The Fundamental Theorem of Calculus and the Newton-Leibniz Formula

The fundamental theorem of calculus links differentiation and integration: if f is continuous on [a, b] and F(x) = ∫_a^x f(t) dt, then F′(x) = f(x), and ∫_a^b f(x) dx = G(b) − G(a) for any antiderivative G of f (the Newton–Leibniz formula). In brief: ∫₁³ x² dx = [x³/3]₁³ = 9 − 1/3 = 26/3, and d/dx ∫₀^x sin(t²) dt = sin(x²). To evaluate a definite integral, find an antiderivative and subtract its values at the endpoints.

Before the theorem, areas were computed by limits of sums, which is laborious. The theorem shows that the integral of a rate of change is the total change, and converts the evaluation of integrals into a question about antiderivatives. It is the central result of calculus. This guide states both parts of the theorem, gives a proof idea, shows how to evaluate definite integrals, how to differentiate integrals with variable limits, how to handle substitution and parts in definite integrals and gives a table of examples.

What is the first part of the theorem?

If f is continuous on [a, b], then the function F(x) = ∫_a^x f(t) dt is differentiable on (a, b) and F′(x) = f(x). The function F accumulates the area under f from a to x, and its rate of change at x is the height f(x). So every continuous function has an antiderivative, namely the area function.

Proof idea: [F(x + h) − F(x)]/h = (1/h) ∫_x^x+h f(t) dt. By the mean value theorem for integrals this equals f(c) for some c between x and x + h, and as h → 0, c → x, so f(c) → f(x) by continuity.

What is the second part (Newton–Leibniz formula)?

If f is continuous on [a, b] and G is any antiderivative of f, then ∫_a^b f(x) dx = G(b) − G(a) = [G(x)]_a^b.

Proof: F(x) = ∫_a^x f and G are both antiderivatives of f, so they differ by a constant: F = G + C. F(a) = 0 gives C = −G(a). Then ∫_a^b f = F(b) = G(b) − G(a). The constant of integration cancels, so it is not written in definite integrals.

How do you evaluate definite integrals?

1. Find an antiderivative G.
2. Evaluate G at the upper limit and at the lower limit.
3. Subtract: G(b) − G(a).

Example 1

∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3 ≈ 8.667. And ∫₀⁴ √x dx = [(2/3)x^3/2]₀⁴ = (2/3)(8) = 16/3.

Example 2

∫₀^π sin x dx = [−cos x]₀^π = −cos π + cos 0 = 1 + 1 = 2. The area under one arch of the sine is 2. And ∫₀^2π sin x dx = 0, since the positive and negative areas cancel.

Example 3

∫₁^e dx/x = [ln x]₁^e = 1. And ∫₀¹ e^2x dx = [e^2x/2]₀¹ = (e² − 1)/2 ≈ 3.195.

Example 4: with substitution

∫₀¹ 2x e^x2 dx: u = x², limits 0 to 1: ∫₀¹ e^u du = e − 1 ≈ 1.718.

Example 5: by parts

∫₀^π/2 x cos x dx = [x sin x]₀^π/2 − ∫₀^π/2 sin x dx = π/2 − [−cos x]₀^π/2 = π/2 − 1 ≈ 0.571.

Example 6: a piecewise function

∫₋₁² |x| dx = ∫₋₁⁰ (−x) dx + ∫₀² x dx = 1/2 + 2 = 5/2. Split at points where the formula changes.

Example 7: the area between curves

The area between y = x and y = x² on [0, 1] is ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.

How do you differentiate integrals with variable limits?

By part 1 and the chain rule: d/dx ∫_a^g(x) f(t) dt = f(g(x)) · g′(x). For two variable limits: d/dx ∫_h(x)^g(x) f(t) dt = f(g(x))g′(x) − f(h(x))h′(x).

Example 8

d/dx ∫₀^x sin(t²) dt = sin(x²). And d/dx ∫₁^x2 ln t dt = ln(x²) · 2x = 4x ln x. And d/dx ∫_x^x2 e^t dt = e^x2 · 2x − e^x.

Example 9: a limit with an integral

lim_x→0 (1/x) ∫₀^x cos(t²) dt = cos(0) = 1, by the definition of the derivative of F at 0.

Example 10: where is the function increasing?

F(x) = ∫₀^x (t² − 4) dt has F′(x) = x² − 4, so F is decreasing on (−2, 2) and increasing outside. Its local extrema are at x = ±2: F(2) = [t³/3 − 4t]₀² = 8/3 − 8 = −16/3.

What does the theorem say about net change?

Since F(b) − F(a) = ∫_a^b F′(x) dx, the integral of a rate of change is the net change. If v(t) is the velocity, ∫_a^b v(t) dt is the displacement, and ∫_a^b |v(t)| dt is the total distance. If the marginal cost is C′(q), the cost of producing from q₁ to q₂ units is ∫ C′(q) dq.

Example 11

A particle has velocity v(t) = t² − 4t + 3 = (t − 1)(t − 3) m/s on [0, 4]. The displacement: ∫₀⁴ v dt = [t³/3 − 2t² + 3t]₀⁴ = 64/3 − 32 + 12 = 4/3 m. The total distance: v > 0 on [0, 1) and (3, 4], v < 0 on (1, 3). ∫₀¹ v = 4/3, ∫₁³ v = −4/3, ∫₃⁴ v = 4/3. The distance is 4/3 + 4/3 + 4/3 = 4 m.

Table of definite integrals

Where does the theorem come from?

Isaac Barrow, Newton’s teacher, proved a geometric version of the first part in the 1660s. Newton and Leibniz, in the 1660s and 1670s, recognised the general principle and used it to calculate areas, and the integral sign and the notation are due to Leibniz. The modern rigorous statement and proof appeared in Cauchy’s work (1823) and in the nineteenth century treatments of Riemann and Darboux.

Where is it used?

The theorem is used in every application of integration: the work done by a force is the integral of force over distance; the total charge is the integral of current; the probability is the integral of a density. Computer algebra systems evaluate definite integrals by finding antiderivatives. When no elementary antiderivative exists, as for e^−x2, numerical methods based on Riemann sums are used.

Which mistakes are common?

Forgetting to subtract the lower limit

The result is G(b) − G(a), not only G(b).

Applying the formula where f is not continuous

∫₋₁¹ dx/x² is not [−1/x]₋₁¹ = −2, since 1/x² has a pole at 0. The integral diverges.

Including +C in the definite integral

The constant cancels in G(b) − G(a). Do not write it.

Mixing the variable of integration with the limit

In ∫₀^x f(t) dt, the variable t is a dummy. Use a different letter from the limit x.