Module 1: Mathematical Foundations · Theory · assumes Foundations 7 and 8
Adding up what changes continuously — and discovering that this is differentiation run backwards.
Mahavakya
To add up something that never holds still, cut it into pieces so small that on each one it does.
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Learning objectives
After this reading you should be able to:
- Build a Riemann sum and say what taking the limit achieves.
- State both parts of the Fundamental Theorem and use the second to evaluate integrals.
- Integrate by substitution and by parts, and recognise which situation calls for which.
- Fix the constant of integration from an initial condition, and say what it means physically.
- Compute displacement, work by a variable force, impulse and potential energy by accumulation.
Notation. The sign ∫ is an elongated S, for summa — a sum. The dx is the width of each slice and is part of the notation, not decoration: it tells you which variable you are accumulating over. Bounds are written ∫ab, and F(x) denotes an antiderivative, meaning dF/dx = f(x).
1. Slicing, and the Riemann sum
You have already done this once. In Foundation 7 you found a displacement by reading the area under a velocity–time graph: a rectangle of 12 m plus a triangle of 4 m, giving 16 m. That worked because the graph was made of straight lines, so two pieces were enough to capture it exactly.
Curves do not cooperate. When velocity varies continuously there is no interval over which it holds still, so there is nothing to multiply by anything. The way out is to stop demanding that it hold still for long — only for a moment.
Thinner slices, smaller error. The integral is what the estimates are approaching.
Cut the interval into N strips. Across each strip pretend the quantity is constant, take its value somewhere inside, multiply by the width, and add everything up. That total is a Riemann sum:
SN = Σ f(xi) Δx
It is wrong, and it is wrong in a controlled way: each strip has a flat top where the real curve slopes, so each contributes a small error, and narrower strips make each error smaller. Let the strips shrink and the number of them grow, and the sum settles on a single value. That value is the definite integral:
∫ab f(x) dx = limN→∞ Σ f(xi) Δx
Notice the structure is the same one Foundation 8 used. There the chord never became the tangent; here the sum of rectangles never becomes the area. In both cases a limit is what we mean by arriving. Calculus is built twice over on that single move.
Geometrically the integral is net signed area. Where the curve dips below the axis, the strips have negative height and subtract. That is not a technicality — a velocity–time graph that goes negative describes motion backwards, and the integral correctly returns a displacement reduced by the return journey.
2. The Fundamental Theorem
Everything so far says integration is about accumulation, and nothing in it mentions derivatives. Foundation 8 was about rates, and nothing in it mentioned areas. They look like separate subjects.
They are not, and the connection is not a small one.
First part — accumulating, then differentiating
(d/dx) ∫ax f(t) dt = f(x)
Accumulate up to a moving endpoint, then ask how fast the total is growing, and the answer is the value of the function at that endpoint. Which makes sense once you picture it: the rate at which the area grows is exactly the height of the strip you are currently adding.
Second part — how integrals are actually evaluated
∫ab f(x) dx = F(b) − F(a)
Find any function whose derivative is f, evaluate it at both ends, subtract. No limits, no summing, no strips. An infinite process is replaced by two substitutions and one subtraction.
Contemplation
Sit with how strange this is before learning to use it. There are two natural questions you can ask about a quantity that changes. How fast is it changing right now? And how much has piled up so far? A rate and a total. Nothing about either question suggests the other; you could imagine a universe where they had entirely separate answers, requiring separate mathematics, taught in separate books.
The Fundamental Theorem says the two operations undo each other. Ask the rate question, then the total question, and you are back where you started. This is why the same subject contains both, and why finding areas — a problem the Greeks worked at for centuries with enormous ingenuity and case-by-case cleverness — became, after Newton and Leibniz, a matter of running differentiation in reverse.
It is called fundamental for a reason. Not because it is the hardest thing here, or the most used, but because without it calculus would be two unrelated techniques sharing a name.
3. The toolbox
Integration is harder than differentiation in an unfair way. Differentiation is mechanical: apply the rules and an answer comes out. Integration is a search — you are looking for a function whose derivative is the one in front of you, and sometimes no such function can be written down at all. What follows are the standard results and the two techniques that cover most of what JEE asks.
Standard antiderivatives
∫ xn dx = xn+1/(n+1) + C (n ≠ −1)
∫ (1/x) dx = ln|x| + C
∫ ekx dx = (1/k) ekx + C
∫ cos x dx = sin x + C ∫ sin x dx = −cos x + C
The exception at n = −1 is not an oversight. The power rule would divide by zero there, and the logarithm steps in to fill the gap — which is why work done in an isothermal expansion comes out as a logarithm rather than a power.
Substitution reverses the chain rule. When the integrand contains a function and also its derivative, name the inner function u and the integral collapses.
Worked example — substitution
Evaluate ∫ x cos(x²) dx.
The awkward part is x² inside the cosine — and conveniently, its derivative 2x is sitting outside. Put u = x², so du = 2x dx, giving x dx = du/2.
The integral becomes ½ ∫ cos u du = ½ sin u + C = ½ sin(x²) + C. Differentiate that back if you want to be convinced.
Integration by parts reverses the product rule. Rearranging (uv)′ = uv′ + vu′ and integrating both sides gives:
∫ u dv = uv − ∫ v du
Worked example — by parts
Evaluate ∫ t e−bt dt, the shape that appears when a force decays exponentially and you want the impulse.
Choose u = t, because differentiating it makes it simpler. Then dv = e−bt dt, so v = −(1/b) e−bt.
Applying the formula: −(t/b) e−bt + (1/b) ∫ e−bt dt = −(t/b) e−bt − (1/b²) e−bt + C.
The whole method rests on one choice: which factor becomes u. Pick the one that gets simpler when differentiated, since the remaining integral must be easier than the original or you have gone backwards.
4. The constant of integration is a physical fact
Differentiation destroys information. Any two functions differing by a constant have identical derivatives, so running the process backwards cannot recover which one you started from. That is what the C records:
∫ f(x) dx = F(x) + C
In mechanics this is never an abstraction. Integrate acceleration and you get velocity up to a constant — and the constant is the velocity you started with, which the acceleration alone could never have told you. Integrate again and the new constant is the starting position. Two integrations, two constants, and they are precisely the two initial conditions Newton’s laws require before a trajectory is determined.
Fixing the constant
A body falls with a = −g. Integrating: v(t) = −gt + C.
At t = 0 the body was moving at v0. Substituting: v0 = 0 + C, so C = v0.
Hence v(t) = v0 − gt, which you have known since school — now derived rather than quoted.
Definite integrals need no constant, and the reason is worth seeing rather than accepting. Write F(b) + C − (F(a) + C) and the constants cancel. A definite integral asks how much accumulated between two points, and that difference never depended on where you started counting.
5. Where it appears in mechanics
Rebuilding motion
v(t) = v0 + ∫ a dt, then x(t) = x0 + ∫ v dt. Foundation 8 climbed down this ladder from position to acceleration; integration climbs back up. Two rungs, two constants, one trajectory.
Work by a variable force
W = ∫ F(x) dx. Foundation 5 gave W = F · d and stated its boundary: constant force, straight displacement. This is the general form promised there. By the work–energy theorem the result equals the change in kinetic energy.
Impulse
J = ∫ F(t) dt, equal to the change in momentum. This is how collisions are handled — the force during an impact is violent and unknown in detail, but its accumulated effect is measurable, and the accumulation is all the physics needs.
Potential energy is defined by accumulating work against a conservative force. Take gravity, and let Fr denote the radial component of the force on m, which is negative because the pull is inward:
Fr = −GMm/r²
U(r) = −∫∞r Fr dr′ = ∫∞r (GMm/r′²) dr′
= [−GMm/r′]∞r = −GMm/r
Two things are worth pausing on. The minus sign in front of the integral is the definition of potential energy, and the minus inside Fr is the direction of the force — keep them separate or the signs will fight. And the answer is negative, which is not a mistake: it says that bringing a mass in from infinity releases energy, so a bound object sits below the zero we chose at infinity. Choosing that zero was our decision, as Foundation 7’s vertical shifts showed. Only differences in U mean anything.
Worked example
A probe of mass 500 kg enters a dust cloud at 400 m/s along the x-axis. The drag on it is modelled as F(x) = −bx² with b = 3.0 × 10−4 N/m², for 0 ≤ x ≤ 1000 m. Find the work done by drag, and the exit speed.
Work. W = ∫01000 (−bx²) dx = −b [x³/3]01000 = −(3.0 × 10−4)(109/3) = −1.0 × 105 J, or −100 kJ. Negative, because drag removes energy.
Exit speed. By the work–energy theorem, −100 000 = ½(500)vf² − ½(500)(400)². Dividing by 250: −400 = vf² − 160 000, so vf² = 159 600 and vf ≈ 399.5 m/s.
Is that answer reasonable? Losing half a metre per second over a kilometre sounds like nothing, and the numbers say why: the probe carries ½(500)(400)² = 40 MJ of kinetic energy, and drag took 100 kJ — one part in four hundred. Always compare a change to the quantity it changed; a bare number tells you much less than a ratio.
Boundary. Real drag depends on speed, not on position — roughly v or v² depending on the regime. Writing it as bx² makes the integral straightforward and the problem solvable at this level, but it is a modelling convenience, not a law of nature. A drag that depends on v couples the force to the motion it is changing, and that needs a differential equation rather than a single integral.
Key insight
Displacement, work, impulse and potential energy look like four separate formulas to remember. They are one operation applied to four different pairs of quantities — something varying, accumulated over the thing it varies with. Learn the operation and the four formulas stop needing to be memorised.
Common misconceptions
1. An integral is an area, so it cannot be negative
Geometric area is positive; a definite integral is net signed area. Below the axis, strips subtract. Negative work done by friction is exactly this, and taking the modulus to make it look tidy destroys the physics.
2. ∫v dt gives the distance travelled
It gives displacement. A body that goes out and comes back has a positive stretch and a negative stretch that cancel. For distance you must integrate the speed, ∫|v| dt, which means splitting the interval wherever the velocity changes sign.
3. The + C is a formality
It is the initial condition. Drop it from v = −gt + C and you have claimed the body started from rest — a different physical situation, silently assumed. Marks are lost here every year.
4. Variables can be pulled outside the integral
Only constants can. ∫x² dx is not x∫x dx, because x is the very thing being varied across the slices. The test is simple: does this quantity change as you move along the interval? If yes, it stays inside.
5. Every function can be integrated in closed form
Differentiation always succeeds; integration often does not. e−x² is perfectly well behaved and has no antiderivative expressible in elementary functions. The integral exists — the area is there — but no formula names it. This is why physics leans on numerical methods and approximations, which is where Foundation 10 goes next.
Reflection
- If F is in newtons and x in metres, what are the units of ∫F dx? What does that tell you about what dx contributes to an integral?
- If f is odd, what is ∫−aa f(x) dx, and can you see the answer from the graph before doing any algebra?
- Differentiation throws away a constant; integration cannot recover it without being told. Is there a sense in which the past of a system is less knowable than its future?
Key takeaways
- An integral is the limit of a Riemann sum — slices thin enough that the quantity holds still on each.
- It measures net signed area, so contributions below the axis subtract.
- The Fundamental Theorem says differentiation and integration undo each other; this is not a convenience but the reason they are one subject.
- Substitution reverses the chain rule; integration by parts reverses the product rule.
- The constant of integration is the initial condition, and two integrations need two of them.
- Displacement, work, impulse and potential energy are one operation applied four times.
- Not every function has an elementary antiderivative — which is why approximation matters.
Prerequisites: Foundations 7 and 8 · Reading: 20 min · Practice: 25 min · Difficulty: Intermediate