Recover a final amount
A hiker's altitude
A hiker starts at an altitude of 1200 meters. The top graph shows her climbing rate \(h'\), in meters per minute. A negative rate means that she goes down.
Drag the orange edge to the right. The shaded area is the net change in altitude since the start, the meters climbed minus the meters descended. At \(t = 4\) it is 32 meters.
The lower graph shows her altitude. It starts at 1200, not at 0, because she was already there. At \(t = 4\) it reads \(1200 + 32 = 1232\).
After \(t = 6\) the rate is negative and its area is shaded red, so the altitude falls. At \(t = 10\) the net change is 20 meters, so the altitude is 1220 meters.
Start plus change
With the numbers removed, the top graph shows a rate and the lower graph shows the amount, which starts on the dashed line.
Drag the edge. The lower graph rises where the shaded area is teal and falls where it is red.
The gap between the lower graph and the dashed line is the signed area under the rate graph. This area is only the change, so the amount at the edge is the starting amount plus the area.
Notation
The amount is \(F\), and its rate of change is the derivative \(F'\). The integral of the rate from \(a\) to \(\cy{b}\) is the net change in the amount.
\[\int_a^b F'(t)\,dt = F(b) - F(a)\]Add \(F(a)\) to both sides. The final amount is the starting amount plus the net change.
\[\cy{F(b)} = \cq{F(a)} + \cb{\int_a^b F'(t)\,dt}\]The problem must give the starting amount \(F(a)\), the amount at the lower limit \(a\), because the integral cannot supply it.
When it fails
Amounts that start at different values can have the same rate graph. Press the button to try three starting amounts.
The lower graph moves up or down but keeps its shape, because the area under the rate graph does not change. The integral alone cannot tell the three amounts apart.
Example 1: An ant colony
A colony has 400 ants at \(t = 0\). It grows at the rate \(20 + 10t\) ants per day. How many ants are in the colony at \(t = 8\)?
The colony starts with 400 ants, so the lower graph starts at 400, on the dashed line.
The net change from 0 to 8 is the integral of the rate, the shaded region. It is a trapezoid with parallel sides \(\cy{20}\) and \(\cy{100}\) and width 8 days.
\[\begin{aligned} \cb{\int_0^8 (20 + 10t)\,dt} &= \frac{20 + 100}{2} \times 8 \\ &= 480 \end{aligned}\]Add the net change to the starting amount.
\[400 + 480 = 880\]The colony has 880 ants at \(t = 8\).
Example 2: A tank being emptied
A tank holds 500 liters. A pump removes water at the rate \(10 + 2t\) liters per minute. How much water is in the tank after 6 minutes?
The pump removes water, so the rate of change of the water is negative, \(-(10 + 2t)\). The red trapezoid below the axis has parallel sides 10 and 22 and width 6. Its area is 96, and it counts as negative.
\[-\frac{10 + 22}{2} \times 6 = -96\]A tempting wrong move: The tank starts with 500 liters, so add the amount pumped out.
\[500 + 96 = 596\]The pump only removes water, so the water graph falls along the orange line. The red line rises instead, and 596 lies above the dashed line, where the water never goes.
The net change is \(-96\), so adding it subtracts 96 liters from the start.
\[500 - 96 = 404\]The tank holds 404 liters after 6 minutes.
Practice
A savings account holds 800 dollars. Money is deposited at the rate \(50 + 6t\) dollars per week. How much is in the account after 5 weeks?
The shaded area is only the change. Add the starting balance on the dashed line.
Solution
The net change is the area under the deposit rate from 0 to 5, a trapezoid with parallel sides 50 and 80 and width 5.
\[\begin{aligned} \cb{\int_0^5 (50 + 6t)\,dt} &= \frac{50 + 80}{2} \times 5 \\ &= 325 \end{aligned}\]Add the starting balance.
\[800 + 325 = 1125\]The account holds 1125 dollars after 5 weeks.
Questions
A reservoir holds 10 million liters at \(t = 0\). The graph shows the rate of change of its volume, in millions of liters per day.
At which time is the volume of water greatest?
- \(t = 4\)
- \(t = 2\)
- \(t = 0\)
- \(t = 6\)
Answer
\(t = 4\)
What is the volume of water at \(t = 6\)?
- 18 million liters
- 8 million liters
- 22 million liters
Answer
18 million liters
A colony has 150 birds at \(t = 0\). It changes at the rate \(6t - 9\) birds per year. How many birds are in the colony at \(t = 4\)?
- 162 birds
- 12 birds
- 165 birds
Answer
162 birds
Summary
To recover a final amount, you add the starting amount to the integral of the rate.
The integral gives only the net change, because the rate graph does not show where the amount began.
You use it when a problem gives a starting amount and a rate, and asks for the amount at a later time.
This is one of four skills in topic 6.1 of Which & Why: AP Calculus.