Interpret the definite result
Water in and out of a tank
The top graph shows \(r(t) = 6 - 2t\), the rate at which water flows into a tank, in liters per minute. When \(r\) is negative, water flows out.
The teal region lies above the t-axis, where water enters, so it counts as a gain. The red region lies below it, where water leaves, so it counts as a loss.
Drag the orange edge to \(t = 5\). The lower graph shows the change in the amount of water since \(t = 0\). It rises 9 liters and then falls 4, so the net change is \(\cg{9} - \cr{4} = \cb{5}\) liters.
Press the button to flip the red region up. The total area, 9 + 4 = 13 liters, counts all the water that moved in or out, not the change in the tank.
Net change, area and the starting amount
The top graph shows a rate, teal above the axis and red below it. The lower graph shows the amount, which rises where the rate is positive and falls where it is negative.
The orange bracket runs from the amount at a to the amount at b. Its length, the change from a to b, is the definite integral of the rate.
Press the button to raise the starting amount. The whole lower graph moves up, but the change from a to b stays the same, because both ends rise equally.
The integral gives a change, not an amount.
Notation
The definite integral of a rate is the net change of the quantity. A region above the axis adds and a region below it subtracts.
\[\int_a^b r(t)\,dt = \cg{A_{\text{above}}} - \cr{A_{\text{below}}}\]Its unit is the unit of the rate times the unit of the input.
\[\frac{\text{liters}}{\text{minute}} \times \text{minutes} = \text{liters}\]The total area counts every region as positive: cut the interval at the zeros of the rate and add the sizes.
\[\text{total area} = \cg{A_{\text{above}}} + \cr{A_{\text{below}}}\]A constant C in an antiderivative appears at both bounds, so it cancels, and a definite integral needs no C.
\[\begin{aligned} &\big(F(b) + C\big) - \big(F(a) + C\big) \\ &= \cb{F(b) - F(a)} \end{aligned}\]The integral equals the area only when the rate is never negative between a and b.
When it fails
Press the button to use \(t\) from 3 to 5. The rate is negative there, so the region is red and the integral is \(\cr{-4}\) liters.
An area cannot be negative, so the integral is not the area. The red region has area 4, and the minus sign means the tank lost 4 liters.
Example 1: what the number means
Oil leaks from a tank at the rate \(L(t) = 6t\) liters per hour, for \(0 \le t \le 4\). Find \(\displaystyle\int_0^4 L(t)\,dt\), give its units and say what it means.
An antiderivative of \(6t\) is \(F(t) = 3t^2\).
Substitute the bounds and subtract.
\[F(4) - F(0) = 48 - 0 = \cb{48}\]Liters per hour times hours gives liters, so 48 liters leaked in the first 4 hours. The rate is never negative, so this is also the area of the teal region.
Example 2: displacement and distance
A cart moves along a track with velocity \(v(t) = t^2 - 6t + 8\) meters per second. Find its displacement and the distance it travels for \(0 \le t \le 3\).
An antiderivative of the velocity is \(F(t) = \dfrac{t^3}{3} - 3t^2 + 8t\).
The integral from 0 to 3 is the displacement, the net change in position, shown by the teal arrow under the number line.
\[F(3) - F(0) = 6 - 0 = 6\]A tempting wrong move: The distance traveled is the integral, 6 meters.
\[\int_0^3 v(t)\,dt = 6\]The velocity is negative from \(t = 2\) to \(t = 3\), so the cart moves backward there, along the red arrow. The integral subtracts that part, so the cart travels more than 6 meters.
Distance counts the forward and backward parts both as positive. Because \(v(t) = (t - 2)(t - 4)\), the velocity is 0 at \(t = 2\), so cut the interval there and find \(F(2)\).
\[F(2) = \tfrac{8}{3} - 12 + 16 = \tfrac{20}{3}\]From 0 to 2 the cart moves forward \(\tfrac{20}{3}\) meters. From 2 to 3 it moves from \(\tfrac{20}{3}\) back to 6, which is \(\tfrac{20}{3} - 6 = \tfrac23\) meters. Add the two sizes.
\[\tfrac{20}{3} + \tfrac{2}{3} = \cb{\tfrac{22}{3}}\]
Practice
The balance of an account changes at the rate \(b'(t) = 2t - 8\) hundred dollars per month, for \(0 \le t \le 3\). Find \(\displaystyle\int_0^3 b'(t)\,dt\), give its units and say what its sign means.
Move the edge to 3. The rate is negative on the whole interval, so the region is red and the integral is negative.
Solution
An antiderivative of \(2t - 8\) is \(F(t) = t^2 - 8t\).
Substitute the bounds and subtract.
\[F(3) - F(0) = -15 - 0 = \cy{-15}\]The integral is −15 hundred dollars, so the balance falls by 1,500 dollars in 3 months, and the red region has area 15.
Questions
The top graph shows \(g(t) = 4 - 2t\), the rate at which the temperature of a cup of tea changes, in degrees Celsius per minute. The lower graph shows the net change since the start of the region.
What is \(\displaystyle\int_0^6 g(t)\,dt\)?
- \(-12\)
- 20
- 4
- 16
Answer
\(-12\)
What does \(\displaystyle\int_2^6 g(t)\,dt\) tell you about the temperature?
- It falls 16 degrees from t = 2 to t = 6
- It rises 16 degrees from t = 2 to t = 6
- It falls 12 degrees from t = 2 to t = 6
- It is 16 degrees at t = 6
Answer
It falls 16 degrees from t = 2 to t = 6
A student evaluates \(\displaystyle\int_1^3 2x\,dx\) with the antiderivative \(F(x) = x^2 + 5\). What does the student find?
- 8
- 13
- 14
- \(-8\)
Answer
8
Summary
You evaluate a definite integral with any antiderivative and state the result as a net change, with its units.
It works because regions above the axis add, regions below subtract, and any constant in the antiderivative cancels.
To find total area or distance instead, you cut where the rate is 0 and add the sizes.
This is one of four skills in topic 6.7 of Which & Why: AP Calculus.