate at the upper limit t = 3.

F(3) = 2(9) + 2(3) = 18 + 6 = 24

Step 3. Evaluate at the lower limit t = 0.

F(0) = 2(0) + 2(0) = 0

Step 4. Subtract.

s = F(3) - F(0) = 24 - 0 = 24 \text{ m}

The velocity $v = 4t + 2$ is a straight line. The shaded region between the line, the $t$-axis, and the vertical dashed lines at $t = 0$ and $t = 3$ has area 24 — the displacement in metres. This area is a trapezoid: parallel sides of 2 and 14, width 3, so area $= \frac{1}{2}(2 + 14)(3) = 24$. The integral gives the same answer.

Result: The autorickshaw travels 24 m in 3 seconds.

What this shows: For a linearly increasing velocity, the integral gives the area of a trapezoid. The integral works for any velocity function — linear, quadratic, sinusoidal — not just the cases where you can see the geometric shape. That is its power.

Quick check. The average velocity is \frac{v(0) + v(3)}{2} = \frac{2 + 14}{2} = 8 m/s. Displacement = average velocity \times time = 8 \times 3 = 24 m. This shortcut works because the velocity is linear — the average is the midpoint. For non-linear velocity functions, you must integrate; the shortcut fails.

Example 2: Work done by a variable force

A force F(x) = 6x^2 N acts on a block along the x-direction. The block moves from x = 0 to x = 2 m. Find the work done by the force.

Setup. The force is not constant — it grows as the square of the position. At x = 0, the force is zero. At x = 1 m, the force is 6 N. At x = 2 m, the force is 24 N. You cannot use W = Fd because F keeps changing.

The work done by a variable force is:

W = \int_0^2 F(x)\,dx = \int_0^2 6x^2\,dx

Why: work is force times displacement, but only when the force is constant. When the force varies, you chop the path into infinitesimal segments dx, compute the tiny bit of work F(x)\,dx for each, and sum them all — that sum is the integral.

Step 1. Find the antiderivative of 6x^2.

\int 6x^2\,dx = 6 \cdot \frac{x^3}{3} = 2x^3

Why: the power rule gives \int x^2\,dx = \frac{x^3}{3}, and the factor of 6 pulls out in front.

Step 2. Evaluate at the limits.

W = \left[2x^3\right]_0^2 = 2(2)^3 - 2(0)^3 = 2(8) - 0 = 16 \text{ J}

The force $F = 6x^2$ is a parabola. The area under the curve from $x = 0$ to $x = 2$ gives the work done: 16 J. Notice how the force is small near $x = 0$ and large near $x = 2$ — most of the work is done in the last portion of the displacement.

Result: The work done is 16 J.

What this shows: If you had naively used the force at x = 2 (which is 24 N) and multiplied by the distance 2 m, you would get 48 J — three times too large. If you had used the force at x = 1 (which is 6 N) and multiplied by 2 m, you would get 12 J — too small. The integral correctly accounts for the fact that the force starts at zero and grows, giving the precise total of 16 J.

Common confusions

"The integral is just the area under a curve." It is — but only when f(t) \geq 0. When the function dips below the t-axis, the integral counts that area as negative. If a velocity is negative (the object is moving backward), the integral correctly gives a negative displacement. The integral is a signed area.
"I can always use s = vt and W = Fd." Only when v or F is constant. If the quantity varies — and in most real physics problems it does — you must integrate. The simple formulas are special cases of integration where the integrand happens to be constant.
"The constant of integration C is always zero." It is not. In an indefinite integral, C is genuinely unknown until you supply an initial condition — a known value of the function at some point. For instance, if you know the object was at x = 5 m when t = 0, then C = 5. Forgetting C in indefinite integrals is a common source of errors.
"Differentiation and integration always cancel perfectly." They do cancel in the sense that \frac{d}{dt}\int f\,dt = f. But be careful: \int \frac{d}{dt}f\,dt = f + C, not exactly f. Integration recovers the function only up to a constant. Differentiation destroys the constant; integration cannot recover what was destroyed.
"Integration is just anti-differentiation." For the functions you meet in introductory physics, yes — finding an antiderivative is how you evaluate integrals. But the definition of the integral is the limit of a sum (the rectangles getting thinner and thinner). The fact that this limit equals F(b) - F(a) is the fundamental theorem — a deep result, not an obvious one.

If you came here to understand what integration is and how to use it in physics problems, you have everything you need. What follows is for readers who want to see integration at work in more advanced settings: non-constant acceleration, the spring force, and continuous mass distributions.

Work done by a spring — the elastic potential energy

A spring obeys Hooke's law: the restoring force is F = -kx, where k is the spring constant and x is the displacement from the natural (unstretched) length. The negative sign means the force always opposes the displacement — stretch the spring, and it pulls back; compress it, and it pushes out.

How much work do you do when you stretch a spring from x = 0 (natural length) to x = x_0?

W = \int_0^{x_0} kx\,dx

Why: you pull the spring in the +x direction with a force equal to kx (to keep it quasi-static — moving so slowly that the spring is always in equilibrium with your pull). The work done by your hand is \int kx\,dx, not \int (-kx)\,dx, because your force is +kx while the spring's force is -kx.

W = k \cdot \frac{x^2}{2}\Big|_0^{x_0} = \frac{1}{2}kx_0^2

This is the elastic potential energy stored in the spring. The x^2 dependence means doubling the stretch quadruples the stored energy — which is why it takes much more effort to stretch a rubber band the last centimetre than the first.

Displacement under non-constant acceleration

An ISRO sounding rocket experiences a thrust that decreases as fuel burns off. Suppose the acceleration is a(t) = 30 - 2t m/s² for the first 10 seconds (the thrust drops linearly as fuel is consumed). The rocket starts from rest. Find the velocity and displacement at t = 10 s.

Velocity is the integral of acceleration:

v(t) = \int_0^t a(\tau)\,d\tau = \int_0^t (30 - 2\tau)\,d\tau = 30t - t^2

Why: at t = 0, v = 0 (starts from rest), so the constant of integration is zero.

At t = 10 s: v(10) = 300 - 100 = 200 m/s.

Displacement is the integral of velocity:

s(t) = \int_0^t v(\tau)\,d\tau = \int_0^t (30\tau - \tau^2)\,d\tau = 15t^2 - \frac{t^3}{3}

At t = 10 s: s(10) = 1500 - \frac{1000}{3} = 1500 - 333.3 = 1166.7 m.

The rocket climbs about 1.17 km in the first 10 seconds. Notice how you used integration twice — once to go from acceleration to velocity, once from velocity to displacement. Each integration reconstructs one layer of the hierarchy:

a(t) \xrightarrow{\int} v(t) \xrightarrow{\int} x(t)

Centre of mass of a non-uniform rod

A thin rod of length L = 1 m has a linear mass density that increases along its length: \lambda(x) = 2x kg/m, where x is measured from the left end. Find the centre of mass.

The total mass is:

M = \int_0^L \lambda(x)\,dx = \int_0^1 2x\,dx = \left[x^2\right]_0^1 = 1 \text{ kg}

The centre of mass position is:

x_{\text{cm}} = \frac{1}{M}\int_0^L x\,\lambda(x)\,dx = \frac{1}{1}\int_0^1 x \cdot 2x\,dx = \int_0^1 2x^2\,dx = \left[\frac{2x^3}{3}\right]_0^1 = \frac{2}{3} \text{ m}

Why: the rod is denser at the right end (larger x), so the centre of mass is pulled toward the right — it sits at x = 2/3 m, to the right of the geometric centre at x = 1/2 m. If the density were uniform, the centre of mass would be at exactly L/2. The non-uniform density shifts it toward the heavier end.

This is a pattern you will see throughout mechanics, electrostatics, and gravitation: whenever a quantity (mass, charge, force) is distributed continuously, integration replaces summation.

Where this leads next

Work and Energy — the integral W = \int F\,dx is the foundation of the work-energy theorem, connecting force to kinetic energy.
Centre of Mass — integration over continuous mass distributions, and the centre-of-mass frame for multi-body problems.
Uniformly Accelerated Motion — the kinematic equations are special cases of integration when acceleration is constant.
Calculus in Physics — Differentiation — the reverse operation: how derivatives give you rates of change from position, velocity, and energy functions.
Dimensional Analysis — check your integrals by verifying that the units of \int f(t)\,dt are the units of f times the units of t.