Integrals and the Fundamental Theorem of Calculus
Integrals and the Fundamental Theorem of Calculus
Definition: The definite integral of a function f from a to b, written ∫ₐᵇ f(x) dx, is the signed area between the curve y = f(x) and the x-axis over that interval; the Fundamental Theorem of Calculus shows it can always be computed as F(b) - F(a), where F is any antiderivative of f.
How It Works
- A Riemann sum approximates the area under a curve by slicing the interval [a, b] into n thin strips of width Δx, building a rectangle on each strip whose height is the function’s value there, and adding up all the rectangles’ areas.
- As the number of rectangles n grows toward infinity, each rectangle’s width Δx shrinks toward zero, and the Riemann sum converges to a single exact value: the definite integral, written ∫ₐᵇ f(x) dx.
- The definite integral measures signed area: regions where f(x) lies above the x-axis contribute positively, regions where it dips below contribute negatively, so a definite integral can come out negative, zero, or positive.
- An antiderivative of f is any function F such that F’(x) = f(x); finding one reverses differentiation, which is why the process is also called integration.
- Because the derivative of any constant is zero, if F(x) is one antiderivative of f(x), then F(x) + C is an antiderivative too, for any constant C. This entire family of functions is the indefinite integral, written ∫f(x) dx = F(x) + C.
- The Fundamental Theorem of Calculus (FTC) is the bridge between these two seemingly unrelated ideas — area as a limit of sums, and antiderivatives as reversed differentiation — and proves they are, remarkably, the same operation viewed two ways.
- The FTC states that if F is any antiderivative of a continuous function f, the definite integral of f from a to b equals F(b) - F(a). A hard geometric limit collapses into two function evaluations and a subtraction.
- In practice, the FTC means a definite integral is almost never computed by literally summing rectangles: find one antiderivative F, evaluate it at the upper and lower bounds, and subtract.
- The power rule for integration says ∫xⁿ dx = xⁿ⁺¹/(n+1) + C for any n ≠ -1: raise the exponent by one, then divide by that new exponent, the mirror image of the power rule for derivatives.
- The restriction n ≠ -1 exists because n+1 would be zero in the denominator when n = -1; ∫x⁻¹ dx is instead ln|x| + C, a genuinely different function that has to be handled as a special case.
- Since indefinite integration produces a whole family of curves differing only by a vertical shift, one extra piece of information — such as a single known point the curve passes through — is enough to pin down the exact value of C.
Illustration
var bIn = document.getElementById(‘int-b’); var bOut = document.getElementById(‘int-b-out’); var resetBtn = document.getElementById(‘int-reset’);
function fmt(n) { return (Math.round(n * 100) / 100).toFixed(2).replace(/.00$/, ‘.0’); }
function pathD(pts) { var d = ‘M’ + pts[0][0].toFixed(1) + ’,’ + pts[0][1].toFixed(1); for (var i = 1; i < pts.length; i++) { d += ’ L’ + pts[i][0].toFixed(1) + ’,’ + pts[i][1].toFixed(1); } return d; }
// f(x) = x² never changes shape, so it only needs sampling once, using // the same step-and-join technique every curve in this domain uses. (function drawCurve() { var pts = []; for (var x = 0; x <= 4.0001; x += 0.05) { var xc = x > 4 ? 4 : x; pts.push(toPx(xc, xc * xc)); } curve.setAttribute(‘d’, pathD(pts)); })();
function update() { var b = parseFloat(bIn.value); bOut.textContent = fmt(b);
// Shaded region: start at the origin, run along the x-axis to (b, 0),
// jump up to the curve at x = b, then trace the curve back down from
// b to 0, which closes the shape exactly back at the origin.
var pts = [toPx(0, 0), toPx(b, 0)];
for (var x = b; x > 0; x -= 0.05) {
pts.push(toPx(x, x * x));
}
pts.push(toPx(0, 0));
areaPath.setAttribute('d', pathD(pts) + ' Z');
var top = toPx(b, b * b);
var bottom = toPx(b, 0);
boundLine.setAttribute('x1', bottom[0]); boundLine.setAttribute('y1', bottom[1]);
boundLine.setAttribute('x2', top[0]); boundLine.setAttribute('y2', top[1]);
point.setAttribute('cx', top[0]); point.setAttribute('cy', top[1]);
var height = b * b;
var exactArea = (b * b * b) / 3; // closed form via the antiderivative x³/3, not a Riemann approximation
bTxt.textContent = 'b = ' + fmt(b);
heightTxt.textContent = 'f(b) = b² = ' + fmt(height);
areaTxt.textContent = 'Area = b³/3 = ' + fmt(exactArea);}
bIn.addEventListener(‘input’, update); resetBtn.addEventListener(‘click’, function () { bIn.value = 2; update(); });
update(); })();
Under the Hood
∫ₐᵇ f(x) dx = lim(n→∞) Σ f(xᵢ)·Δx (definition: the definite integral as a limit of Riemann sums, Δx = (b-a)/n)∫ₐᵇ f(x) dx = F(b) - F(a) (Fundamental Theorem of Calculus, where F is any antiderivative of f, i.e. F' = f)∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (power rule for integration, n ≠ -1)∫ x⁻¹ dx = ln|x| + C (special case, n = -1, where the power rule above breaks down)Worked Example 1: Definite integral via the FTC Given: ∫ from 1 to 3 of x² dx. Step 1: find an antiderivative of x² with the power rule: F(x) = x³/3 (n = 2, so n+1 = 3). Step 2: apply the FTC, ∫₁³ x² dx = F(3) - F(1). Step 3: F(3) = 27/3 = 9, and F(1) = 1/3. Answer: ∫₁³ x² dx = 9 - 1/3 = 26/3 ≈ 8.67.
Worked Example 2: Indefinite integral with the +C Given: find ∫(3x² + 4x) dx. Step 1: integrate term by term with the power rule: ∫3x² dx = 3·(x³/3) = x³. Step 2: ∫4x dx = 4·(x²/2) = 2x². Answer: ∫(3x² + 4x) dx = x³ + 2x² + C (check by differentiating: d/dx[x³ + 2x² + C] = 3x² + 4x, back to the start).
Worked Example 3: Area under a line, two ways Given: the area under f(x) = 2x + 1 from x = 0 to x = 3, a straight line, so the region is a trapezoid sitting on the x-axis. Geometric method — Step 1: the parallel sides are f(0) = 1 and f(3) = 7, with width 3. Geometric method — Step 2: Area = ½(b₁ + b₂)·h = ½(1 + 7)(3) = 12. Integral method — Step 1: an antiderivative of 2x + 1 is F(x) = x² + x (power rule on each term). Integral method — Step 2: ∫₀³ (2x + 1) dx = F(3) - F(0) = (9 + 3) - (0 + 0) = 12. Answer: both methods give Area = 12, confirming the FTC agrees with elementary geometry whenever the curve happens to be a straight line.
History
- Archimedes developed the method of exhaustion around 250 BCE, inscribing and circumscribing polygons with more and more sides to close in on the area of circles, parabolic segments, and spheres — the earliest known systematic ancestor of integration.
- For nearly 1,900 years after Archimedes, finding the area under a new curve stayed a one-off geometric puzzle, solved from scratch with clever, curve-specific tricks rather than any general method.
- Isaac Newton and Gottfried Wilhelm Leibniz independently formulated the Fundamental Theorem of Calculus in the late 17th century, each arriving at it through different reasoning and notation within a few years of the other.
- The Fundamental Theorem is what elevated area-finding from a case-by-case geometric puzzle into a systematic, mechanical procedure: once a function’s antiderivative was known, any definite integral could be evaluated by simple substitution.
- The near-simultaneous discovery led to a bitter priority dispute between British and continental mathematicians that dragged on for decades; today it is Leibniz’s notation, ∫ and dx, that survives in everyday use.
- Bernhard Riemann formalized the rigorous limit-of-sums definition of the definite integral in the 19th century, the version described above and now named the Riemann integral in his honor.
Why It Matters
- Physics computes the work done by a variable force as the integral of force over distance, since work isn’t simply force times distance once the force itself changes along the way.
- Given a velocity function, integrating it over time recovers total distance traveled, exactly reversing the way differentiating a position function gives velocity.
- In probability, the area under a probability density curve between two values gives the probability that a random outcome falls in that range, with the total area under the whole curve always equal to 1.
- Engineers integrate to find volumes, centers of mass, and moments of inertia of irregularly shaped objects, treating them as infinitely many infinitesimally thin slices, exactly like a Riemann sum extended to three dimensions.
- Economists compute consumer and producer surplus as the area between a supply or demand curve and the market price, a direct application of area-under-a-curve to a market.
- Signal processing, pharmacology (drug concentration over time), and epidemiology (cumulative infections) all recover a total quantity by integrating a rate.
Common Pitfalls
- Forgetting the +C on an indefinite integral: without it, the answer is only one member of an entire family of valid antiderivatives, not the general solution.
- Assuming a definite integral always gives a physically positive area. Where the curve dips below the x-axis, that portion is subtracted, since a definite integral gives signed area, not the total physical area between curve and axis.
- Misapplying the power rule to n = -1: ∫x⁻¹ dx is not x⁰/0 (undefined). It is ln|x| + C, a genuinely different function that has to be memorized as a special case.
- Confusing definite and indefinite integral notation: ∫ₐᵇ f(x) dx, with bounds, is a single number, while ∫f(x) dx, with no bounds, is a function. Dropping or adding bounds carelessly changes what kind of answer is even being asked for.
- Wanting total physical (unsigned) area when the curve crosses the x-axis, but forgetting that this requires splitting the integral at each crossing and adding the absolute value of each piece, not just evaluating one integral across the whole interval.
- Treating the Riemann sum as merely an approximation technique, rather than recognizing that the definite integral is defined as its limit; the sum only becomes exact in that limit, never at any finite n.
Comparison
| Definite Integral | Indefinite Integral | |
|---|---|---|
| Notation | ∫ₐᵇ f(x) dx | ∫f(x) dx |
| Produces | A single number (signed area) | A family of functions, F(x) + C |
| Needs bounds a, b? | Yes | No |
| Depends on | The function and the specific interval [a, b] | Only the function f(x) itself |
| Answers the question | “How much signed area lies between the curve and the x-axis from a to b?” | “What function, when differentiated, gives back f(x)?” |
| Computed via | The FTC: F(b) - F(a) | Reversing differentiation rules, like the power rule |
FAQ
Why does area relate to the ANTIderivative? That seems backwards. It only looks backwards because area (a Riemann sum) and antiderivatives (reversed differentiation) seem to come from two different worlds, one geometric and one algebraic. The FTC is precisely the non-obvious proof that they’re secretly the same thing: define an “area so far” function A(x) = ∫ₐˣ f(t) dt, and it turns out A’(x) = f(x), so the area function is itself an antiderivative of f. That is exactly why evaluating any antiderivative at the two endpoints hands back the area.
Do I always need a Riemann sum to compute a definite integral? No, and that is the entire point of the FTC. Riemann sums are how the definite integral is defined, and they matter for building intuition and for functions with no elementary antiderivative, but in practice almost every definite integral is computed by finding an antiderivative and subtracting, never by summing rectangles by hand.
Can the “area” from a definite integral actually be negative? Yes. Because the integral gives signed area, a function that stays entirely below the x-axis on [a, b] produces a negative definite integral, even though no physical area is truly negative. This is intentional: it is what makes integrals add up consistently when a region includes parts above and below the axis.
Example
A tank is being filled at a rate of r(t) = 3t² liters per minute, t minutes after the valve opens. Since total volume added is the integral of the rate, the amount added during the first 2 minutes is ∫₀² 3t² dt = [t³]₀² = 8 - 0 = 8 liters, found instantly via the Fundamental Theorem rather than by summing up infinitely many infinitesimal splashes.