Calculus 1

The net change theorem in context

A meter reads only how fast something is changing, such as cars per hour through a gate, kilowatts into a battery or barrels per hour through a pipe. How much did the amount itself change, and how much is there now?

You need

The derivative as a rate of changeCalculus 1 The definite integral as a limit of Riemann sumsCalculus 1

Where you are 3 in Accumulation and net change

Calculus8

Applications of integrals10

Accumulation and net change3

Graphing an accumulation function Position, displacement and distance from velocity

The net change theorem in context

When you reach for it

A rate is known and an amount is wanted. Wulan runs a parking garage and counts cars per hour at its gates. Ratri charges electric buses and reads kilowatts on the charger. Bagus collects rain from his roof and reads liters per hour off the downspout. The amount (cars, kilowatt-hours, liters) is what matters, yet the instruments report how fast it changes.

The derivative as a rate of change read a rate as the derivative of an amount, in units of amount per time. This lesson runs the other way, from the rate back to the amount. The tool is the theorem of The second fundamental theorem of calculus, read in the language of the situation. Day 1 takes rates given by formulas. Day 2 takes rates given by tables and graphs, and densities, where the variable is a position instead of a time.

The procedure

Day 1: the net change theorem

The theorem. If F' is continuous on [a, b], then

\int_a^b F'(t)\,dt = F(b) - F(a).

This is the fundamental theorem of The second fundamental theorem of calculus, with F as the antiderivative of F', which it is by definition. Nothing new is proved here; what is new is the reading. If F(t) is an amount at time t, then F'(t) is the rate at which the amount changes, and the theorem says: the integral of a rate over a time interval is the net change in the amount over that interval.

Why the units multiply. The integral is a limit of Riemann sums \sum F'(t_k)\,\Delta t (The definite integral as a limit of Riemann sums). Each term is a rate times a short time, so it is about the change in the amount over that short time: kilowatts (kilowatt-hours per hour) times hours is kilowatt-hours. Adding the pieces adds the changes. So the units of \int_a^b r(t)\,dt are the units of r times the units of t, which are the units of the amount.

The amount at a later time. Moving F(a) across gives

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

The amount at time b is the starting amount plus the net change. The rate fixes only the change. The starting amount has to come from somewhere else, a reading or a count, just as a derivative fixes a function only up to +\,C.

Net. The word means gains minus losses. Where F'(t) < 0 the amount falls, the Riemann terms there are negative, and they cancel part of the gains (Signed area and integrals from geometry). A negative integral says the amount ended lower than it started.

In and out. When an amount has an inflow at rate I(t) and an outflow at rate O(t), its rate of change is the net rate F'(t) = I(t) - O(t), so

\begin{aligned} F(b) - F(a) &= \int_a^b \big(I(t) - O(t)\big)\,dt \\ &= \int_a^b I(t)\,dt - \int_a^b O(t)\,dt. \end{aligned}

The first integral is everything that came in and the second everything that went out. Each answers its own question.

Saying it in a sentence. An integral in context is reported as a sentence with the interval, the quantity, the direction and the units: “between 7 a.m. and 11 a.m., 1360 cars entered the garage.” A bare number is half an answer. A formula for the rate is integrated with an antiderivative; when the rate is a composition, the antiderivative comes from u-substitution first.

Day 2: tables, graphs and density

Rates from a table. A few readings of a rate give no formula to find an antiderivative of, but the integral is still the net change, and a Riemann sum estimates it (Approximating area with Riemann sums). The trapezoid sum usually does better, since it uses the reading at both ends of each piece (The trapezoidal rule and Simpson's rule). The result is an estimate of the net change, in the same units.

Rates from a graph. The integral is the signed area between the graph of the rate and the t-axis: area above counts as gain, area below as loss. Areas of triangles, rectangles and trapezoids are read off the grid.

Density. A rod whose material changes along its length has a linear density \rho(x), its mass per unit length at position x. A short piece from x to x + \Delta x has mass about \rho(x)\,\Delta x, and adding the pieces gives a Riemann sum whose limit is the mass:

M = \int_a^b \rho(x)\,dx.

This is the net change theorem again. If M(x) is the mass of the rod from its end up to position x, then M'(x) = \rho(x), and the integral of \rho over [a, b] is M(b) - M(a). Kilograms per meter times meters is kilograms. Anything spread along a line works the same way: people per kilometer along a road integrate to people.

Net change hides back-and-forth. The integral of a rate that changes sign gives only the net change: ten barrels in and ten out net to zero. When positive and negative mean flow in two directions through one channel, the total that moved is \int_a^b |r(t)|\,dt, the sum of the areas with no signs. For motion along a line, net change is displacement and the total is distance, the subject of Position, displacement and distance from velocity.

Worked examples

Example 1

Day 1: Ratri’s bus battery

Ratri plugs an electric bus into the depot charger at 10 p.m., when its battery holds 85 kilowatt-hours. For the next three hours the battery charges at

r(t) = 120 - 20t \text{ kilowatts},

t hours after 10 p.m. A kilowatt is one kilowatt-hour per hour. Find \int_0^3 r(t)\,dt and the charge at 1 a.m. Then say what \int_1^2 r(t)\,dt means and find it.

Solution

The rate r(t) is the derivative of the charge Q(t), so the integral is the net change in charge. Its units are kilowatts times hours, which is kilowatt-hours. An antiderivative of 120 - 20t is 120t - 10t^2, so

\begin{aligned} \int_0^3 (120 - 20t)\,dt &= (360 - 90) - 0 \\ &= 270 \text{ kilowatt-hours}. \end{aligned}

Between 10 p.m. and 1 a.m. the battery gained 270 kilowatt-hours. The charge at 1 a.m. is the starting charge plus the change:

\begin{aligned} Q(3) &= Q(0) + \int_0^3 r(t)\,dt \\ &= 85 + 270 = 355 \text{ kilowatt-hours}. \end{aligned}

The integral from 1 to 2 is the charge gained during the second hour, from 11 p.m. to midnight:

\begin{aligned} \int_1^2 (120 - 20t)\,dt &= (240 - 40) - (120 - 10) \\ &= 90 \text{ kilowatt-hours}. \end{aligned}

The battery gained 90 kilowatt-hours in that hour. The rate falls as the battery fills, so each hour adds less: the first hour adds 110 and the third adds 70, and 110 + 90 + 70 = 270 checks the total.

Example 2

Day 1: Wulan’s garage, in and out

Wulan’s parking garage holds 150 cars at 7 a.m. For the next four hours, t hours after 7 a.m., cars enter at E(t) = 180 + 240t - 60t^2 cars per hour and leave at L(t) = 60 + 30t cars per hour. How many cars entered and how many left between 7 and 11 a.m.? How many cars are in the garage at 11 a.m.? Was the count rising the whole morning?

Solution

Each rate integrates to a number of cars, since cars per hour times hours is cars. The cars that entered:

\begin{aligned} \int_0^4 E(t)\,dt &= \Big[180t + 120t^2 - 20t^3\Big]_0^4 \\ &= 720 + 1920 - 1280 \\ &= 1360 \text{ cars}. \end{aligned}

The cars that left:

\begin{aligned} \int_0^4 L(t)\,dt &= \Big[60t + 15t^2\Big]_0^4 \\ &= 240 + 240 = 480 \text{ cars}. \end{aligned}

Between 7 and 11 a.m., 1360 cars entered and 480 left, a net gain of 880 cars. At 11 a.m. the garage holds 150 + 880 = 1030 cars.

The count rises exactly where the net rate is positive. That rate is

\begin{aligned} E(t) - L(t) &= 120 + 210t - 60t^2 \\ &= -30(2t + 1)(t - 4), \end{aligned}

which is positive for 0 \le t < 4 and zero at t = 4. So the garage gained cars the whole morning and stopped gaining at 11 a.m. Whether 11 a.m. is the fullest moment depends on the rates afterward; Graphing an accumulation function finds the largest value of an accumulated amount.

Example 3

Day 1: rain from Bagus’s roof

During a storm, rain runs off Bagus’s roof into a tank at

R(t) = 80t\,e^{-t^2/4} \text{ liters per hour},

t hours after the storm begins, while a garden tap lets water out at a steady 20 liters per hour. The tank holds 50 liters at t = 0. Find how much rain runs in during the first two hours, and the amount in the tank at t = 2 and at t = 6.

Solution

The factor t is, up to a constant, the derivative of the exponent \displaystyle -\frac{t^2}{4}, so substitution finds an antiderivative. With \displaystyle u = -\frac{t^2}{4}, \displaystyle du = -\frac{t}{2}\,dt, so t\,dt = -2\,du and

\begin{aligned} \int 80t\,e^{-t^2/4}\,dt &= \int -160\,e^{u}\,du \\ &= -160\,e^{-t^2/4} + C. \end{aligned}

Differentiating -160e^{-t^2/4} gives \displaystyle -160 \cdot \left(-\frac{t}{2}\right)e^{-t^2/4} = 80t\,e^{-t^2/4}, as it should. Then

\begin{aligned} \int_0^2 R(t)\,dt &= -160e^{-1} + 160 \\ &= 160\big(1 - e^{-1}\big) \approx 101.14 \text{ liters}. \end{aligned}

About 101.14 liters of rain ran into the tank in the first two hours. The tap let out \int_0^2 20\,dt = 40 liters, so the volume V at t = 2 is

\begin{aligned} V(2) &= 50 + \int_0^2 \big(R(t) - 20\big)\,dt \\ &\approx 50 + 101.14 - 40 = 111.14 \text{ liters}. \end{aligned}

Over six hours the rain brings 160\big(1 - e^{-9}\big) \approx 159.98 liters and the tap removes 120, so V(6) \approx 50 + 159.98 - 120 = 89.98 liters. The tank holds less at t = 6 than at t = 2: by then the runoff has dropped below 20 liters per hour (at t = 4 it is 320e^{-4} \approx 5.9), and the tap wins.

Example 4

Day 2: Bagus’s city from a table

Bagus studies a city whose population changes only by births and deaths (people moving in or out are left aside). The records give the birth rate b(t) and the death rate d(t), in people per year, t years after 2010 (the city and its numbers are made up):

t	b(t)	d(t)
0	5200	3100
2	5400	3200
4	5500	3400
6	5450	3700
8	5300	4000
10	5100	4300

The population in 2010 was 248{,}000. Estimate the population in 2020.

Solution

The population P has P'(t) = b(t) - d(t), so P(10) = P(0) + \int_0^{10} \big(b(t) - d(t)\big)\,dt. There is no formula, so the integral is estimated from the readings. The net rate at the six readings is 2100, 2200, 2100, 1750, 1300 and 800 people per year. With \Delta t = 2 the trapezoid sum is

\begin{aligned} T_5 &= \tfrac{2}{2}\big(2100 + 2(7350) + 800\big) \\ &= 17{,}600 \text{ people}, \end{aligned}

since 2200 + 2100 + 1750 + 1300 = 7350.

\begin{aligned} P(10) &\approx 248{,}000 + 17{,}600 \\ &= 265{,}600. \end{aligned}

People per year times years is people, so the population grew by about 17{,}600 from 2010 to 2020. The left sum gives 2(9450) = 18{,}900 and the right sum 2(8150) = 16{,}300; the trapezoid sum is their average. How good the estimate is depends on how the rates behave between readings, which the table cannot show.

Example 5

Day 2: Ratri’s pipeline terminal, from a graph

At a pipeline terminal, one pipe joins a storage tank to the main line, and oil can flow through it either way. The graph shows the net flow rate r(t) into the tank, in hundreds of barrels per hour, t hours after midnight; negative values mean oil is flowing out. The tank held 3000 barrels at midnight. Find the amount at noon, the largest amount during the twelve hours, and the total oil that passed through the pipe.

Figure 1The net flow rate into Ratri’s tank, in hundreds of barrels per hour. Area above the axis is oil gained, area below is oil lost.

Solution

Hundreds of barrels per hour times hours is hundreds of barrels, so each area on the graph is an amount of oil. Above the axis the pieces are a triangle, a rectangle and a triangle, with areas \displaystyle \frac{1}{2}(2)(6) = 6, 2 \cdot 6 = 12 and 6. Below it the areas are \displaystyle \frac{1}{2}(2)(4) = 4, 2 \cdot 4 = 8 and 4, counted as negative. So

\begin{aligned} \int_0^{12} r(t)\,dt &= (6 + 12 + 6) - (4 + 8 + 4) \\ &= 24 - 16 = 8 \text{ hundred barrels}. \end{aligned}

The tank gained a net 800 barrels from midnight to noon and holds 3000 + 800 = 3800 barrels at noon.

The amount rises while r(t) > 0 and falls while r(t) < 0, so it is largest at t = 6, when the flow turns around. At 6 a.m. the tank holds 3000 + 2400 = 5400 barrels.

The net change hides the traffic. Oil flowed in for six hours (2400 barrels) and out for six (1600 barrels), so 4000 barrels passed through the pipe:

\begin{aligned} \int_0^{12} |r(t)|\,dt &= 24 + 16 = 40 \text{ hundred barrels} \\ &= 4000 \text{ barrels}. \end{aligned}

Example 6

Day 2: the mass of Wulan’s rod

Wulan’s workshop casts a metal rod 4 meters long whose alloy gets denser along its length. At x meters from the light end its linear density is

\rho(x) = 5 + 3\sqrt{x} \text{ kilograms per meter}.

Find the mass of the rod and the mass of each half.

Figure 2A short piece of the rod has nearly constant density, so its mass is about the density times its length.

Solution

A short piece from x to x + \Delta x has nearly constant density, so its mass is about \rho(x)\,\Delta x kilograms. Adding the pieces gives a Riemann sum for \int_0^4 \rho(x)\,dx, and its limit is the mass. An antiderivative of 5 + 3x^{1/2} is 5x + 2x^{3/2}, so

\begin{aligned} M &= \int_0^4 \big(5 + 3\sqrt{x}\big)\,dx \\ &= 20 + 2 \cdot 8 = 36 \text{ kilograms}. \end{aligned}

The light half, 0 \le x \le 2, has mass \displaystyle 10 + 2 \cdot 2^{3/2} = 10 + 4\sqrt{2} \approx 15.66 kilograms. The heavy half has the rest, \displaystyle 36 - (10 + 4\sqrt{2}) = 26 - 4\sqrt{2} \approx 20.34 kilograms.

The value \rho(1) = 8 is not a mass. It says that near x = 1 a centimeter of rod, 0.01 meter, has mass about 0.08 kilograms.

Example 7

Day 2: Bagus counts people along a highway

A highway runs 20 kilometers out of a city. A survey gives the population density \delta(x) along it, in people per kilometer of highway, counting everyone who lives within a kilometer of the road, at x kilometers from the city center (the readings are made up):

x	\delta(x)
0	1800
2	1650
5	1300
10	800
16	350
20	200

Estimate how many people live along the highway.

Solution

People per kilometer times kilometers is people, so the number of people is \int_0^{20} \delta(x)\,dx: the net change theorem with position in place of time. The readings are unevenly spaced, so each trapezoid takes its own width:

\begin{aligned} &2 \cdot \tfrac{1800 + 1650}{2} + 3 \cdot \tfrac{1650 + 1300}{2} \\ &\quad + 5 \cdot \tfrac{1300 + 800}{2} + 6 \cdot \tfrac{800 + 350}{2} \\ &\quad + 4 \cdot \tfrac{350 + 200}{2} \\ &= 3450 + 4425 + 5250 + 3450 + 1100 \\ &= 17{,}675 \text{ people}. \end{aligned}

About 17{,}675 people live along the highway. Since the density between survey points is unknown, 17{,}700 is a more honest way to report it.

Boundaries

The theorem needs F' continuous on [a, b], so that F is an antiderivative of F' on the whole interval. An amount that jumps, such as a tank emptied into a truck in a moment, has no rate at the jump, and the integral of the recorded rate misses the jump.
The integral gives a change, never an amount. The amount at time b needs the amount at some time a.
Units must agree. A rate in kilowatts integrated over time in minutes gives kilowatt-minutes, which must be divided by 60 to give kilowatt-hours.
A table or a sketch gives an estimate. Nothing in the data says what the rate did between readings.
When the rate changes sign, the integral is the net change, not the total that moved.
A linear density is mass per unit length along the variable of integration. Mass spread over an area or through a volume needs a density per unit area or volume and an integral over that region.

Where this goes next

Position, displacement and distance from velocity applies the same reading to velocity, where the net change is displacement and the integral of speed is distance. Graphing an accumulation function asks when an accumulated amount is largest or smallest. The pattern used for the rod, cutting a quantity into small pieces, approximating each and adding, is the one later applications use to find areas between curves, volumes and work.

Exercises

Day 1: meaning and units

Q(t) is the charge of a drone’s battery, in watt-hours, t minutes after takeoff. Give the units of \int_0^{12} Q'(t)\,dt, and say in a sentence what \int_0^{12} Q'(t)\,dt = -38 means.
P(t) is the population of a town t years after 2000. Say in words what \int_5^8 P'(t)\,dt and P(5) + \int_5^8 P'(t)\,dt mean, with units.

Day 1: rates given by formulas

A delivery van’s battery holds 64 kilowatt-hours when it is plugged in, and then charges at r(t) = 50 - 8t kilowatts for 0 \le t \le 4 hours. How much charge does it gain, and what does it hold at t = 4?
Oil flows into a tank holding 1200 barrels at \displaystyle f(t) = 30 + 6\sqrt{t} barrels per hour for 0 \le t \le 9 hours. How much oil flows in, and how much is in the tank at t = 9?
Rain falls at \displaystyle 6\sin\left(\frac{\pi t}{4}\right) millimeters per hour for 0 \le t \le 4 hours. How much rain falls, measured as depth in millimeters?
A parking garage holds 220 cars at t = 0. Cars enter at E(t) = 400 - 50t and leave at L(t) = 100 + 40t cars per hour for 0 \le t \le 6. Find how many cars entered and how many left, the count at t = 6, and the change in the count from t = 5 to t = 6.
A town grows at \displaystyle P'(t) = \frac{6000t}{t^2 + 9} people per year, t years after a census. By how many people does it grow in the first four years?
Rain runs into a tank holding 200 liters at R(t) = 90e^{-t/2} liters per hour while a valve lets water out at 15 liters per hour. Find the net change from t = 0 to t = 4 hours and the amount at t = 4.

Day 1: reasoning

Explain, using a Riemann sum, why the units of \int_a^b r(t)\,dt are the units of r times the units of t.
Bagus writes V(6) = \int_0^6 V'(t)\,dt for the water in his tank. When is this right, and what is the correct statement in general?
Two batteries are charged by the same rate r(t) for 0 \le t \le 2. One starts with 10 kilowatt-hours and the other with 30. Compare their net changes and their charges at t = 2.

Day 2: rates from tables

Oil flows into a tank, and the rate is read every hour for 0 \le t \le 4: 120, 150, 160, 140, 100 barrels per hour. Estimate the oil that flowed in with the trapezoid sum.
A bus’s battery holds 300 kilowatt-hours at t = 0. The power it supplies to the motor, read at t = 0, 15, 30, 45, 60 minutes, is 80, 110, 95, 120, 70 kilowatts. Estimate the energy used in kilowatt-hours with the trapezoid sum, and the charge left at t = 60.
A city of 86{,}400 people grows only by births minus deaths. Its net growth rate, read at t = 0, 3, 5, 10 years, is 900, 1100, 1300, 1000 people per year. Estimate the population at t = 10 with trapezoids.

Day 2: rates from graphs

A home battery is charged by solar panels and supplies the house through one connection. The power into it, r(t) kilowatts, t hours after noon (negative while the battery supplies the house), is piecewise linear, with straight segments joining (0, 3), (2, 3), (4, 0), (6, -2) and (8, -2). At noon it holds 4 kilowatt-hours. Find the net change from noon to 8 p.m., the charge at 8 p.m., the largest charge and when it occurs, and the total energy that passed through the connection.
For the battery in Exercise 15, find \int_3^7 r(t)\,dt and say what it means.

Day 2: density

A rod 3 meters long has linear density \rho(x) = 4 + 0.6x^2 kilograms per meter at x meters from one end. Find its mass.
Along a road, the population density is \delta(x) = 2000e^{-0.2x} people per kilometer at x kilometers from a town center. About how many people live along the first 10 kilometers?
A beam 2 meters long has linear density read every half meter: 12, 10, 9, 8.5, 8 kilograms per meter. Estimate its mass with trapezoids.
A rod has linear density \rho(x) = 3 + 2x kilograms per meter. How long a piece, measured from x = 0, has mass 10 kilograms?

Day 2: reasoning

Over one day the net flow rate into a tank has \int_0^{24} r(t)\,dt = 0. Does that mean no liquid moved? What does it say about the tank?
A rate is read at t = 0, 2, 4 and all three readings are 10. Explain why \int_0^4 r(t)\,dt need not be 40, and say what extra fact would make 40 exact.
A rod has positive density \rho(x), and M(x) is the mass from 0 to x. Explain why M is increasing, and why the graph of M is concave up where the rod gets denser.

Answers

Watt-hours. During the first 12 minutes of the flight, the battery’s charge fell by 38 watt-hours.
\int_5^8 P'(t)\,dt is the net change in the town’s population from 2005 to 2008, in people. Adding P(5) gives P(8), the population in 2008, in people.
\int_0^4 (50 - 8t)\,dt = 200 - 64 = 136 kilowatt-hours gained, so it holds 64 + 136 = 200 kilowatt-hours.
An antiderivative is 30t + 4t^{3/2}, so \displaystyle \int_0^9 (30 + 6\sqrt{t})\,dt = 270 + 108 = 378 barrels flow in, so the tank holds 1200 + 378 = 1578 barrels.
An antiderivative is \displaystyle -\frac{24}{\pi}\cos\left(\frac{\pi t}{4}\right), so the depth is \displaystyle \frac{24}{\pi} + \frac{24}{\pi} = \frac{48}{\pi} \approx 15.28 millimeters.
Entered: \big[400t - 25t^2\big]_0^6 = 2400 - 900 = 1500. Left: \big[100t + 20t^2\big]_0^6 = 600 + 720 = 1320. Count: 220 + 1500 - 1320 = 400. From t = 5 to t = 6: \int_5^6 (300 - 90t)\,dt = 300 - 45(36 - 25) = -195, so the count fell by 195 cars.
With u = t^2 + 9, an antiderivative is 3000\ln(t^2 + 9), so the growth is \displaystyle 3000\ln 25 - 3000\ln 9 = 3000\ln\frac{25}{9} \approx 3065 people.
An antiderivative of 90e^{-t/2} - 15 is -180e^{-t/2} - 15t, so the net change is (-180e^{-2} - 60) - (-180) = 120 - 180e^{-2} \approx 95.64 liters, so the tank holds about 200 + 95.64 \approx 295.64 liters.
A Riemann sum adds terms r(t_k)\,\Delta t, each a rate times a time, so each term and the whole sum carry the units of r times the units of t. A limit of such sums keeps those units.
It is right only when V(0) = 0, an empty tank at the start. In general V(6) = V(0) + \int_0^6 V'(t)\,dt.
Both net changes are \int_0^2 r(t)\,dt, the same number. The charges at t = 2 differ by 20 kilowatt-hours, the same gap they started with.
With \Delta t = 1 the trapezoid sum is \displaystyle \frac{1}{2}(120 + 900 + 100) = 560 barrels.
With \Delta t = 15 minutes the trapezoid sum is \displaystyle \frac{15}{2}(80 + 650 + 70) = 6000 kilowatt-minutes, and 6000 \div 60 = 100 kilowatt-hours. About 300 - 100 = 200 kilowatt-hours are left.
The three trapezoids are \displaystyle 3 \cdot \frac{900 + 1100}{2} = 3000, \displaystyle 2 \cdot \frac{1100 + 1300}{2} = 2400 and \displaystyle 5 \cdot \frac{1300 + 1000}{2} = 5750. They add to 3000 + 2400 + 5750 = 11{,}150, so about 86{,}400 + 11{,}150 = 97{,}550 people.
Area above the axis: 6 + 3 = 9; below: 2 + 4 = 6. Net change 9 - 6 = 3 kilowatt-hours, so at 8 p.m. it holds 4 + 3 = 7 kilowatt-hours. The charge is largest when r turns negative, at 4 p.m.: 4 + 9 = 13 kilowatt-hours. Total through the connection: 9 + 6 = 15 kilowatt-hours.
On [3, 4], r falls from 1.5 to 0, area 0.75; on [4, 6] and [6, 7] the areas are -2 and -2. So \int_3^7 r(t)\,dt = 0.75 - 2 - 2 = -3.25: from 3 p.m. to 7 p.m. the battery’s charge fell by 3.25 kilowatt-hours.
\int_0^3 (4 + 0.6x^2)\,dx = 12 + 0.2 \cdot 27 = 17.4 kilograms.
An antiderivative is -10{,}000e^{-0.2x}, so \int_0^{10} 2000e^{-0.2x}\,dx = 10{,}000\big(1 - e^{-2}\big) \approx 8647 people.
With \Delta x = 0.5 the trapezoid sum is \displaystyle \frac{0.5}{2}(12 + 55 + 8) = 0.25(75) = 18.75 kilograms.
\int_0^a (3 + 2x)\,dx = 3a + a^2 = 10, so (a + 5)(a - 2) = 0 and, since a length is positive, a = 2 meters.
No. Liquid may have flowed in and out in equal amounts; \int_0^{24} |r(t)|\,dt measures how much moved. The integral says only that the tank ends the day holding what it started with.
The readings say nothing about the rate between them. For example \displaystyle r(t) = 10 + 6\sin^2\left(\frac{\pi t}{2}\right) equals 10 at all three times, yet its integral over [0, 4] is 52. If the rate were known to be constant on [0, 4] (or linear on each piece), 40 would be exact.
M'(x) = \rho(x) > 0 by the net change theorem, so M is increasing. M''(x) = \rho'(x), which is positive where the density increases, so there M is concave up: each further meter of rod adds more mass than the one before.