Evaluating Definite Integrals

The definite integral is one of the most powerful tools in calculus, allowing us to calculate exact accumulated quantities — areas, distances, volumes, and much more. While the idea of accumulation stretches back to ancient Greece, it was the Fundamental Theorem of Calculus (FTC) that turned integration from a geometric puzzle into an algebraic procedure. The FTC connects the two great ideas of calculus — differentiation and integration — revealing them as inverse operations. Before the FTC, computing a definite integral meant taking the limit of an infinite sum of rectangles (a Riemann sum), which is tedious even for simple functions. The FTC provides a shortcut: if you can find an antiderivative F(x) of a function f(x), then the exact value of the definite integral from a to b is simply F(b) − F(a). This transforms an infinite process into a two-step evaluation. In this article you will learn the two parts of the FTC, a clear step-by-step process for evaluating definite integrals, and how to apply it to polynomial, trigonometric, and exponential functions. Understanding this concept is the foundation for every application of integration — from computing net displacement in physics to calculating net change in economics.

Key takeaways

FTC Part 1 – The Antiderivative Connection
Find the Antiderivative (Indefinite Integral)
Apply the Evaluation Notation
Interpret the Result
FTC Part 2 – Differentiating an Integral (Bonus)

FTC Part 1 – The Antiderivative Connection

FTC Part 1 states: if f is continuous on [a, b] and F is any antiderivative of f (meaning F'(x) = f(x)), then the definite integral equals F(b) − F(a). In notation: ∫ from a to b of f(x) dx = F(b) − F(a). This result is remarkable — it tells us that accumulated change equals net change in the antiderivative.

Find the Antiderivative (Indefinite Integral)

The first task is to identify a function F(x) such that F'(x) = f(x). Use standard antiderivative rules: the power rule gives ∫ x^n dx = x^(n+1)/(n+1) + C for n ≠ −1; trigonometric rules give ∫ cos(x) dx = sin(x) + C and ∫ sin(x) dx = −cos(x) + C; and the exponential rule gives ∫ e^x dx = e^x + C. You do NOT need to include the constant C when evaluating a definite integral because it cancels: [F(b) + C] − [F(a) + C] = F(b) − F(a).

Apply the Evaluation Notation

Write F(x) evaluated from a to b using bracket notation: [F(x)] from a to b. This means substitute the upper limit b, then subtract the result of substituting the lower limit a. Be careful with negatives — especially when the lower limit substitution produces a negative number. Always compute F(b) first, then F(a), then subtract.

Interpret the Result

The number you obtain is the net signed area between the curve y = f(x) and the x-axis on [a, b]. If f(x) > 0 everywhere on [a, b], the integral equals the geometric area. If f(x) < 0 on part of the interval, those regions contribute negative values. The total result is the net accumulation — very useful in physics (net displacement), biology (net population growth), and finance (net profit).

FTC Part 2 – Differentiating an Integral (Bonus)

FTC Part 2 states that if g(x) = ∫ from a to x of f(t) dt, then g'(x) = f(x). In other words, differentiating an integral with a variable upper limit recovers the original integrand. This is used in more advanced problems involving accumulation functions, and it completes the picture of differentiation and integration as inverse processes.

The Key Equation

\int_{a}^{b} f(x)\, dx = F(b) - F(a), \quad \text{where } F'(x) = f(x)

\int_{a}^{b} — Definite integral (depends on context): The accumulated signed area under the curve f(x) from x = a to x = b.
f(x) — Integrand (depends on context): The function being integrated; it must be continuous on [a, b] for the FTC to apply directly.
F(x) — Antiderivative (depends on context): Any function satisfying F'(x) = f(x). The constant of integration cancels in definite integrals.
a,\, b — Limits of integration (x-units): The lower limit a and upper limit b define the interval of accumulation. The value a can be greater than b, which reverses the sign of the result.

Worked Example

Problem. Evaluate the definite integral ∫ from 1 to 3 of (2x + 4) dx.

\int_{1}^{3}(2x+4)\,dx = \left[x^2 + 4x\right]_{1}^{3} = 21 - 5 = 16

Answer. 16

Figure

Diagram

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · English

Newton developed calculus (which he called 'the method of fluxions') independently during the 1660s–1670s. He recognized the inverse relationship between differentiation and integration — the core insight of the Fundamental Theorem of Calculus. His work allowed integration to be computed algebraically rather than geometrically for the first time. He and Gottfried Wilhelm Leibniz are jointly credited with the invention of calculus, though they developed it independently. Newton reportedly kept his discovery of calculus secret for nearly 20 years before publishing, which sparked one of history's most bitter priority disputes with Leibniz over who deserved credit for inventing calculus.

Riemann Sums — Riemann sums are the limiting definition of the definite integral; understanding them explains why the FTC is such a powerful shortcut.
Antiderivatives and Indefinite Integrals — The FTC requires finding an antiderivative F(x); mastering antiderivative rules is an essential prerequisite for evaluating definite integrals.
Accumulation Functions and FTC Part 2 — FTC Part 2 shows how differentiating an integral with a variable upper limit recovers the integrand, deepening understanding of the FTC.
Area Between Curves — A direct application of definite integrals; once you can evaluate ∫f(x)dx, computing areas between two curves follows naturally.
U-Substitution — U-substitution is the primary technique for finding antiderivatives of composite functions, extending the range of definite integrals you can evaluate with the FTC.