Difficulty: Intermediate | Prerequisites: Familiarity with antiderivatives, basic integration rules, and the Fundamental Theorem of Calculus Part 1.
The Fundamental Theorem of Calculus (FTC) is the bridge between differential and integral calculus. Part 2 gives you the practical tool for evaluating definite integrals: instead of computing limits of Riemann sums, you find an antiderivative and subtract. This is the result that makes most of the integral problems in your course actually solvable by hand. You should already be comfortable finding antiderivatives of basic functions (polynomials, exponentials, 1/x) before working through this material.
If F is any antiderivative of f, then the definite integral from a to b equals F(b) minus F(a). This lets you evaluate definite integrals without Riemann sums. Find an antiderivative, plug in the endpoints, subtract.
Definite integral
The integral of f(x) from a to b, written as the integral from a to b of f(x) dx. It represents the net signed area between the curve y = f(x) and the x-axis over the interval [a, b].
In simple terms, this means: the total accumulated quantity between two endpoints.
Antiderivative (indefinite integral)
A function F(x) whose derivative equals f(x), that is, F'(x) = f(x). There are infinitely many antiderivatives for a given function, each differing by a constant C.
Think of it as: the reverse of differentiation. If differentiating F gives you f, then F is an antiderivative of f.
Fundamental Theorem of Calculus, Part 2 (FTC Part 2, evaluation theorem)
States that if f is continuous on [a, b] and F is any antiderivative of f, then the integral from a to b of f(x) dx equals F(b) minus F(a).
Think of it as: the shortcut that lets you skip Riemann sums entirely and just plug in the endpoints.
Evaluation bar notation
The vertical bar notation F(x) evaluated from a to b, written F(x)|_a^b, means F(b) minus F(a). A compact way to show the substitution step.
In simple terms, this means: "plug in the top number, then subtract what you get when you plug in the bottom number."
Net displacement
When v(t) is velocity and s(t) is position, the integral from a to b of v(t) dt gives s(b) minus s(a), the net change in position over that time interval.
Think of it as: how far you end up from where you started, accounting for direction.
If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then:
The integral from a to b of f(x) dx = F(b) – F(a)
The constant of integration C cancels out (it appears in both F(b) and F(a)), so you can use any antiderivative you like.
If v(t) is the velocity of an object and s(t) is its position at time t, then v(t) = s'(t).
s is an antiderivative of v, so FTC Part 2 gives:
The integral from a to b of v(t) dt = s(b) – s(a)
This means: integrating velocity over a time interval gives the net change in position (displacement) over that interval.
Evaluate the integral from 1 to 3 of e^x dx.
The antiderivative of e^x is e^x (plus C, but we drop C for definite integrals).
F(3) = e^3
F(1) = e^1 = e
Result: e^3 – e, which is approximately 17.37.
Find the area under y = x^2 from x = 0 to x = 1.
Set up: A = integral from 0 to 1 of x^2 dx.
The antiderivative of x^2 is (1/3)x^3.
Evaluate: (1/3)(1)^3 – (1/3)(0)^3 = 1/3 – 0 = 1/3.
The area under the parabola on [0, 1] is 1/3.
Evaluate the integral from 3 to 6 of (1/x) dx.
Rewrite dx/x as (1/x) dx.
The antiderivative of 1/x is ln|x|.
Evaluate: ln|6| – ln|3| = ln(6/3) = ln 2, which is approximately 0.693.
Note the use of the logarithm quotient rule: ln(a) – ln(b) = ln(a/b).
\int_a^b f(x)\,dx = F(b) - F(a)where F is any antiderivative of f (i.e. F'(x) = f(x)).
Velocity-position form:
\int_a^b v(t)\,dt = s(b) - s(a)Key antiderivatives used in the examples:
Antiderivative of e^x is e^x
Antiderivative of x^n is x^(n+1)/(n+1), provided n is not equal to -1
Antiderivative of 1/x is ln|x|
The FTC Part 2 is behind any calculation where you need the total accumulated effect of a rate. Engineers use it to find the total distance an object travels from its velocity profile, the total charge that flows through a circuit from a current function, or the total energy transferred from a power curve. In economics, integrating a marginal cost function over a production range gives total cost, which is a direct application of this theorem.
Students often forget that the constant of integration C cancels out in a definite integral. You do not need to include +C when evaluating F(b) – F(a).
Students sometimes confuse the integral from a to b of f(x) dx with finding "the antiderivative." The definite integral is a number, not a function. The indefinite integral (no limits) is a family of functions.
When the integrand is 1/x, some students write the antiderivative as ln(x) instead of ln|x|. The absolute value matters because ln is only defined for positive arguments, and x could be negative on the interval.
A common error is subtracting in the wrong order: computing F(a) – F(b) instead of F(b) – F(a). The upper limit always goes first.
⚠️ FTC Part 2 is tested directly on nearly every integral calculus exam. Expect to evaluate definite integrals of polynomial, exponential, and rational functions.
⚠️ The velocity/position interpretation is a favourite source of application problems. Be ready to interpret a definite integral of v(t) as net displacement.
⚠️ Know your log rules. Problems involving the integral of 1/x often test whether you can simplify ln(a) – ln(b) into ln(a/b).
⚠️ Watch the order of evaluation: F(b) – F(a), never the reverse. One mark lost for a sign error is the most common way students drop points on these questions.
True or false: When evaluating a definite integral using FTC Part 2, you must include the constant C in your antiderivative.
False. The constant cancels when you subtract F(a) from F(b).
Fill in the blank: The integral from 0 to 2 of 3x^2 dx = ___.
Answer: 8. The antiderivative is x^3, and (2)^3 – (0)^3 = 8.
True or false: The integral from a to b of v(t) dt gives the total distance travelled.
False. It gives the net displacement (signed). Total distance requires integrating |v(t)|.
Fill in the blank: ln(6) – ln(3) = ln() = ln()
Answer: ln(6/3) = ln(2).
Q: Evaluate the integral from 2 to 5 of (2x + 1) dx.
A: The antiderivative is x^2 + x. Evaluate: (25 + 5) – (4 + 2) = 30 – 6 = 24.
Q: Find the area under y = e^x from x = 0 to x = ln(3).
A: The antiderivative of e^x is e^x. Evaluate: e^(ln 3) – e^0 = 3 – 1 = 2.
Q: Evaluate the integral from 1 to 4 of (1/x) dx.
A: The antiderivative of 1/x is ln|x|. Evaluate: ln(4) – ln(1) = ln(4) – 0 = ln(4), which equals 2 ln(2), approximately 1.386.
Q: A car has velocity v(t) = 3t^2 metres per second. What is its net displacement from t = 1 to t = 3?
A: The integral from 1 to 3 of 3t^2 dt. Antiderivative is t^3. Evaluate: 27 – 1 = 26 metres.
Q: Evaluate the integral from 0 to pi of sin(x) dx.
A: The antiderivative of sin(x) is –cos(x). Evaluate: –cos(pi) – (–cos(0)) = –(–1) – (–1) = 1 + 1 = 2.
FTC Part 2 pairs directly with FTC Part 1, which defines a function as an integral with a variable upper limit and shows that its derivative recovers the integrand. Together the two parts establish the inverse relationship between differentiation and integration.
This theorem is the foundation for every integration technique you will learn next: substitution, integration by parts, and partial fractions all produce antiderivatives that you then evaluate using F(b) – F(a).
The velocity/displacement interpretation connects to kinematics in physics. Later in the course, the same logic extends to work done by a variable force and to computing volumes of solids of revolution.
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