31 Integral calculus

Accumulation

Your car’s speedometer shows 80 km/h. Two minutes later it shows 95 km/h. How far have you travelled? You cannot just multiply speed by time — the speed was changing the whole way. What you need is a way to add up all those instantaneous velocities over the interval. That is integration.

A water tank fills at a rate that varies with time — fast at first, slower as back-pressure builds. You want to know how much water is in the tank after an hour. You know the rate at each moment. You need the total.

These are the same problem. This chapter builds the tool.

31.1 What this chapter helps you do

Symbols to keep handy

These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.

f(x),dx: the integral of f of x with respect to x
_a^b f(x),dx: the definite integral of f from a to b
F(x): an antiderivative of f — F prime equals f
_a^b: F evaluated from a to b — F(b) minus F(a)

Definitions to keep handy

These are the words we keep coming back to. If one feels slippery, come back here and steady it before you push on.

definite integral: A total accumulated amount over an interval, often pictured as area under a curve.
indefinite integral / antiderivative: A family of functions whose derivative is the given function (plus a constant).
Riemann sum: An approximation to total accumulation by adding many small rate x width slices.
Fundamental Theorem of Calculus: The bridge that says accumulation (definite integrals) and antiderivatives are two views of the same object.

This chapter is about turning rates into totals and thin slices into whole quantities. We begin with the geometric picture of accumulation and the Riemann-sum idea that builds area out of many small contributions. Then we use the Fundamental Theorem of Calculus to connect that picture to antiderivatives. After that, the main techniques become usable: integration rules, substitution, and area calculations. The goal is not only to compute an integral, but to understand what has accumulated and why the integral represents that total.

Watch for this

Students often blur two related but different ideas:

the definite integral, which gives a total over an interval
the indefinite integral, which gives a family of antiderivatives

The Fundamental Theorem connects them, but they are not the same object.

31.2 What the notation is saying

How to read integral notation

Symbol: \int_a^b f(x)\,dx
Reads as: “the integral of f(x) with respect to x, from a to b”
Means: the total accumulated amount (often area) built from values of f between a and b
Use when: you want a total over an interval (distance from velocity, mass from density, probability from a PDF)
Common misread: the result is a number, not a function
Symbol: \int f(x)\,dx
Reads as: “the integral of f(x) with respect to x”
Means: a family of functions whose derivative is f(x) (an antiderivative), written with a +C
Use when: you are recovering a function from its rate of change
Common misread: forgetting the constant of integration

The integral sign \int is an elongated S — it stands for sum. The dx at the end names the variable we are integrating with respect to. The expression

\int_a^b f(x)\,dx

reads: “the integral of f(x) with respect to x, from a to b.” Picture many thin rectangles, each with height f(x) and width \Delta x — we’ll shrink \Delta x to zero in the limit, so the strips get thinner and thinner until they fit exactly under the curve. The integral adds up the areas of all of them. The result is the total area between the curve y = f(x) and the x-axis, between x = a and x = b. This means a definite integral is a limit of many small contributions added together.

That is the geometric picture. The computational meaning follows from Riemann sums.

31.2.1 Riemann sums

Partition the interval [a, b] into n equal subintervals, each of width

\Delta x = \frac{b - a}{n}

Pick a sample point x_i in each subinterval — say the left endpoint, the right endpoint, or the midpoint. At each sample point, the rectangle has height f(x_i) and width \Delta x. Sum all n rectangles:

\text{Area} \approx \sum_{i=1}^{n} f(x_i)\,\Delta x

This is a Riemann sum — you saw the idea of summing up many small terms in Volume 4, and the same question of what happens as the number of terms grows without bound applies here too. The definite integral is the limit of this sum as the rectangles get infinitely thin: This means the approximation becomes exact by refining the partition until the rectangles capture the curve arbitrarily well.

\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x

For well-behaved functions, this limit exists and is a single, precise number. This means integration is where the ideas of area, summation, and limits come together.

Interactive: Riemann sum builder. Adjust the number of rectangles and choose a sum type to see how the approximation converges to the exact value.

Code

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  const innerW = W - margin.left - margin.right;
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  const a = 0, b = 3;
  const f = x => x * x + 1;
  const exactArea = 12; // ∫₀³ (x²+1) dx = [x³/3 + x]₀³ = 9 + 3 = 12

  const n = Math.max(1, Math.round(riemannN));
  const type = riemannType;
  const dx = (b - a) / n;

  // Compute approximate area
  let approxArea = 0;
  for (let i = 0; i < n; i++) {
    let xSample;
    if (type === "Left")     xSample = a + i * dx;
    else if (type === "Right") xSample = a + (i + 1) * dx;
    else                     xSample = a + (i + 0.5) * dx; // Midpoint
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    .text("f(x) = x² + 1");

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  const error = Math.abs(approxArea - exactArea);

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31.3 The Fundamental Theorem of Calculus

The Riemann sum definition is conceptually important but computationally brutal — you would need to take a limit every time you evaluate an integral. The Fundamental Theorem of Calculus (FTC) is the result that makes integration tractable. It has two parts.

31.3.1 Part 1 — Differentiation undoes integration

Define a function F by accumulating area from a fixed point a:

F(x) = \int_a^x f(t)\,dt

Part 1 says: differentiate F and you get f back.

F'(x) = f(x)

In words: the rate at which accumulated area grows at x is exactly f(x). If f(x) is large and positive, area accumulates quickly; if f(x) is zero, the accumulated area is momentarily flat; if f(x) is negative, the running total is decreasing. The integral of a rate is an accumulation function, and its derivative returns the original rate. This means accumulation and rate are paired concepts: one builds totals, the other reads off how fast those totals are changing.

This is the theoretical bridge between differentiation and integration. They are inverse operations — like squaring and taking a square root — but with a subtlety: integration produces a family of antiderivatives differing by a constant. We’ll see why immediately.

31.3.2 Part 2 — Evaluating definite integrals

Part 2 is the computational workhorse. If F is any antiderivative of f (meaning F'(x) = f(x)), then:

\int_a^b f(x)\,dx = F(b) - F(a)

The shorthand \bigl[F(x)\bigr]_a^b means evaluate F at the top limit, subtract F at the bottom limit:

\bigl[F(x)\bigr]_a^b = F(b) - F(a)

In words: to evaluate a definite integral, find any antiderivative of the integrand, plug in the upper and lower limits, and subtract. You never need to sum up rectangles explicitly. This means the Fundamental Theorem turns a hard limiting process into an antiderivative calculation.

A quick example. Evaluate \displaystyle\int_0^3 2x\,dx.

An antiderivative of 2x is x^2 (because the derivative of x^2 is 2x). Apply Part 2:

\int_0^3 2x\,dx = \bigl[x^2\bigr]_0^3 = 3^2 - 0^2 = 9

The area under the line y = 2x between x = 0 and x = 3 is exactly 9. This means the theorem recovers a geometric quantity without estimating every strip separately.

31.4 Integration rules

Building antiderivatives is the reverse of differentiation. Where differentiation had rules for breaking functions apart, integration has rules for putting them back together. These five cover most of first-year calculus.

1. Power rule (reversed)

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

Check it: differentiate \dfrac{x^{n+1}}{n+1} using the power rule and you recover x^n. This means antidifferentiation is usually “undo the derivative rule” rather than invent a completely new method.

2. Exponential

\int e^x\,dx = e^x + C

The exponential function is its own derivative, so it is also its own antiderivative.

3. Reciprocal

\int \frac{1}{x}\,dx = \ln|x| + C

The absolute value handles negative x. The derivative of \ln|x| is 1/x for all x \neq 0. This means the reciprocal function is the special case where the ordinary power rule breaks and the logarithm takes over.

4. Sine

\int \sin(x)\,dx = -\cos(x) + C

Check: the derivative of -\cos(x) is \sin(x).

5. Cosine

\int \cos(x)\,dx = \sin(x) + C

Check: the derivative of \sin(x) is \cos(x).

Why the constant of integration?

Every antiderivative of a function f differs from every other by a constant. If F'(x) = f(x), then (F(x) + 7)' = f(x) too. And (F(x) - 3)' = f(x). They all work.

Writing +C after an indefinite integral captures the entire family of valid antiderivatives at once. The +C is not optional decoration — omitting it loses real solutions. For example, if you are solving a differential equation and need to apply an initial condition, the value of C is determined by that condition. Dropping C before you apply the condition makes the problem unsolvable. This means an indefinite integral does not produce one function, but every function with the same derivative.

For definite integrals the constant cancels. When you compute F(b) - F(a), the +C appears in both terms and vanishes: (F(b) + C) - (F(a) + C) = F(b) - F(a). The area between two limits is uniquely defined.

31.5 Integration by substitution

Substitution is the chain rule running in reverse. Recall that the chain rule for differentiation says:

\frac{d}{dx}\bigl[f(g(x))\bigr] = f'(g(x)) \cdot g'(x)

If you see a product where one factor looks like the derivative of what is inside a function, you can reverse this process. This means substitution is the integration version of recognising an inner function and its derivative as a pair.

The method, step by step:

Identify a substitution u = g(x) — usually the “inside” function.
Differentiate to find du = g'(x)\,dx, then solve for dx.
Rewrite the entire integrand in terms of u and du.
Integrate with respect to u.
Substitute back to express the answer in terms of x.

Worked example: Evaluate \displaystyle\int 2x\cos(x^2)\,dx.

Step 1. The inside function is x^2. Let u = x^2.

Step 2. Differentiate: \dfrac{du}{dx} = 2x, so du = 2x\,dx.

Step 3. Notice that 2x\,dx appears in the integrand. Substitute:

\int 2x\cos(x^2)\,dx = \int \cos(u)\,du

Step 4. Integrate using the cosine rule:

\int \cos(u)\,du = \sin(u) + C

Step 5. Substitute back u = x^2:

\int 2x\cos(x^2)\,dx = \sin(x^2) + C

Check: Differentiate \sin(x^2) using the chain rule: \dfrac{d}{dx}\bigl[\sin(x^2)\bigr] = \cos(x^2) \cdot 2x = 2x\cos(x^2). Correct.

The substitution worked because the factor 2x outside the cosine was exactly the derivative of x^2 inside it. When that match exists, substitution unravels the integral cleanly. This means a good substitution turns a complicated-looking integrand into a standard form you already know how to integrate.

Interactive: Area under a curve with sign. Move the sliders to set the limits a and b. Positive area (above the axis) is shaded blue; negative area (below the axis) is shaded red. The integral value updates live.

Code

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  const xMin = -Math.PI, xMax = 2 * Math.PI;
  const f = x => Math.sin(x);

  // Clamp and order a, b
  const aRaw = Math.max(xMin, Math.min(xMax, sinA));
  const bRaw = Math.max(xMin, Math.min(xMax, sinB));
  const lo = Math.min(aRaw, bRaw);
  const hi = Math.max(aRaw, bRaw);

  // Exact integral: ∫sin(x)dx = -cos(x), so F(b) - F(a) = -cos(b) + cos(a)
  const integralValue = -Math.cos(bRaw) + Math.cos(aRaw);

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31.6 Worked examples

31.6.1 Example 1 — Power rule and the FTC (computing / probability)

Problem. Find the indefinite integral \displaystyle\int (3x^2 - 4x + 1)\,dx, then evaluate the definite integral from x = 0 to x = 2.

A probability density function for a random variable on [0, 2] might have this polynomial form; the definite integral would give the total probability mass over that interval.

Indefinite integral. Apply the power rule term by term:

\int (3x^2 - 4x + 1)\,dx = \frac{3x^3}{3} - \frac{4x^2}{2} + x + C = x^3 - 2x^2 + x + C

Definite integral. Apply FTC Part 2 with F(x) = x^3 - 2x^2 + x:

\int_0^2 (3x^2 - 4x + 1)\,dx = \bigl[x^3 - 2x^2 + x\bigr]_0^2

Evaluate at the upper limit:

F(2) = 2^3 - 2(2^2) + 2 = 8 - 8 + 2 = 2

Evaluate at the lower limit:

F(0) = 0 - 0 + 0 = 0

The definite integral equals F(2) - F(0) = 2 - 0 = 2. This means the same antiderivative can answer both the indefinite question and the definite one.

31.6.2 Example 2 — FTC Part 2 (science / measurement)

Problem. A radioactive tracer is administered to a patient and its concentration in the bloodstream decays in a way modelled by c(x) = x^{-1/2} (in appropriate units, for x > 0). Compute the total exposure — the area under the concentration curve — between x = 1 and x = 4.

\int_1^4 \frac{1}{\sqrt{x}}\,dx = \int_1^4 x^{-1/2}\,dx

Apply the power rule with n = -\tfrac{1}{2}:

\int x^{-1/2}\,dx = \frac{x^{-1/2 + 1}}{-1/2 + 1} = \frac{x^{1/2}}{1/2} = 2\sqrt{x} + C

Apply the limits:

\bigl[2\sqrt{x}\bigr]_1^4 = 2\sqrt{4} - 2\sqrt{1} = 2(2) - 2(1) = 4 - 2 = 2

Total exposure over the interval is 2 (in the model’s units). This means a definite integral turns a changing concentration into one total accumulated exposure.

31.6.3 Example 3 — Substitution (engineering / signals)

Problem. In signal processing, an engineer needs to evaluate \displaystyle\int x\,e^{x^2}\,dx as part of computing the energy in a modulated waveform.

Let u = x^2, so du = 2x\,dx, which gives x\,dx = \dfrac{du}{2}.

\int x\,e^{x^2}\,dx = \int e^u \cdot \frac{du}{2} = \frac{1}{2}\int e^u\,du = \frac{1}{2}e^u + C

Substitute back u = x^2:

\int x\,e^{x^2}\,dx = \frac{1}{2}e^{x^2} + C

Check: Differentiate \tfrac{1}{2}e^{x^2} with the chain rule: \dfrac{d}{dx}\bigl[\tfrac{1}{2}e^{x^2}\bigr] = \tfrac{1}{2} \cdot e^{x^2} \cdot 2x = x\,e^{x^2}. Correct. This means the factor outside the exponential is exactly what makes the substitution succeed.

31.6.4 Example 3b — Area between two curves

Problem. Find the area enclosed between f(x) = x + 3 and g(x) = x^2 + 1.

Step 1 — Find intersections:

x + 3 = x^2 + 1 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0

Intersections at x = -1 and x = 2.

Step 2 — Identify which is on top: At x = 0, f(0) = 3 and g(0) = 1, so f > g on [-1, 2].

Step 3 — Integrate:

\int_{-1}^{2} \bigl[(x+3) - (x^2+1)\bigr]\,dx = \int_{-1}^{2} (2 + x - x^2)\,dx

= \left[2x + \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{2}

= \left(4 + 2 - \frac{8}{3}\right) - \left(-2 + \frac{1}{2} + \frac{1}{3}\right)

= \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}

The enclosed area is \dfrac{9}{2}. This means area between curves is found by integrating “top minus bottom” over the interval where that ordering stays true.

31.6.5 Example 4 — Area between two curves (business / economic surplus)

Problem. In a competitive market, the demand curve is p_D(q) = 8 - q^2 and the supply curve is p_S(q) = 2q. Find the consumer surplus — the area between the demand curve and the equilibrium price, from zero to the equilibrium quantity.

Step 1. Find the equilibrium. Set p_D = p_S:

8 - q^2 = 2q \implies q^2 + 2q - 8 = 0 \implies (q+4)(q-2) = 0

Equilibrium is at q^* = 2 (taking the positive root), equilibrium price p^* = 4.

Step 2. Consumer surplus is the area between the demand curve and the equilibrium price from q = 0 to q = 2:

\text{CS} = \int_0^2 \bigl[(8 - q^2) - 4\bigr]\,dq = \int_0^2 (4 - q^2)\,dq

Step 3. Evaluate:

\int_0^2 (4 - q^2)\,dq = \bigl[4q - \tfrac{q^3}{3}\bigr]_0^2 = \left(8 - \frac{8}{3}\right) - 0 = \frac{16}{3} \approx 5.33

Consumer surplus is \tfrac{16}{3} (in the model’s currency units per unit quantity). This is the aggregate benefit consumers receive by paying the equilibrium price rather than their maximum willingness to pay. This means an economic surplus is literally an area interpretation of value above or below a market price.

31.7 Where this goes

This chapter built integration from limits, then showed how the Fundamental Theorem connects accumulation back to derivatives. The next uses of these ideas are broader settings, not different principles.

Multivariable calculus (Vol 7). A single integral accumulates a function over an interval — a one-dimensional slice. In two and three dimensions you can integrate over areas, volumes, and surfaces — the same idea of accumulation, applied to more variables. Those tools (Vol 7) are the language of electromagnetism, fluid dynamics, and heat flow.

Ordinary differential equations (Vol 7). A differential equation relates a function to its own derivatives. The simplest case — separable equations — is solved by moving all the y-terms to one side and all the x-terms to the other, then integrating both sides. Every technique in this chapter applies directly. The ability to integrate is the ability to solve differential equations.

31.8 Applications

Distance from velocity (physics). Integrate v(t) over a time interval to recover displacement. If v(t) = 9.8t (free fall from rest), then \int_0^T 9.8t\,dt = 4.9T^2 — the familiar kinematic formula, derived by integration.
Work by a variable force (mechanics). Work is W = \int_a^b F(x)\,dx when force F varies with position x. Compressing a spring, pumping water from a tank — both are definite integrals.
Consumer and producer surplus (economics). The area between a demand (or supply) curve and the equilibrium price is a definite integral. It measures welfare.
Probability from a density function (statistics). If f(x) is a probability density function, then P(a \leq X \leq b) = \int_a^b f(x)\,dx. The cumulative distribution function is F(x) = \int_{-\infty}^x f(t)\,dt — literally the FTC in action.
Average value of a function. The average value of f over [a,b] is \dfrac{1}{b-a}\int_a^b f(x)\,dx — the integral normalised by the length of the interval.

31.9 What you can do now

You can now read an integral as accumulation rather than as a decorative symbol. You should be able to move between area, total change, and antiderivative language; use the Fundamental Theorem to evaluate definite integrals; and recognise when substitution is the right structural move. After computing an integral, the key habit is interpretive: say what total quantity it represents.

31.10 Exercises

Code

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Exercise 1. Find the indefinite integral:

\int \left(5x^4 - 3x^2 + 2\right)dx

Include the constant of integration in your answer.

Code

makeStepperHTML(1, [
  { op: "Apply power rule term by term",       eq: "\\int 5x^4\\,dx - \\int 3x^2\\,dx + \\int 2\\,dx", note: "Integration is linear: split across addition and pull out constants." },
  { op: "Integrate each term",                 eq: "\\frac{5x^5}{5} - \\frac{3x^3}{3} + 2x + C" },
  { op: "Simplify coefficients",               eq: "x^5 - x^3 + 2x + C" },
  { op: "Check by differentiating",            eq: "\\frac{d}{dx}\\left(x^5 - x^3 + 2x + C\\right) = 5x^4 - 3x^2 + 2 \\checkmark" }
])

Exercise 2. Evaluate the definite integral using the Fundamental Theorem of Calculus Part 2:

\int_1^9 \frac{3}{\sqrt{x}}\,dx

Code

makeStepperHTML(2, [
  { op: "Rewrite as a power",                  eq: "\\int_1^9 3x^{-1/2}\\,dx", note: "1/√x = x^{-1/2} — apply the power rule." },
  { op: "Find antiderivative",                 eq: "3 \\cdot \\frac{x^{1/2}}{1/2} = 6x^{1/2} = 6\\sqrt{x}" },
  { op: "Apply FTC Part 2",                    eq: "\\left[6\\sqrt{x}\\right]_1^9 = 6\\sqrt{9} - 6\\sqrt{1}" },
  { op: "Evaluate at limits",                  eq: "6(3) - 6(1) = 18 - 6 = 12" },
  { op: "Answer",                              eq: "\\int_1^9 \\frac{3}{\\sqrt{x}}\\,dx = 12" }
])

Exercise 3. Use u-substitution to evaluate:

\int 3x^2\sin(x^3)\,dx

Code

makeStepperHTML(3, [
  { op: "Choose substitution",                 eq: "u = x^3", note: "The inside function of sin is x³. Its derivative 3x² appears outside." },
  { op: "Differentiate",                       eq: "\\frac{du}{dx} = 3x^2 \\implies du = 3x^2\\,dx" },
  { op: "Substitute into integral",            eq: "\\int \\sin(u)\\,du" },
  { op: "Integrate",                           eq: "-\\cos(u) + C" },
  { op: "Substitute back",                     eq: "-\\cos(x^3) + C" },
  { op: "Check by differentiating",            eq: "\\frac{d}{dx}\\left[-\\cos(x^3)\\right] = \\sin(x^3) \\cdot 3x^2 = 3x^2\\sin(x^3) \\checkmark" }
])

Exercise 4. Find the area under the curve f(x) = x^2 + 2x between x = 0 and x = 3. Sketch the region (or describe it in words) before computing.

Code

makeStepperHTML(4, [
  { op: "Set up the definite integral",        eq: "\\int_0^3 (x^2 + 2x)\\,dx", note: "f(x) ≥ 0 on [0,3], so the integral gives the geometric area." },
  { op: "Find the antiderivative",             eq: "F(x) = \\frac{x^3}{3} + x^2" },
  { op: "Apply FTC Part 2",                    eq: "\\left[\\frac{x^3}{3} + x^2\\right]_0^3 = F(3) - F(0)" },
  { op: "Evaluate at upper limit",             eq: "F(3) = \\frac{27}{3} + 9 = 9 + 9 = 18" },
  { op: "Evaluate at lower limit",             eq: "F(0) = 0 + 0 = 0" },
  { op: "Compute the area",                    eq: "18 - 0 = 18" }
])

Exercise 5. Find the area enclosed between f(x) = 4 - x^2 and g(x) = x + 2.

Hint: find the intersection points first. The area between two curves is \int_a^b \bigl[f(x) - g(x)\bigr]dx when f(x) \geq g(x) on [a, b].

Code

makeStepperHTML(5, [
  { op: "Find intersections",                  eq: "4 - x^2 = x + 2 \\implies x^2 + x - 2 = 0 \\implies (x+2)(x-1) = 0", note: "Intersections at x = −2 and x = 1." },
  { op: "Check which is on top",               eq: "\\text{At } x=0:\\; f(0)=4,\\; g(0)=2 \\implies f > g \\text{ on } [-2,1]" },
  { op: "Set up integral of difference",       eq: "\\int_{-2}^{1} \\bigl[(4-x^2) - (x+2)\\bigr]\\,dx = \\int_{-2}^{1} (2 - x - x^2)\\,dx" },
  { op: "Find antiderivative",                 eq: "F(x) = 2x - \\frac{x^2}{2} - \\frac{x^3}{3}" },
  { op: "Evaluate at upper limit",             eq: "F(1) = 2 - \\tfrac{1}{2} - \\tfrac{1}{3} = \\tfrac{12-3-2}{6} = \\tfrac{7}{6}" },
  { op: "Evaluate at lower limit",             eq: "F(-2) = -4 - 2 + \\tfrac{8}{3} = \\tfrac{-18+8}{3} = -\\tfrac{10}{3}" },
  { op: "Compute the area",                    eq: "\\tfrac{7}{6} - \\left(-\\tfrac{10}{3}\\right) = \\tfrac{7}{6} + \\tfrac{20}{6} = \\tfrac{27}{6} = \\tfrac{9}{2}" }
])

Exercise 6 — Accumulation in context. A solar panel generates electricity at a rate of P(t) = 4\sin\!\left(\dfrac{\pi t}{12}\right) kilowatts, where t is hours after sunrise (0 \leq t \leq 12).

Find the total energy (in kilowatt-hours) generated over the full 12-hour day: \displaystyle\int_0^{12} P(t)\,dt.
Interpret the result: what does the definite integral represent physically?

(Your answer will be in terms of \pi — that is expected.)

Code

makeStepperHTML(6, [
  { op: "Set up the integral",                 eq: "\\int_0^{12} 4\\sin\\!\\left(\\frac{\\pi t}{12}\\right)dt" },
  { op: "Choose substitution",                 eq: "u = \\frac{\\pi t}{12} \\implies du = \\frac{\\pi}{12}\\,dt \\implies dt = \\frac{12}{\\pi}\\,du", note: "u-substitution clears the coefficient inside sin." },
  { op: "Change limits",                       eq: "t=0 \\Rightarrow u=0;\\quad t=12 \\Rightarrow u=\\pi" },
  { op: "Substitute and simplify",             eq: "4 \\cdot \\frac{12}{\\pi} \\int_0^{\\pi} \\sin(u)\\,du = \\frac{48}{\\pi}\\int_0^{\\pi} \\sin(u)\\,du" },
  { op: "Integrate sin",                       eq: "\\frac{48}{\\pi}\\left[-\\cos(u)\\right]_0^{\\pi}" },
  { op: "Evaluate at limits",                  eq: "\\frac{48}{\\pi}\\left(-\\cos(\\pi) + \\cos(0)\\right) = \\frac{48}{\\pi}(1+1) = \\frac{96}{\\pi}" },
  { op: "Interpret the result",                eq: "\\text{Total energy} = \\frac{96}{\\pi} \\approx 30.6 \\text{ kWh}", note: "The definite integral sums the instantaneous power output over the full day — the accumulated energy." }
])