Chapter 11 — Front Matter
Purpose and Overview
This chapter sits at the heart of the calculus course. Everything that came before it — limits, derivatives, antiderivatives, Riemann sums, and definite integrals — has been building toward a single, extraordinary discovery: differentiation and integration are not two separate subjects. They are inverse processes, two sides of the same mathematical structure, connected through the concept of accumulation.
The Fundamental Theorem of Calculus is not a computational trick. It is a deep truth about how rates of change and accumulated quantities relate to each other. When students understand this theorem, they understand why calculus is unified, why it is powerful, and why it became the foundation of modern mathematics, physics, engineering, economics, and data science.
Chapter 11 develops both parts of the Fundamental Theorem carefully. It begins by asking why differentiation and integration should be connected at all, then builds the idea of accumulation functions, derives both parts of the theorem conceptually and computationally, and applies the theorem to motion, science, economics, and modeling. It also establishes the Net Change Theorem and the Average Value of a function as important consequences.
Students should leave this chapter understanding not just how to use the Fundamental Theorem, but what it means.
Opening Question
A car travels along a highway. Its velocity at every moment is known — perhaps from a speedometer reading recorded as a function of time. You want to know the total distance the car traveled between two moments in time.
You already know one approach: approximate the distance by multiplying velocity by small time intervals and summing — a Riemann sum. Make the time intervals smaller, take the limit, and you get a definite integral.
But here is a different question. If you happen to know a position function — a function whose derivative gives the velocity — can you find the exact distance without summing anything? Can you compute a definite integral simply by evaluating an antiderivative at two points?
The answer is yes. The Fundamental Theorem of Calculus explains exactly why. And once you see why, the mechanics of integration become simple and the meaning becomes profound.
14Sections
11Objectives
16Key Terms
12+Worked Examples
Learning Objectives
- 1Explain why differentiation and integration are related as inverse processes.
- 2Define an accumulation function and explain what it measures.
- 3State and interpret the Fundamental Theorem of Calculus, Part 1.
- 4Differentiate accumulation functions using the Fundamental Theorem, Part 1.
- 5State and apply the Fundamental Theorem of Calculus, Part 2.
- 6Evaluate definite integrals using antiderivatives.
- 7Apply the Net Change Theorem to interpret definite integrals as total change.
- 8Compute the average value of a function over an interval.
- 9Interpret the units of a definite integral in context.
- 10Apply the Fundamental Theorem to motion, science, and economics problems.
- 11Identify and correct common Fundamental Theorem mistakes.
Sections 11.1 – 11.3
The Connection Between Differentiation and Integration
Why They Are Inverses · Accumulation Functions · The Area Function Idea
Section 11.1
Why Differentiation and Integration Are Connected
Before stating the Fundamental Theorem of Calculus, it is worth asking why anyone would expect differentiation and integration to be connected at all. They appear, on the surface, to be quite different operations.
Differentiation zooms in. It asks: at this precise moment, how fast is this quantity changing? It produces a rate. The derivative of a position function is velocity. The derivative of a cost function is marginal cost. The derivative of a temperature function is the rate at which temperature is changing. Differentiation is a local operation — it describes behavior at a single point.
Integration accumulates. It asks: over this entire interval, how much of this quantity has built up? It produces a total. The integral of a velocity function over a time interval gives distance traveled. The integral of a rate of flow gives total volume. The integral of a marginal cost function gives total cost over a range of production. Integration is a global operation — it describes accumulated behavior over an interval.
These two ideas seem opposite in character. And yet they undo each other. This is not a coincidence. It is a consequence of the way continuous change and continuous accumulation are related.
Think about it this way. Suppose you are watching a quantity accumulate. At every moment, something is being added to the total — perhaps water flowing into a tank, money accumulating in an account, or distance building up as a car moves. The rate at which the total is growing at any moment is exactly the rate of flow at that moment. The total accumulated so far is the integral of the rate. But the derivative of the total — how fast the total is growing — is the rate. So differentiation undoes integration. And integration undoes differentiation. This is the Fundamental Theorem.
| Operation | Character | What it produces | Example |
|---|
| Differentiation | Local — zooms in at one point | A rate of change | \(v(t) = s'(t)\): velocity from position |
| Integration | Global — accumulates over an interval | A total quantity | \(\int_a^b v(t)\,dt\): displacement from velocity |
Section 11.2
Accumulation Functions
To understand the Fundamental Theorem, it helps to think carefully about what happens when the upper limit of integration is a variable.
Suppose \(f\) is a continuous function on an interval \([a, b]\), and let \(x\) be any value in \([a, b]\). Define a new function \(A(x)\) as:
$$A(x) = \int_a^x f(t)\, dt$$
This function \(A(x)\) is called an accumulation function. It tells you the total signed area under the graph of \(f\) from \(a\) up to \(x\).
Notice several things about this definition. First, \(x\) is the variable — it is the upper limit of integration, and it changes. Second, the variable inside the integral is \(t\), not \(x\). This is important. The variable of integration must be different from the limit of integration. The letter \(t\) is a dummy variable — it represents the input values of \(f\) as the integration runs from \(a\) to \(x\). Third, \(A(a) = \int_a^a f(t)\,dt = 0\) — starting and ending at the same point accumulates nothing. Fourth, if \(f(t) > 0\) throughout, then \(A(x)\) is positive and increasing.
Example 11.2.1
Evaluating an Accumulation Function
- ProblemLet \(f(t) = 3\). Define \(A(x) = \int_0^x 3\,dt\). Find \(A(0)\), \(A(2)\), and \(A(5)\).
- ThinkSince \(f(t) = 3\) is a constant, the integral from 0 to \(x\) is the area of a rectangle with height 3 and width \(x\).
- Calculate
\[A(x) = \int_0^x 3\, dt = 3x\]
\[A(0) = 0, \quad A(2) = 6, \quad A(5) = 15\]
- CheckTaking the derivative: \(A'(x) = 3 = f(x)\). The derivative of the accumulation function gives back the integrand. This is exactly what the Fundamental Theorem will say.
Answer: \(A(0) = 0\), \(A(2) = 6\), \(A(5) = 15\). The accumulation function is \(A(x) = 3x\), a linear function whose slope equals the integrand value.
Graph Instruction
Draw the graph of the constant function \(f(t) = 3\) on the horizontal axis from \(t = 0\) to \(t = x\). Shade the rectangular region beneath it. Label its width as \(x\) and its height as 3. Show that as \(x\) increases from 0 to 5, the shaded area grows linearly from 0 to 15. This illustrates that the accumulation function \(A(x)\) is a linear function whose slope equals the integrand value.
Section 11.3
The Area Function Idea
To build deeper intuition for the Fundamental Theorem, consider what happens to the accumulation function \(A(x) = \int_a^x f(t)\,dt\) when \(x\) increases by a small amount \(h\). The additional accumulation from \(x\) to \(x + h\) is:
$$A(x+h) - A(x) = \int_a^{x+h} f(t)\, dt - \int_a^x f(t)\, dt = \int_x^{x+h} f(t)\, dt$$
This is the signed area of a thin vertical strip of width \(h\) under the graph of \(f\), starting at \(x\). If \(f\) is continuous and \(h\) is very small, this thin strip is approximately a rectangle with height \(f(x)\) and width \(h\):
$$\int_x^{x+h} f(t)\, dt \approx f(x) \cdot h$$
$$\frac{A(x+h) - A(x)}{h} \approx f(x)$$
As \(h \to 0\), this approximation becomes exact:
$$A'(x) = \lim_{h \to 0} \frac{A(x+h) - A(x)}{h} = f(x)$$
This is the key insight. The rate at which the accumulation function grows at any point \(x\) is exactly \(f(x)\) — the value of the function being integrated. The accumulation function is an antiderivative of the integrand. This is the heart of the Fundamental Theorem of Calculus, Part 1.
Graph Instruction
Draw a continuous curve \(y = f(t)\) above the horizontal axis. Mark a point \(x\) on the horizontal axis and a nearby point \(x + h\). Shade the full region under the curve from \(a\) to \(x + h\). Then separately outline the thin strip from \(x\) to \(x + h\). Label the strip's width as \(h\) and approximate its height as \(f(x)\). The strip's area is approximately \(f(x) \cdot h\), which means the slope of \(A\) at \(x\) is approximately \(f(x)\). This picture is what the derivative definition says, applied to the accumulation function.
Sections 11.4 – 11.9
The Fundamental Theorem: Both Parts
FTC Part 1 · Derivatives of Accumulation Functions · FTC Part 2 · Evaluating Integrals · Net Change · Average Value
Section 11.4
Fundamental Theorem of Calculus, Part 1
Theorem
Fundamental Theorem of Calculus, Part 1
Let \(f\) be a continuous function on \([a, b]\). Define the accumulation function:
\[A(x) = \int_a^x f(t)\, dt\]
Then \(A\) is differentiable on \((a, b)\), and:
\[A'(x) = f(x)\]
In other words, the accumulation function is an antiderivative of \(f\).
This statement says something remarkable. When you build a function by integrating \(f\) from a fixed starting point to a variable upper limit, the result is a function whose derivative is \(f\) itself. Integration, followed by differentiation, gives you back the original function.
Part 1 of the Fundamental Theorem has two important consequences. First, it guarantees that every continuous function has an antiderivative — the accumulation function \(A(x) = \int_a^x f(t)\,dt\) is always a valid antiderivative of \(f\). Second, it gives a direct method for differentiating accumulation functions, which we develop in the next section.
Section 11.5
Derivatives of Accumulation Functions
The Rule — Two Forms
Basic form (upper limit is \(x\)):
\[\frac{d}{dx} \int_a^x f(t)\, dt = f(x)\]
Chain rule form (upper limit is \(g(x)\)):
\[\frac{d}{dx} \int_a^{g(x)} f(t)\, dt = f(g(x)) \cdot g'(x)\]
When the variable is in the lower limit, reverse the limits and introduce a negative sign before applying the rule.
Example 11.5.1
Basic Accumulation Function Derivative
- ProblemFind \(\dfrac{d}{dx} \int_1^x (t^2 + 3t)\, dt\).
- ThinkThe upper limit is \(x\) and the lower limit is the constant 1. Apply Part 1 directly — replace \(t\) with \(x\) in the integrand.
- Differentiate\[\frac{d}{dx} \int_1^x (t^2 + 3t)\, dt = x^2 + 3x\]
- CheckThe result is the integrand evaluated at \(t = x\). No integration is needed — just substitution.
- InterpretThe accumulation function \(A(x) = \int_1^x(t^2+3t)\,dt\) grows at rate \(x^2 + 3x\) at each value of \(x\). At \(x = 2\), area is accumulating at a rate of \(4 + 6 = 10\) square units per unit of \(x\).
Answer: \(x^2 + 3x\)
Example 11.5.2
Accumulation Function with Chain Rule
- ProblemFind \(\dfrac{d}{dx} \int_0^{x^3} \sin(t)\, dt\).
- ThinkThe upper limit is \(g(x) = x^3\), not just \(x\). Apply the chain rule version of Part 1.
- Differentiate\[\frac{d}{dx} \int_0^{x^3} \sin(t)\, dt = \sin(x^3) \cdot \frac{d}{dx}[x^3] = \sin(x^3) \cdot 3x^2\]
- CheckThe integrand evaluated at the upper limit is \(\sin(x^3)\). Then multiply by the derivative of the upper limit, \(3x^2\). Both factors are needed.
Answer: \(3x^2\sin(x^3)\)
Example 11.5.3
Reversed Limits
- ProblemFind \(\dfrac{d}{dx} \int_x^5 e^{t^2}\, dt\).
- ThinkThe variable \(x\) is in the lower limit. Reverse the limits and introduce a negative sign: \(\int_x^5 e^{t^2}\,dt = -\int_5^x e^{t^2}\,dt\).
- Differentiate\[\frac{d}{dx}\left[-\int_5^x e^{t^2}\, dt\right] = -e^{x^2}\]
Answer: \(-e^{x^2}\)
Conceptual Note
Many students want to compute the integral before differentiating. This defeats the purpose. Part 1 is a shortcut: the derivative of an accumulation function is simply the integrand evaluated at the upper limit, times the derivative of the upper limit if needed. No integration required.
Section 11.6
Fundamental Theorem of Calculus, Part 2
Theorem
Fundamental Theorem of Calculus, Part 2
Let \(f\) be a continuous function on \([a, b]\), and let \(F\) be any antiderivative of \(f\) — that is, any function with \(F'(x) = f(x)\). Then:
\[\int_a^b f(x)\, dx = F(b) - F(a)\]
This is the theorem that makes calculus computationally powerful. To find the exact value of a definite integral, you no longer need Riemann sums. You find an antiderivative, evaluate it at the upper limit \(b\) and the lower limit \(a\), and subtract.
Why any antiderivative works. If \(F\) and \(G\) are both antiderivatives of \(f\), they differ only by a constant: \(G(x) = F(x) + C\). Then \(G(b) - G(a) = [F(b)+C] - [F(a)+C] = F(b) - F(a)\). The constant cancels.
Bracket Notation
The expression \(F(b) - F(a)\) is often written using bracket notation:
\[\int_a^b f(x)\, dx = \Big[F(x)\Big]_a^b = F(b) - F(a)\]
This bracket notation signals: evaluate the antiderivative at the upper limit, then subtract its value at the lower limit.
Section 11.7
Evaluating Definite Integrals with Antiderivatives
With Part 2 in hand, evaluating definite integrals becomes a three-step process: (1) find an antiderivative \(F(x)\), (2) evaluate \(F\) at \(b\) and \(a\), (3) compute \(F(b) - F(a)\). No Riemann sums. No limits. Just antidifferentiation and subtraction.
Example 11.7.1
Evaluating a Polynomial Integral
- ProblemEvaluate \(\int_1^4 (3x^2 - 2x + 1)\, dx\).
- ThinkFind the antiderivative using the power rule, then apply Part 2.
- Antiderivative\[F(x) = x^3 - x^2 + x\]
- Evaluate
\[F(4) = 64 - 16 + 4 = 52\]
\[F(1) = 1 - 1 + 1 = 1\]
\[\int_1^4 (3x^2 - 2x + 1)\, dx = 52 - 1 = 51\]
- CheckVerify \(F'(x) = 3x^2 - 2x + 1\). Signs and powers match the integrand. ✓
- InterpretThe net signed area under \(f(x) = 3x^2 - 2x + 1\) from \(x=1\) to \(x=4\) is 51 square units.
Answer: 51
Example 11.7.2
Evaluating a Trigonometric Integral
- ProblemEvaluate \(\int_0^{\pi} \sin(x)\, dx\).
- Antiderivative\(F(x) = -\cos(x)\)
- Evaluate
\[F(\pi) = -\cos(\pi) = -(-1) = 1\]
\[F(0) = -\cos(0) = -1\]
\[\int_0^{\pi} \sin(x)\, dx = 1 - (-1) = 2\]
- CheckThe graph of \(\sin(x)\) lies entirely above the horizontal axis from 0 to \(\pi\), so the area is positive. A value of 2 is reasonable.
- InterpretThe total signed area under one arch of the sine curve, from \(x = 0\) to \(x = \pi\), is exactly 2.
Answer: 2
Example 11.7.3
Evaluating an Exponential Integral
- ProblemEvaluate \(\int_0^2 e^x\, dx\).
- Antiderivative\(F(x) = e^x\)
- Evaluate\[\int_0^2 e^x\, dx = e^2 - e^0 = e^2 - 1 \approx 6.389\]
- InterpretThe area under \(y = e^x\) from \(x = 0\) to \(x = 2\) is \(e^2 - 1\), an exact result that cannot be simplified further.
Answer: \(e^2 - 1 \approx 6.389\)
Example 11.7.4
A Negative Integrand
- ProblemEvaluate \(\int_0^1 (x^2 - 3)\, dx\).
- ThinkOn \([0,1]\), \(x^2 \leq 1 < 3\), so \(x^2 - 3 < 0\) throughout. The result should be negative (signed area below the axis).
- Antiderivative\(F(x) = \dfrac{x^3}{3} - 3x\)
- Evaluate
\[F(1) = \frac{1}{3} - 3 = -\frac{8}{3}, \quad F(0) = 0\]
\[\int_0^1 (x^2 - 3)\, dx = -\frac{8}{3}\]
- CheckThe function is negative throughout \([0,1]\), so a negative signed area confirms our setup is correct.
Answer: \(-\dfrac{8}{3}\)
Section 11.8
Net Change Theorem
Theorem
Net Change Theorem
If \(F'(x) = f(x)\), then the integral of the rate of change over an interval equals the net change in the quantity:
\[\int_a^b F'(x)\, dx = F(b) - F(a)\]
The net change in \(F\) from \(x = a\) to \(x = b\) equals the integral of its derivative \(F'\).
| Rate of Change \(f(t)\) | Integral | Meaning |
|---|
| \(v(t)\) — velocity (ft/s) | \(\int_a^b v(t)\,dt\) | Net displacement (ft) |
| \(r(t)\) — flow rate (liters/hr) | \(\int_a^b r(t)\,dt\) | Total volume (liters) |
| \(C'(x)\) — marginal cost ($/unit) | \(\int_{x_1}^{x_2} C'(x)\,dx\) | Additional cost ($) |
| \(P'(t)\) — population growth (people/yr) | \(\int_a^b P'(t)\,dt\) | Net population change (people) |
Example 11.8.1
Net Displacement from Velocity
- ProblemA particle moves along a number line. Its velocity is \(v(t) = t^2 - 4t + 3\) ft/s. Find the net displacement from \(t = 0\) to \(t = 3\).
- ThinkNet displacement = \(\int_0^3 v(t)\,dt\). The velocity may be positive or negative on this interval, but the integral gives net signed displacement.
- Antiderivative\(F(t) = \dfrac{t^3}{3} - 2t^2 + 3t\)
- Evaluate
\[F(3) = 9 - 18 + 9 = 0, \quad F(0) = 0\]
\[\int_0^3 v(t)\, dt = 0\]
- Check\(v(t) = (t-1)(t-3)\) — positive on \([0,1)\), negative on \((1,3)\). The particle moved forward, then reversed, returning to its starting point. ✓
- InterpretThe net displacement is 0 feet. The particle returned to its starting position after 3 seconds. It moved forward and backward, but the signed areas cancelled.
Answer: Net displacement = 0 feet
Net Displacement vs. Total Distance
The net displacement integral allows positive and negative velocity to cancel. Total distance does not allow cancellation. To find total distance, split the integral at points where \(v(t) = 0\) and add the absolute values: \(\int_0^1 |v(t)|\,dt + \int_1^3 |v(t)|\,dt\).
Example 11.8.2
Accumulated Flow
- ProblemWater flows into a reservoir at \(r(t) = 200 + 50\sin(t)\) liters per hour. How much water enters from \(t = 0\) to \(t = 4\)?
- Antiderivative\(F(t) = 200t - 50\cos(t)\)
- Evaluate
\[F(4) = 800 - 50\cos(4), \quad F(0) = -50\cos(0) = -50\]
\[\int_0^4 r(t)\,dt = [800 - 50\cos(4)] - (-50) = 850 - 50\cos(4)\]
- CalculateUsing \(\cos(4) \approx -0.6536\): \[\approx 850 + 32.68 \approx 882.68 \text{ liters}\]
- InterpretApproximately 882.68 liters enter the reservoir during the first 4 hours. Units check: liters/hr × hr = liters. ✓
Answer: \(850 - 50\cos(4) \approx 882.7\) liters
Section 11.9
Average Value of a Function
Definition
Average Value of a Function
If \(f\) is continuous on \([a, b]\), the average value of \(f\) on \([a, b]\) is:
\[f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x)\, dx\]
The average value is the height of a rectangle with width \(b - a\) that has the same area as the area under the graph of \(f\) from \(a\) to \(b\).
The Mean Value Theorem for Integrals says: if \(f\) is continuous on \([a, b]\), there is at least one point \(c\) in \((a, b)\) where \(f(c) = f_{\text{avg}}\). The function actually attains its average value somewhere in the interval.
Example 11.9.1
Average Value of a Quadratic
- ProblemFind the average value of \(f(x) = x^2\) on \([0, 3]\).
- Calculate
\[f_{\text{avg}} = \frac{1}{3-0}\int_0^3 x^2\, dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \cdot \frac{27}{3} = \frac{1}{3} \cdot 9 = 3\]
- InterpretThe average value of \(x^2\) on \([0,3]\) is 3. The function equals its average at \(x = \sqrt{3} \approx 1.73\), which lies in \((0,3)\) as the Mean Value Theorem promises.
Answer: \(f_{\text{avg}} = 3\)
Example 11.9.2
Average Temperature
- ProblemTemperature in a city over 12 hours: \(T(t) = 60 + 10\sin\!\left(\tfrac{\pi t}{12}\right)\)°F, where \(t\) is hours after midnight. Find the average temperature from \(t = 0\) to \(t = 12\).
- Antiderivative\(F(t) = 60t - \dfrac{120}{\pi}\cos\!\left(\dfrac{\pi t}{12}\right)\)
- Evaluate
\[F(12) = 720 + \frac{120}{\pi}, \quad F(0) = -\frac{120}{\pi}\]
\[\int_0^{12} T(t)\,dt = 720 + \frac{240}{\pi}\]
\[T_{\text{avg}} = \frac{1}{12}\left(720 + \frac{240}{\pi}\right) = 60 + \frac{20}{\pi} \approx 66.4^\circ\text{F}\]
- InterpretThe average temperature over the 12-hour period is approximately 66.4°F — higher than the baseline of 60°F because the sine function lifts the average during the morning hours.
Answer: \(60 + \dfrac{20}{\pi} \approx 66.4°\text{F}\)
Sections 11.10 – 11.14
Units, Motion, Science, and Looking Ahead
Units in Integrals · Motion Problems · Science & Economics · Common Mistakes · Preparing for Chapter 12
Section 11.10
Interpreting Units in Integrals
If \(f(x)\) is measured in units of \(\frac{\text{output}}{\text{input}}\) and \(x\) is measured in input units, then \(\int_a^b f(x)\,dx\) is measured in units of output. The product \(f(x)\,dx\) has units of (output/input) × input = output. Summing these products over the interval gives a total in output units.
| Integrand | Variable | Units of Integral | Meaning |
|---|
| \(v(t)\) in m/s | \(t\) in seconds | meters | Displacement |
| \(C'(x)\) in $/unit | \(x\) in units | dollars | Cost change |
| \(r(t)\) in gal/min | \(t\) in minutes | gallons | Total volume |
| \(P'(t)\) in people/yr | \(t\) in years | people | Population change |
| \(F(x)\) in Newtons | \(x\) in meters | Joules (N·m) | Work |
Example 11.10.1
Unit Check — Factory Production
- ProblemA factory produces widgets at rate \(R(t) = 120 + 5t\) widgets per hour, where \(t\) is hours after opening. How many widgets are produced in the first 8 hours?
- Units\(R(t)\) is widgets/hour, \(dt\) is hours. Product: widgets/hr × hr = widgets. ✓
- Calculate
\[\int_0^8 (120 + 5t)\,dt = \left[120t + \frac{5t^2}{2}\right]_0^8 = 960 + 160 = 1120\]
- InterpretThe factory produces 1120 widgets in the first 8 hours.
Answer: 1120 widgets
Section 11.11
FTC in Motion Problems
Motion Relationships
Position \(s(t)\), velocity \(v(t) = s'(t)\), and acceleration \(a(t) = v'(t)\) are connected by differentiation and integration.
\[v(b) - v(a) = \int_a^b a(t)\, dt \quad\text{(net change in velocity)}\]
\[s(b) - s(a) = \int_a^b v(t)\, dt \quad\text{(net displacement)}\]
\[\text{Total distance} = \int_a^b |v(t)|\, dt \quad\text{(split at sign changes)}\]
Example 11.11.1
Finding Displacement and Position
- ProblemA particle has velocity \(v(t) = 6t - t^2\) ft/s. At \(t = 0\), its position is \(s(0) = 4\) feet. Find the position at \(t = 3\).
- ThinkNet displacement from 0 to 3 = \(\int_0^3 v(t)\,dt\). Then \(s(3) = s(0) + \int_0^3 v(t)\,dt\).
- Calculate
\[\int_0^3 (6t - t^2)\, dt = \left[3t^2 - \frac{t^3}{3}\right]_0^3 = [27 - 9] - 0 = 18\]
\[s(3) = 4 + 18 = 22 \text{ feet}\]
- InterpretAt \(t = 3\) seconds the particle is at position 22 feet. The definite integral gave the net displacement; adding the initial position gave the final position.
Answer: \(s(3) = 22\) feet
Example 11.11.2
Total Distance vs. Net Displacement
- ProblemUsing \(v(t) = 6t - t^2\), find the total distance traveled from \(t = 0\) to \(t = 7\).
- Think\(v(t) = t(6-t)\) is zero at \(t = 0\) and \(t = 6\). Positive on \([0,6]\), negative on \([6,7]\). Split the integral at \(t = 6\) and add absolute values.
- Calculate
\[\int_0^6 (6t-t^2)\,dt = \left[3t^2-\frac{t^3}{3}\right]_0^6 = 108 - 72 = 36\]
\[\int_6^7 (6t-t^2)\,dt = \left[3t^2-\frac{t^3}{3}\right]_6^7 = \left[147-\frac{343}{3}\right] - 36 = \frac{98}{3} - 36 = -\frac{10}{3}\]
\[\text{Total distance} = 36 + \frac{10}{3} = \frac{118}{3} \approx 39.33 \text{ feet}\]
- InterpretThe particle travels 36 ft forward, then \(\tfrac{10}{3}\) ft back, for a total of \(\tfrac{118}{3}\) ft. Net displacement would be \(\tfrac{98}{3}\) ft — less than total distance because backward motion was cancelled.
Answer: Total distance = \(\dfrac{118}{3} \approx 39.33\) feet
Section 11.12
FTC in Science and Economics
The Fundamental Theorem applies across every field where rates of change matter. The Net Change Theorem provides the framework: integrate a rate of change to find the net change in the quantity.
Example 11.12.1
Marginal Cost and Total Additional Cost
- ProblemA manufacturer's marginal cost is \(C'(x) = 0.06x^2 - 4x + 110\) dollars per unit, where \(x\) is units produced. Find the additional cost of increasing production from 20 to 50 units.
- SetupAdditional cost = \(\int_{20}^{50} C'(x)\,dx\)
- Antiderivative\(F(x) = 0.02x^3 - 2x^2 + 110x\)
- Evaluate
\[F(50) = 2500 - 5000 + 5500 = 3000\]
\[F(20) = 160 - 800 + 2200 = 1560\]
\[\int_{20}^{50} C'(x)\,dx = 3000 - 1560 = \$1{,}440\]
- InterpretIt costs an additional \$1,440 to increase production from 20 to 50 units. The marginal cost function, integrated, gives the total incremental cost.
Answer: \$1,440
Example 11.12.2
Population Change
- ProblemA city's population growth rate is \(P'(t) = 1200e^{0.02t}\) people per year, where \(t\) is years after 2020. How much does the population grow from 2020 to 2030?
- Antiderivative\(F(t) = \dfrac{1200}{0.02}e^{0.02t} = 60000e^{0.02t}\)
- Evaluate
\[F(10) = 60000e^{0.2} \approx 73{,}284, \quad F(0) = 60000\]
\[\int_0^{10} 1200e^{0.02t}\,dt \approx 73{,}284 - 60{,}000 = 13{,}284\]
- InterpretThe city's population grows by approximately 13,284 people between 2020 and 2030, given the modeled growth rate.
Answer: approximately 13,284 people
Section 11.13
Common Fundamental Theorem Mistakes
Mistake 1 — Forgetting to subtract F(a)
\(\int_1^3 x^2\,dx = \left[\frac{x^3}{3}\right] = \frac{27}{3} = 9\)
Correct: \(\int_1^3 x^2\,dx = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}\). The formula is \(F(b) - F(a)\), not just \(F(b)\).
Mistake 2 — Including +C in a definite integral
Writing \(\left[\frac{x^3}{3} + C\right]_1^3\) then treating \(C\) as a number
For definite integrals, the constant cancels: \([F(b)+C]-[F(a)+C] = F(b)-F(a)\). Do not include \(+C\) when evaluating definite integrals.
Mistake 3 — Confusing net displacement with total distance
"The total distance = \(\int_0^6 v(t)\,dt = 0\)"
\(\int_a^b v(t)\,dt\) gives net displacement, not total distance. If velocity changes sign, the positive and negative contributions cancel. Total distance requires integrating \(|v(t)|\), splitting at sign changes.
Mistake 4 — Forgetting the chain rule in Part 1
\(\dfrac{d}{dx}\int_0^{x^3}\sin(t)\,dt = \sin(x^3)\)
Correct: \(\sin(x^3) \cdot 3x^2\). When the upper limit is \(g(x)\), multiply by \(g'(x)\). Forgetting this factor is one of the most common Part 1 errors.
Mistake 5 — Mixing up Part 1 and Part 2
Using Part 2 to "evaluate" an accumulation function that should be differentiated with Part 1
Part 1: differentiating an accumulation function with variable upper limit. Part 2: evaluating a definite integral using antiderivatives. These serve different purposes.
Mistake 6 — Assuming the integral always gives unsigned area
"The area under \(f\) is \(\int_a^b f(x)\,dx\) regardless of sign."
The definite integral gives signed area. Regions below the axis contribute negative values. To find total unsigned area, split the integral at sign changes and add absolute values.
Mistake 7 — Using the same variable inside and outside the integral
\(A(x) = \int_a^x f(x)\,dx\)
The variable of integration must be different from the limit. Write \(\int_a^x f(t)\,dt\). Using \(x\) for both the dummy variable and the upper limit creates ambiguity and errors.
Mistake 8 — Evaluating limits in the wrong order
\(\int_a^b f(x)\,dx = F(a) - F(b)\)
The formula is \(F(b) - F(a)\), not \(F(a) - F(b)\). Reversing the order changes the sign of the entire result.
Section 11.14
Preparing for Integration Techniques and Applications
The Fundamental Theorem of Calculus is the foundation, but calculus does not stop here. Chapter 12 applies definite integrals to find areas between curves, volumes of solids of revolution, accumulated change, and physical and economic quantities. All of those applications depend on the ability to set up a definite integral and evaluate it using antiderivatives — exactly what Part 2 enables.
Beyond Chapter 12, more advanced integration techniques — substitution, integration by parts, partial fractions, trigonometric substitution — will be explored in Calculus II. These are methods for finding antiderivatives of more complicated integrands. All of them feed back into Part 2: once you find the antiderivative, evaluate it at the limits and subtract.
The Fundamental Theorem also has deep extensions. In multivariable calculus, it generalizes to Green's Theorem, Stokes' Theorem, and the Divergence Theorem — connecting integrals over regions to integrals over boundaries. The same structure — that integration and differentiation are inverse operations — appears in every dimension.
For now, make sure you understand: \(A(x) = \int_a^x f(t)\,dt\) is a function with \(A'(x) = f(x)\); \(\int_a^b f(x)\,dx = F(b) - F(a)\) for any antiderivative \(F\); the integral of a rate of change gives net change; the average value is the integral divided by the interval length; units matter.
MGU Library Connection
- Calculus II — extends integration with substitution, integration by parts, and improper integrals that build on FTC Part 2
- Physics — work and energy pages use definite integrals computed via FTC Part 2 for variable forces
- Economics — consumer and producer surplus pages apply the Fundamental Theorem to areas between supply and demand curves
- Differential Equations — the Net Change Theorem connects directly to separable equation solutions
- Calculus III — the FTC generalizes to Green's Theorem, Stokes' Theorem, and the Divergence Theorem
Chapter 11 — Back Matter
Chapter Summary
- Sec 11.1Differentiation is local (rates at a point); integration is global (accumulated totals over an interval). They undo each other because the rate at which a total grows is exactly the rate being accumulated.
- Sec 11.2The accumulation function \(A(x) = \int_a^x f(t)\,dt\) measures total signed area from \(a\) to \(x\). The dummy variable \(t\) differs from the limit variable \(x\). \(A(a) = 0\) always.
- Sec 11.3The thin strip argument shows \(A(x+h) - A(x) \approx f(x) \cdot h\), so \(A'(x) = f(x)\). The accumulation function is an antiderivative of the integrand.
- Sec 11.4FTC Part 1: if \(A(x) = \int_a^x f(t)\,dt\), then \(A'(x) = f(x)\). Every continuous function has an antiderivative, even when it cannot be written in closed form.
- Sec 11.5Basic form: \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\). Chain rule form: \(\frac{d}{dx}\int_a^{g(x)}f(t)\,dt = f(g(x))\cdot g'(x)\). Reversed limits require a negative sign.
- Sec 11.6FTC Part 2: \(\int_a^b f(x)\,dx = F(b) - F(a)\) for any antiderivative \(F\). The constant of integration cancels. Bracket notation: \(\big[F(x)\big]_a^b\).
- Sec 11.7Three steps: find an antiderivative, evaluate at \(b\) and \(a\), subtract. Applies to polynomial, trigonometric, exponential, and other integrable functions.
- Sec 11.8Net Change Theorem: \(\int_a^b F'(x)\,dx = F(b) - F(a)\). Integrating a rate of change gives net change. Net displacement and total distance differ when velocity changes sign.
- Sec 11.9Average value: \(f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx\). The Mean Value Theorem for Integrals guarantees the function attains its average value at some point in the interval.
- Sec 11.10Units: (output/input) × input = output. Always check units when setting up an integral; they reveal the meaning and confirm the setup is correct.
- Sec 11.11Motion: \(\int_a^b v(t)\,dt = s(b)-s(a)\) gives net displacement. Total distance requires integrating \(|v(t)|\), splitting the integral at sign changes of velocity.
- Sec 11.12FTC across disciplines: population growth, electrical charge, marginal cost, drug absorption, and production rates all use the Net Change Theorem to find totals from rates.
- Sec 11.13Eight common mistakes: forgetting \(F(a)\), including \(+C\), confusing displacement and distance, missing the chain rule, mixing up Part 1 and 2, assuming positive area, reusing the dummy variable, and reversing the order of evaluation.
- Sec 11.14Chapter 12 applies integrals to area between curves, volumes, and physical applications — all using Part 2. Calculus II will extend integration techniques. The FTC generalizes to higher-dimensional theorems.
Key Terms Review
accumulation function
\(A(x) = \int_a^x f(t)\,dt\). Measures total signed area from \(a\) to \(x\). Its derivative is the integrand: \(A'(x) = f(x)\).
average value
The height of the rectangle with the same area as the region under \(f\): \(f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx\).
dummy variable
The variable of integration inside a definite integral (\(t\) in \(\int_a^x f(t)\,dt\)). It is replaced upon evaluation and is not the same as the limit variable.
FTC, Part 1
If \(A(x) = \int_a^x f(t)\,dt\), then \(A'(x) = f(x)\). Differentiating an accumulation function returns the integrand.
FTC, Part 2
If \(F'(x) = f(x)\), then \(\int_a^b f(x)\,dx = F(b) - F(a)\). Evaluating a definite integral using an antiderivative.
Net Change Theorem
\(\int_a^b F'(x)\,dx = F(b) - F(a)\). The integral of a rate of change over an interval equals the net change in the quantity.
net displacement
Signed difference between final and initial position: \(\int_a^b v(t)\,dt\). Positive and negative velocity contributions may cancel.
signed area
Area measured with sign. Regions below the horizontal axis contribute negative values to a definite integral.
total distance
Unsigned total path length: \(\int_a^b |v(t)|\,dt\). Requires splitting the integral at points where \(v(t) = 0\) changes sign.
bracket notation
\(\big[F(x)\big]_a^b = F(b) - F(a)\). The standard notation for evaluating a definite integral using an antiderivative.
Mean Value Theorem (Integrals)
If \(f\) is continuous on \([a,b]\), there is at least one \(c\) in \((a,b)\) where \(f(c) = f_{\text{avg}}\). The function attains its average value.
net change
The difference between a quantity's final and initial values: \(F(b) - F(a)\). Equal to the integral of the rate of change.
Concept Review Questions
- In your own words, explain why differentiation and integration are inverse processes.
- What does the accumulation function \(A(x) = \int_a^x f(t)\,dt\) measure? How does \(A(x)\) change as \(x\) increases?
- State the Fundamental Theorem of Calculus, Part 1. What does it say about the derivative of an accumulation function?
- What is the difference between FTC Part 1 and Part 2? When do you use each?
- Why does the constant of integration \(C\) not appear when evaluating a definite integral using Part 2?
- State the Net Change Theorem. Give two real-world examples of how it applies.
- A function \(f\) has \(\int_0^5 f(x)\,dx = 12\). What is the average value of \(f\) on \([0, 5]\)?
- How do net displacement and total distance differ for a moving particle? How does the sign of velocity affect each?
- If \(v(t) < 0\) for all \(t\) in \([a,b]\), is the net displacement positive or negative? Is the total distance positive or negative?
- Why is it important to check units when setting up a definite integral in an applied problem?
Evaluate each accumulation function at the given values.
- \(A(x) = \int_0^x (2t + 1)\,dt\). Find \(A(3)\) and \(A(0)\).
- \(A(x) = \int_1^x t^3\,dt\). Find \(A(2)\) and \(A(1)\).
- \(A(x) = \int_0^x e^t\,dt\). Find \(A(1)\) and \(A(2)\).
- \(A(x) = \int_0^x \cos(t)\,dt\). Find \(A(\pi/2)\) and \(A(\pi)\).
- Let \(A(x) = \int_2^x (t^2 - 4)\,dt\). Find \(A(2)\). Explain why the answer is what it is without computing anything.
- \(\dfrac{d}{dx} \int_0^x (t^3 - 2t)\, dt\)
- \(\dfrac{d}{dx} \int_3^x \sqrt{t^2 + 1}\, dt\)
- \(\dfrac{d}{dx} \int_0^x e^{-t^2}\, dt\)
- \(\dfrac{d}{dx} \int_x^7 \sin(t^2)\, dt\)
- \(\dfrac{d}{dx} \int_1^{x^2} \ln(t)\, dt\)
- \(\dfrac{d}{dx} \int_0^{x^3} \cos(t^2)\, dt\)
- \(\dfrac{d}{dx} \int_{\sin(x)}^5 t^4\, dt\)
- \(\dfrac{d}{dx} \int_0^{\sqrt{x}} e^{t}\, dt\)
Evaluate each definite integral exactly.
- \(\int_0^3 (4x - 1)\, dx\)
- \(\int_1^4 x^2\, dx\)
- \(\int_{-1}^{2} (3x^2 - 2x + 5)\, dx\)
- \(\int_0^{\pi/2} \cos(x)\, dx\)
- \(\int_0^{\pi} \sin(x)\, dx\)
- \(\int_1^e \frac{1}{x}\, dx\)
- \(\int_0^2 e^x\, dx\)
- \(\int_0^1 (x^{1/2} + x^{3/2})\, dx\)
- \(\int_1^4 \frac{1}{\sqrt{x}}\, dx\)
- \(\int_0^{\pi/4} \sec^2(x)\, dx\)
- \(\int_0^2 (x^2 - x - 2)\, dx\)
- \(\int_{-2}^{0} (x^3 + x)\, dx\)
- Average value of \(f(x) = 4x\) on \([0, 5]\).
- Average value of \(f(x) = x^2 - 1\) on \([0, 3]\).
- Average value of \(f(x) = \sin(x)\) on \([0, \pi]\).
- Average value of \(f(x) = e^x\) on \([0, 2]\).
- Temperature \(T(t) = 70 + 5\cos\!\left(\frac{\pi t}{12}\right)\)°F. Find average over 24 hours.
- Velocity \(v(t) = 3t^2\) ft/s. Find average velocity from \(t = 0\) to \(t = 4\).
Applications
For each problem, set up and evaluate the integral. State the answer with correct units and interpret the result in context.
- Displacement: A particle's velocity is \(v(t) = 2t - 4\) ft/s. Find the net displacement from \(t = 0\) to \(t = 5\) and from \(t = 0\) to \(t = 2\).
- Total distance: Using \(v(t) = 2t - 4\), find the total distance traveled from \(t = 0\) to \(t = 5\). (Hint: find where \(v = 0\) and split the integral.)
- Population growth: A town's population grows at \(P'(t) = 500 + 20t\) people per year, where \(t\) is years after 2010. By how much does the population grow from 2010 to 2020?
- Water flow: Water flows into a pool at \(r(t) = 50 + 10t\) liters per minute. How much water enters from \(t = 0\) to \(t = 6\) minutes?
- Marginal cost: A company's marginal cost is \(C'(x) = 2x + 30\) dollars per unit. Find the cost of producing units 10 through 25.
- Drug absorption: A medication is absorbed at \(r(t) = 8e^{-0.5t}\) mg/hr. How many milligrams are absorbed during the first 4 hours?
- Velocity and acceleration: A particle starts from rest. Its acceleration is \(a(t) = 6t - 2\) ft/s². Find the velocity at \(t = 3\) and the net displacement from \(t = 0\) to \(t = 3\).
- Average rainfall: Monthly rainfall: \(R(t) = 3 + 2\sin\!\left(\frac{\pi t}{6}\right)\) inches/month, \(t\) = months after January. Find average rainfall over one full year.
Error Analysis
Each problem shows a student's work. Identify the error and write the correct solution.
Problem 1
Student evaluates \(\int_0^4 x^2\,dx\): "Antiderivative is \(\frac{x^3}{3}\). At \(x = 4\), that gives \(\frac{64}{3}\)."
What is the error? What is the correct value?
Problem 2
Student evaluates \(\dfrac{d}{dx}\int_0^{x^2}\sin(t)\,dt\) and writes: \(\sin(x^2)\).
What is the error? What is the correct answer?
Problem 3
Student evaluates \(\int_{-1}^{1} x^2\,dx\): "The function has \((-1)^2 = 1 = 1^2\), so it's symmetric and the integral is 0."
Identify the error. (\(x^2\) is not an odd function.) What is the correct value?
Problem 4
Student finds total distance with \(v(t) = t - 3\) on \([0,6]\): \(\int_0^6(t-3)\,dt = 0\). "Total distance is 0."
Identify the error. The integral gives net displacement, not total distance. Show the correct total distance calculation.
Problem 5
Student evaluates \(\int_0^2 e^x\,dx\): "Antiderivative of \(e^x\) is \(xe^x\). Answer: \(2e^2 - 0 = 2e^2\)."
What is the error? The antiderivative of \(e^x\) is \(e^x\), not \(xe^x\). What is the correct answer?
Problem 6
Student writes: "\(\dfrac{d}{dx}\int_x^5 t^3\,dt = x^3\). The variable is in the lower limit, but it still gives the integrand at \(x\)."
What is the error? When the variable is in the lower limit, a negative sign is needed. What is the correct answer?
Chapter 11 Checkpoint
Chapter 11 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 11
Part A — Conceptual
- In one or two sentences, explain what the Fundamental Theorem of Calculus says. Use plain language.
- Is the accumulation function \(A(x) = \int_2^x f(t)\,dt\) equal to zero at \(x = 2\)? Explain.
- If \(\int_0^5 f(x)\,dx = 30\), what is the average value of \(f\) on \([0, 5]\)?
- A particle moves with positive velocity on \([0, 4]\) and negative velocity on \([4, 7]\). Will the net displacement on \([0, 7]\) be larger or smaller in magnitude than the total distance? Explain.
Part B — Skills
- \(\dfrac{d}{dx} \int_3^x (2t^2 - t + 1)\,dt\)
- \(\dfrac{d}{dx} \int_0^{x^4} e^t\,dt\)
- \(\int_0^4 (3x^2 + 2x)\,dx\)
- \(\int_0^{\pi/2} \sin(x)\,dx\)
- \(\int_1^e \dfrac{3}{x}\,dx\)
- Find the average value of \(f(x) = 4 - x^2\) on \([0, 2]\).
Part C — Applications
- A particle's velocity is \(v(t) = t^2 - 6t + 8\) ft/s. Find the net displacement from \(t = 0\) to \(t = 5\). Then find the total distance traveled on \([0, 5]\). (Hint: factor \(v(t)\) to find where it equals zero.)
- A company's marginal revenue function is \(MR(x) = 80 - 0.4x\) dollars per unit. Find the total revenue generated from selling units 50 through 100.
- Water drains from a tank at \(r(t) = 100 - 10t\) gallons per minute, for \(0 \leq t \leq 10\). How many total gallons leave the tank?
- The temperature in a greenhouse is \(T(t) = 68 + 6\sin\!\left(\frac{\pi t}{12}\right)\)°F, where \(t\) is hours after midnight. Find the average temperature from midnight to noon.
Part D — Bridge
- In Chapter 12, you will compute the area between two curves \(f(x)\) and \(g(x)\) using \(\int_a^b [f(x) - g(x)]\,dx\). Explain why this formula works using the Net Change Theorem or the idea that integrals accumulate quantities. What does \(f(x) - g(x)\) represent geometrically?
Answers may be placed in the answer key.
Looking Ahead
Chapter 12: Applications of Integration
Chapter 11 established the Fundamental Theorem as the foundation of integral calculus. Chapter 12 puts it to work. You will use definite integrals to find areas between curves, volumes of solids generated by revolving regions around an axis, accumulated change from rate functions, and real-world quantities in physics, economics, and engineering. Every technique in Chapter 12 relies on setting up a definite integral and evaluating it using Part 2 — exactly the skill this chapter developed. The conceptual work is done; the next chapter shows what can be built with it.