The FTC and Definite Integrals — AP® Calculus AB Notes

Topic 6.4 showed that an accumulation function $\int_a^x f(t)\,dt$ is an antiderivative of $f$ . This page turns that around into a way to compute definite integrals exactly, and into the idea the exam uses most: the integral of a rate is the net change.

From one antiderivative to the integral

Let $F$ be any antiderivative of a continuous $f$ , so $F' = f$ . The accumulation function $g(x)$ $= \int_a^x f(t)\,dt$ is another one. Two functions with the same derivative on an interval differ by a constant (a consequence of the Mean Value Theorem), so

$g(x) = F(x) + k.$

At $x = a$ : $g(a) = 0$ , so $k = -F(a)$ . Now put $x = b$ :

$\int_a^b f(x)\,dx$ $= F(b) - F(a)$ (the Fundamental Theorem of Calculus, part 2).

The shorthand $F(x)\Big|_a^b$ means $F(b) - F(a)$ . Any antiderivative works; a constant added to $F$ cancels in the subtraction.

Worked example 1. Find $\int_1^3 (3x^2 - 2)\,dx$ .

Find a function whose derivative is $3x^2 - 2$ : $F(x) = x^3 - 2x$ (check: $F'(x) = 3x^2 - 2$ ).

$F(3) = 27 - 6 = 21$ and $F(1) = 1 - 2 = -1$ .

$\int_1^3 (3x^2 - 2)\,dx$ $= 21 - (-1)$ $= 22$ . Keep the parentheses: subtracting a negative is where most slips happen.

Antiderivatives come from reading derivative rules backward: $\sin x$ for $\cos x$ , $\ln x$ for $\frac{1}{x}$ (with $x > 0$ ), $e^x$ for $e^x$ , $\arctan x$ for $\frac{1}{1 + x^2}$ . The next topic organizes these rules.

The integral of a rate is the net change

Top: the graph of f prime of x = 1 - x/4 on [0, 6], with a triangle of area 2 above the axis from 0 to 4 and a triangle of area 0.5 below from 4 to 6. Bottom: f(x) = 1 + x - x squared over 8 rises from (0, 1) to a maximum at x = 4 and falls to (6, 2.5); a bracket shows f(6) - f(0) = 2 - 0.5 = 1.5. — **Figure 1.** The signed area under f prime equals the change in f. LevlPrep original diagram.

Apply part 2 to $f'$ , whose antiderivative is $f$ itself:

$\int_a^b f'(x)\,dx = f(b) - f(a).$

This is the net change theorem: the integral of a rate of change is the total change. If $P(t)$ is a population and $\int_0^{30} P'(t)\,dt$ $= 12$ thousand, the population grew by 12 thousand over those 30 years, whatever it did in between. Rearranged:

$f(b)$ $= f(a)$ $+ \int_a^b f'(x)\,dx$ : value at the end = value at the start + accumulated change.

Worked example 2. The graph of $f'$ is constant 2 on $[0, 3]$ , falls in a straight line from $(3, 2)$ to $(5, -2)$ , and is constant $-2$ on $[5, 8]$ . $f(2) = 5$ . Find $f(5)$ , $f(0)$ and the absolute maximum of $f$ .

$f(5)$ $= f(2)$ $+ \int_2^5 f'(x)\,dx$ $= 5$ $+ (2 + 1 - 1)$ $= 7$ .

$f(0)$ : going backward, $f(0)$ $= f(2) - \int_0^2 f'(x)\,dx$ $= 5 - 4$ $= 1$ .

Maximum: $f'$ changes from positive to negative at $x = 4$ . Candidates: $f(0) = 1$ , $f(4) = 5 + 3 = 8$ , $f(8)$ $= 5 + (2 + 1 - 1 - 6)$ $= 1$ . The absolute maximum is 8, at $x = 4$ .

When there is no formula: the calculator

Some rates, such as $2.5\sin\left(\frac{t^2}{10}\right) - 0.5$ , have no antiderivative made of familiar functions. The net change idea still works; the calculator evaluates the integral. Write the full setup:

$D(8) = D(0) + \int_0^8 D'(t)\,dt = 10 + \int_0^8 \left(2.5\sin\left(\frac{t^2}{10}\right) - 0.5\right)dt \approx 9.413 \text{ inches}.$

Use radian mode, keep full precision until the end, and give three decimal places.

Interpreting in context

A complete interpretation of $\int_a^b r(t)\,dt$ names the quantity, its units and the interval: " $\int_2^6 D'(t)\,dt$ is the change in the depth of the snow, in inches, from $t = 2$ to $t = 6$ hours." Three things to avoid: calling it a rate (inches per hour), calling it the amount at the end (that also needs the starting value), and leaving out the time interval.