AP Calculus AB and BC

Definite Integral Step Tool: FTC, Shown Honestly

To evaluate a definite integral, find any antiderivative F of the integrand, then compute F(top) minus F(bottom). That is the Fundamental Theorem of Calculus: the integral from a to b of f equals F(b) - F(a). The hard part is finding the antiderivative; the evaluation is just substitute, substitute, subtract.

Pick a definite integral to evaluate with the Fundamental Theorem of Calculus:

Evaluate

\int_0^2 x^2\,dx

The shaded region between x = a and x = b is what this integral measures. The theorem turns that area into two evaluations of an antiderivative.

Steps

0 / 4

This tool takes a curated set of definite integrals and shows the evaluation the way you have to reproduce it on the no-calculator sections: find an antiderivative, substitute the top bound, substitute the bottom bound, and subtract. A solver hands you a number and hides all three steps. Here every line stays on the table, so you learn the structure instead of just the answer.

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

The Fundamental Theorem of Calculus (Topic 6.7, essential knowledge FUN-6.B.3) applies whenever $f$ is continuous on the closed interval $[a, b]$ and you can find an antiderivative $F$ . Recognizing that the FTC is the tool for the job is the easy part: numbers on the integral sign mean a definite integral, and the FTC is how you turn that into an exact value. The real work is finding $F$ , which is differentiation run backward.

Why the constant of integration disappears

An indefinite integral needs $+\,C$ , but a definite integral does not. If you carried the constant, you would compute $(F(b) + C) - (F(a) + C)$ , and the $C$ cancels. So pick the simplest antiderivative, the one with $C = 0$ , and evaluate that.

Confirm a definite integral: the sign carries a lower bound $a$ and an upper bound $b$ , and the integrand is continuous on $[a, b]$ .
Find any one antiderivative $F$ of the integrand by reversing your differentiation rules.
Substitute the upper bound to get $F(b)$ .
Substitute the lower bound to get $F(a)$ .
Subtract in that order, $F(b) - F(a)$ : upper minus lower, never the reverse.

The mistake that costs the most points

Reversing the subtraction. $F(a) - F(b)$ gives the negative of the correct answer, and on a free-response question a sign flip can cascade into every later part. The bracket notation $\left[ F(x) \right]_a^b$ exists to remind you: top value first, then subtract the bottom value.

When the antiderivative is not obvious, the FTC still runs these same steps; you just need a technique to find $F$ first. If the integrand is a composition with the inner function's derivative present, try u-substitution (Topic 6.9). For a product of unlike functions, BC students reach for integration by parts (Topic 6.11, BC only). Choosing that technique is a separate decision from the evaluation you drill here, which is why this tool keeps the two apart: the antidifferentiation is kept routine (reverse power rule, or the antiderivative is supplied), so you can focus on the bracket-and-subtract mechanics that every FTC problem shares.

Worked example

Evaluating a definite integral with the FTC

Evaluate $\int_1^3 (3x^2 + 2)\,dx$ .

Check that the theorem applies. The integrand $3x^2 + 2$ is a polynomial, so it is continuous on $[1, 3]$ , and both bounds are finite. This is a definite integral of a continuous function, so the Fundamental Theorem of Calculus gives an exact value: find an antiderivative and evaluate it at the bounds.
Find an antiderivative using the reverse power rule, term by term. An antiderivative of $3x^2$ is $x^3$ , and an antiderivative of the constant $2$ is $2x$ , so take $F(x) = x^3 + 2x$ . Verify it: $F'(x) = 3x^2 + 2$ , which matches the integrand, so $F$ is correct. No $+\,C$ is needed, because it would cancel in the subtraction.
Set up the evaluation bracket. Write the antiderivative in brackets with the bounds attached: $\left[ x^3 + 2x \right]_1^3$ , which is shorthand for $F(3) - F(1)$ .
Substitute the upper bound $x = 3$ : $F(3) = 3^3 + 2(3) = 27 + 6 = 33$ .
Substitute the lower bound $x = 1$ : $F(1) = 1^3 + 2(1) = 1 + 2 = 3$ .
Subtract, upper minus lower: $F(3) - F(1) = 33 - 3 = 30$ .

$\int_1^3 (3x^2 + 2)\,dx = 30$ .

Frequently asked questions

Do you need a +C when evaluating a definite integral?

No. The constant is only for indefinite integrals. In a definite integral you compute $(F(b) + C) - (F(a) + C)$ , and the $C$ cancels, so it never affects the answer. Use the antiderivative with $C = 0$ .

What is the difference between the two parts of the Fundamental Theorem of Calculus?

The first part (Topic 6.4) is about accumulation functions: if $g(x) = \int_a^x f(t)\,dt$ , then $g'(x) = f(x)$ , so differentiating an integral gives the integrand back. The second part (Topic 6.7) is the evaluation rule this tool shows: $\int_a^b f(x)\,dx = F(b) - F(a)$ . This tool teaches the second part.

What is the difference between a definite and an indefinite integral?

A definite integral has bounds and evaluates to a single number, $F(b) - F(a)$ . An indefinite integral has no bounds and evaluates to a whole family of antiderivatives, $F(x) + C$ . The FTC is the bridge: you find the indefinite integral (an antiderivative), then evaluate it at the two bounds to get the definite value.

Can you always use the FTC to evaluate a definite integral?

Only when you can find an antiderivative in closed form. Many continuous functions, such as $e^{-x^2}$ , have no elementary antiderivative (FUN-6.C.3), so you cannot use the FTC to produce an exact value by hand. In those cases you approximate the definite integral with a Riemann sum or another numerical method instead.