Math Core

When you know how fast something is changing, the definite integral tells you how much it changed. That single idea covers a particle moving along a line, water flowing into a tank, and people entering a stadium. AP free-response questions lean on it heavily, often with a rate given by a formula, a graph or a table.

Net change from a rate

The Fundamental Theorem of Calculus says that if $F'(x) = f(x)$ , then

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

Read it from right to left: integrating a rate of change over an interval gives the net change in the amount. Adding the starting amount gives the ending amount.

Accumulation

If $A'(t)$ is the rate at which a quantity $A$ changes, then

A(b) = A(a) + \int_a^b A'(t)\,dt.

Ending amount = starting amount + net change.

The units work out automatically: a rate in gallons per minute, integrated over minutes, gives gallons.

Motion along a line

For a particle moving along a line with position $x(t)$ , velocity $v(t) = x'(t)$ and acceleration $a(t) = v'(t)$ :

Displacement on $[a, b]$ is the net change in position: $\displaystyle\int_a^b v(t)\,dt = x(b) - x(a)$ .
Position at time $b$ is $x(b) = x(a) + \displaystyle\int_a^b v(t)\,dt$ .
Velocity at time $b$ is $v(b) = v(a) + \displaystyle\int_a^b a(t)\,dt$ .
Total distance traveled on $[a, b]$ is $\displaystyle\int_a^b |v(t)|\,dt$ .

Displacement can be negative or zero; distance never is. When the particle moves left ( $v < 0$ ), that motion subtracts from displacement but still adds to distance.

Definition

Total distance traveled

The total distance a particle travels on $[a, b]$ is $\displaystyle\int_a^b |v(t)|\,dt$ . Without a calculator, find where $v(t) = 0$ , split the interval there, integrate $v$ on each piece, and add the absolute values of the results.

v(t) = t² − 4t + 3 on [0, 4]. The signed areas add to a displacement of 4/3; their absolute values add to a distance of 4.Open in grapher →

Common mistake

"Displacement" and "total distance" are different questions. If an AP problem asks how far a particle traveled, integrate the speed $|v(t)|$ , not $v(t)$ . On a calculator, enter the absolute value directly: $\int_a^b |v(t)|\,dt$ .

Rate in, rate out

Many accumulation problems have one rate adding to a quantity and another taking away. If water enters a tank at $R(t)$ and leaves at $D(t)$ , the net rate is $R(t) - D(t)$ , so

W(t) = W(0) + \int_0^t \big(R(s) - D(s)\big)\,ds.

Because $W'(t) = R(t) - D(t)$ , you can use derivative tools on $W$ :

$W$ is increasing when $R(t) > D(t)$ and decreasing when $R(t) < D(t)$ .
$W$ has candidates for an absolute max or min where $R(t) = D(t)$ and at the endpoints. Compare all of them (the Candidates Test).

Worked examples

Worked example: Displacement and distance

A particle moves along the $x$ -axis with velocity $v(t) = t^2 - 4t + 3$ for $0 \le t \le 4$ . At $t = 0$ it is at $x = 5$ . Find its displacement, its position at $t = 4$ , and the total distance it travels.

Solution. Let $F(t) = \dfrac{t^3}{3} - 2t^2 + 3t$ , an antiderivative of $v$ .

Displacement: $\displaystyle\int_0^4 v(t)\,dt = F(4) - F(0) = \dfrac{64}{3} - 32 + 12 = \dfrac{4}{3}$ .

Position: $x(4) = 5 + \dfrac{4}{3} = \dfrac{19}{3}$ .

Distance: $v(t) = (t - 1)(t - 3)$ is zero at $t = 1$ and $t = 3$ . Using $F(1) = \dfrac{4}{3}$ and $F(3) = 0$ :

\int_0^1 v\,dt = \frac{4}{3}, \qquad \int_1^3 v\,dt = -\frac{4}{3}, \qquad \int_3^4 v\,dt = \frac{4}{3}.

Total distance $= \dfrac{4}{3} + \dfrac{4}{3} + \dfrac{4}{3} = 4$ .

Worked example: Velocity from acceleration

A car's acceleration is $a(t) = 2t$ ft/s² and its velocity at $t = 1$ is 3 ft/s. Find $v(4)$ .

Solution.

v(4) = v(1) + \int_1^4 2t\,dt = 3 + \Big[t^2\Big]_1^4 = 3 + 15 = 18 \text{ ft/s}.

Worked example: A tank that fills and drains

A tank holds 100 gallons at $t = 0$ . Water flows in at $R(t) = 10 + 2t$ gallons per minute and drains at a constant 16 gallons per minute, for $0 \le t \le 10$ . How much water is in the tank at $t = 10$ ? At what time is the amount of water least, and how much is there then?

Solution. The net rate is $R(t) - 16 = 2t - 6$ , so

W(t) = 100 + \int_0^t (2s - 6)\,ds = 100 + t^2 - 6t.

At $t = 10$ : $W(10) = 100 + 100 - 60 = 140$ gallons.

$W'(t) = 2t - 6 = 0$ at $t = 3$ . Compare the candidates: $W(0) = 100$ , $W(3) = 100 + 9 - 18 = 91$ , $W(10) = 140$ . The least amount is 91 gallons, at $t = 3$ minutes. Before then the tank drains faster than it fills; after, it fills faster.

Tip

On the AP exam, write the integral expression with the starting value before you evaluate it, such as $W(10) = 100 + \int_0^{10} (R(t) - 16)\,dt$ . The setup earns points even if an arithmetic slip costs you the final number.

Practice

Practice 1

A particle moves along a line with velocity $v(t) = 3t^2 - 6t$ for $0 \le t \le 3$ . Find the total distance it travels.