Integration as Accumulation: Not Just Reverse Differentiation

A car accelerates from rest. Its velocity at each moment is known. How far has it travelled after 10 seconds?

If the velocity were constant, the answer is easy: distance = velocity × time.

But velocity is changing. Every second, the car is moving faster. The formula distance = velocity × time uses a single velocity — which one do you pick? The velocity at the start? The end? The middle?

None of them is exactly right. The correct answer requires adding up infinitely many tiny contributions — each one the velocity at a particular instant, multiplied by the vanishingly small time during which that velocity held.

That infinite sum is an integral. And the remarkable fact is: you can compute it exactly, without actually summing infinitely many terms, by reversing differentiation.

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The idea: adding up tiny pieces

Suppose a car's velocity is v(t) = 2t m/s — starting from rest, accelerating uniformly.

How far does it travel between t = 0 and t = 4 seconds?

Divide the 4 seconds into small intervals of width Δt. During each interval, the velocity is approximately constant at its value at the start of that interval. The distance covered in each interval is approximately v(t) · Δt.

Sum all these small distances:

$\text{Total distance} \approx \sum v(t) \cdot \Delta t = \sum 2t \cdot \Delta t$

With Δt = 1 second: 2(0)(1) + 2(1)(1) + 2(2)(1) + 2(3)(1) = 0 + 2 + 4 + 6 = 12 m

With Δt = 0.5 second: the sum has more terms but each is smaller. Try it — the answer approaches 16 m.

As Δt → 0, this sum approaches the exact answer. That limit of the sum is the definite integral:

$\text{Distance} = \int_0^4 v(t)\, dt = \int_0^4 2t\, dt$

The ∫ symbol is an elongated S — for sum. The dt at the end indicates the variable and the width of each infinitesimal piece.

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The fundamental theorem: the shortcut

Evaluating the integral by summing tiny pieces directly is tedious. The fundamental theorem of calculus provides a shortcut:

If F(t) is a function whose derivative is f(t) — that is, dF/dt = f(t) — then:

$\int_a^b f(t)\, dt = F(b) - F(a)$

Find a function whose derivative is the integrand. Evaluate it at the upper limit, subtract its value at the lower limit. Done.

For our velocity problem: we need a function whose derivative is 2t. From the derivative rules in the previous article, d/dt(t²) = 2t. So F(t) = t².

$\int_0^4 2t\, dt = [t^2]_0^4 = 4^2 - 0^2 = 16 \text{ m}$

The exact answer is 16 m. The approximate sum with Δt = 0.5 was already approaching this.

This is the power of the fundamental theorem: the problem of summing infinitely many tiny pieces reduces to finding one function and evaluating it twice.

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Integration rules

Since integration reverses differentiation, every differentiation rule gives an integration rule.

Power rule for integration:

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

Raise the exponent by one, divide by the new exponent. The opposite of differentiation.

The constant C is the constant of integration — it appears because differentiating any constant gives zero, so infinitely many functions have the same derivative. When you reverse the process, you cannot recover which constant was there.

Constant multiple and sum rules work the same as for differentiation:

$\int cf(t)\, dt = c\int f(t)\, dt$

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

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Definite vs indefinite integrals

An indefinite integral has no limits:

$\int 2t\, dt = t^2 + C$

The result is a family of functions — all differing by a constant. This is used when you want the general form of the position or velocity function, and will determine C from an initial condition.

A definite integral has limits:

$\int_0^4 2t\, dt = 16$

The result is a number — the accumulated quantity between two specific moments. The constant C cancels when you subtract F(a) from F(b).

In physics, you use definite integrals to find distances, areas, work done, impulse — whenever you need a total accumulated over an interval. You use indefinite integrals when building the position or velocity function from scratch.

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The kinematic chain in both directions

The previous article showed differentiation going forward:

$x(t) \xrightarrow{\text{differentiate}} v(t) \xrightarrow{\text{differentiate}} a(t)$

Integration goes backward:

$a(t) \xrightarrow{\text{integrate}} v(t) \xrightarrow{\text{integrate}} x(t)$

> > **Backward (integrate):** > Acceleration → Velocity → Position > > Given acceleration as a function of time, integrate once to get velocity (use initial velocity to find C), integrate again to get position (use initial position to find C).

This is the general method for kinematics when acceleration is not constant. The standard equations of motion (v = u + at, s = ut + ½at²) are the special case when acceleration is constant — they are already-integrated results for a = constant.

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The geometric meaning: area under the curve

The definite integral ∫ₐᵇ f(t) dt is the area under the curve y = f(t) between t = a and t = b.

This is not just a geometric curiosity — it is physically meaningful.

Velocity-time graph: the area under the curve is displacement. A car moving with varying velocity — the area under the v-t graph from t = 0 to t = T gives the displacement in that time.

Acceleration-time graph: the area under the curve is change in velocity.

Force-time graph: the area under the curve is impulse — change in momentum.

{/* DIAGRAM: v-t graph with shaded area representing displacement */}

When the curve dips below the t-axis, that area is negative — displacement in the negative direction, decrease in velocity, or negative impulse. Always account for sign when reading area from a graph.

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Determining the constant of integration

When you integrate to find a velocity or position function, the constant C must be determined from initial conditions.

Example: a particle has acceleration a(t) = 6t − 4 m/s². At t = 0, its velocity is 3 m/s and its position is 2 m. Find x(t).

Step 1 — Integrate acceleration to get velocity:

$v(t) = \int (6t - 4)\, dt = 3t^2 - 4t + C_1$

Apply initial condition v(0) = 3:

$3 = 3(0)^2 - 4(0) + C_1 \implies C_1 = 3$

$v(t) = 3t^2 - 4t + 3$

Step 2 — Integrate velocity to get position:

$x(t) = \int (3t^2 - 4t + 3)\, dt = t^3 - 2t^2 + 3t + C_2$

Apply initial condition x(0) = 2:

$2 = 0 - 0 + 0 + C_2 \implies C_2 = 2$

$x(t) = t^3 - 2t^2 + 3t + 2$

Every integration step requires one initial condition to fix the constant. Two integrations require two initial conditions — typically initial velocity and initial position.

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Common mistakes

> > **Using the wrong power rule direction.** Differentiation: bring exponent down, reduce by one. Integration: raise exponent by one, divide by new exponent. Students mix these up under pressure. > > **Forgetting to subtract F(a) in definite integrals.** ∫ₐᵇ f dt = F(b) − F(a), not just F(b). The lower limit contributes. > > **Treating area below the axis as positive.** Area below the t-axis in a v-t graph is negative displacement — the object moved backward. Always respect the sign. > > **Not using initial conditions to find C.** The constant of integration is not optional decoration. It carries physical information — the starting state of the system.

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A worked example

A ball is thrown downward from a height of 80 m with initial velocity 10 m/s. Taking downward as positive and g = 10 m/s², find (a) velocity as a function of time, (b) position as a function of time, (c) when it hits the ground.

Given: a = 10 m/s² (constant), v(0) = 10 m/s, x(0) = 0 (taking the starting point as origin, ground at x = 80 m).

(a) Velocity:

$v(t) = \int 10\, dt = 10t + C_1$

v(0) = 10: C₁ = 10

$v(t) = 10t + 10$

(b) Position:

$x(t) = \int (10t + 10)\, dt = 5t^2 + 10t + C_2$

x(0) = 0: C₂ = 0

$x(t) = 5t^2 + 10t$

(c) Hits ground when x = 80:

$5t^2 + 10t = 80$ $t^2 + 2t - 16 = 0$ $t = \frac{-2 + \sqrt{4 + 64}}{2} = \frac{-2 + \sqrt{68}}{2} \approx \frac{-2 + 8.25}{2} \approx 3.1 \text{ s}$

(Taking the positive root since t > 0.)

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What to remember

An integral is an accumulated sum — infinitely many infinitely small pieces added together.

The fundamental theorem converts this into a practical calculation: find a function whose derivative is the integrand, evaluate at limits, subtract.

Integration reverses differentiation. The kinematic chain runs both ways — differentiate forward, integrate backward.

The area under a curve is the definite integral — and in physics, that area is always something physical: displacement, velocity change, impulse, work.

Initial conditions determine the constant of integration. Always apply them.

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The next article completes the mathematical tools chapter: dimensional analysis — the technique of checking and discovering formulas using only the units involved. It requires no calculus and no vectors, but it is the most underused tool in physics.