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Calculus I  ·  Chapter 5  ·  Section 5.3

The Definite Integral &
The Fundamental Theorem

From infinite slices of area to the most powerful bridge in mathematics — connecting differentiation and integration into one breathtaking theorem.

Opening Story

A Monsoon, a River, and
a Question of Survival

It is July 2022. The monsoon has arrived in Sindh, and the Indus River is rising. Engineers at the Tarbela Dam need to answer one urgent question: how much water will flow through this point in the next 72 hours?

They have sensors reading the flow rate $R(t)$ in cubic metres per second, updated every hour. The rate changes constantly — faster after heavy rain, slower in the early morning. Nobody can memorise a continuously changing function. But every engineer in that room knows one thing: the total volume of water is the area under the $R(t)$ curve.

They need to compute $\displaystyle\int_0^{72} R(t)\,dt$. The decision to open the spillway — potentially displacing millions of people — depends on getting this number right. This is not a textbook exercise. This is the definite integral saving lives.

This story captures exactly why the definite integral exists. Whenever a quantity accumulates continuously — water, money, distance, electrical charge, heat — the tool that measures the total accumulation is the definite integral. And the tool that lets us compute it efficiently, without adding millions of rectangles by hand, is the Fundamental Theorem of Calculus.

The Big Picture: The Fundamental Theorem reveals that differentiation and integration — which appear to be completely different operations — are actually inverse processes of each other. This is one of the most surprising and beautiful results in all of mathematics.
Learning Objectives

What You Will Master
in This Section

1

Show how area under a curve can be expressed as the limit of a Riemann sum.

2

Define the definite integral and explore its algebraic properties and rules.

3

State the Fundamental Theorem of Calculus and use it to evaluate definite integrals efficiently.

4

Use the FTC to solve applied problems involving net change and accumulation.

5

Provide a geometric justification of the Fundamental Theorem of Calculus.

§ 1 — Area as the Limit of a Sum

From One Rectangle
to Infinite Precision

Let us find the area under $f(x) = x^2$ from $x = 0$ to $x = 2$. We will start crudely and progressively improve.

Approximation 1
One Rectangle — Very Rough

Use a single rectangle spanning the entire interval $[0, 2]$. Take the height as $f(2) = 4$ (right endpoint).

Area $\approx f(2) \cdot 2 = 4 \times 2 = 8$
1230014xyArea ≈ 8f(x) = x²

The true area is $\frac{8}{3} \approx 2.667$. Our estimate of $8$ is a big overestimate — the rectangle towers over the curve everywhere. We can do better by using more rectangles.

Approximation 2
Two Rectangles — Getting Closer

Divide $[0, 2]$ into 2 equal subintervals of width $\Delta x = 1$. Use the right endpoint of each.

Area $\approx f(1)\cdot 1 + f(2)\cdot 1 = 1 + 4 = 5$
01214xy1×1=14×1=4f(x) = x²
Δx = (2 − 0) / 2 = 1
Rectangle 1 → right endpoint x = 1, f(1) = 1² = 1 → area = 1 × 1 = 1
Rectangle 2 → right endpoint x = 2, f(2) = 2² = 4 → area = 4 × 1 = 4
Total ≈ 1 + 4 = 5

The true area is $\frac{8}{3} \approx 2.667$. Two rectangles give us $5$ — still an overestimate, but already much better than $8$. More rectangles means a closer approximation.

Approximation 3
Four Rectangles — Much Better

Divide $[0,2]$ into 4 equal sub-intervals, each of width $\Delta x = 0.5$. Use right endpoints for height.

Sub-intervalRight endpoint $x_i$Height $f(x_i) = x_i^2$Area of rectangle
$[0, 0.5]$$0.5$$0.25$$0.25 \times 0.5 = 0.125$
$[0.5, 1]$$1.0$$1.00$$1.00 \times 0.5 = 0.500$
$[1, 1.5]$$1.5$$2.25$$2.25 \times 0.5 = 1.125$
$[1.5, 2]$$2.0$$4.00$$4.00 \times 0.5 = 2.000$
Total Area $\approx 0.125 + 0.500 + 1.125 + 2.000 = \mathbf{3.75}$
00.511.52Area ≈ 3.75f(x) = x²xy

With 4 rectangles we get $3.75$ vs. the true $\frac{8}{3} \approx 2.667$. Still overshooting (right endpoints overestimate for increasing functions), but much closer than before. The key insight: more rectangles means more accuracy.

General Formula
$n$ Rectangles — The Riemann Sum

Divide $[a, b]$ into $n$ equal sub-intervals, each of width $\Delta x = \dfrac{b-a}{n}$. The right endpoint of the $i$-th sub-interval is $x_i = a + i\Delta x$.

Riemann Sum

$$S_n = \sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n} f\!\left(a + i\cdot\frac{b-a}{n}\right)\cdot\frac{b-a}{n}$$

For our example $f(x)=x^2$, $[0,2]$, using right endpoints:

$$S_n = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n} = \frac{8}{n^3}\sum_{i=1}^{n}i^2 = \frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{4(n+1)(2n+1)}{3n^2}$$

$n$$S_n$Error from $\frac{8}{3}$
$1$$8.000$$5.333$
$4$$3.750$$1.083$
$10$$2.960$$0.293$
$100$$2.6934$$0.0267$
$1000$$2.6693$$0.0027$
$\infty$$\dfrac{8}{3} \approx 2.6\overline{6}$$0$
§ 2 — Interactive Exploration

Play With It:
The Riemann Sum Explorer

Drag the slider to increase the number of rectangles and watch the approximation converge to the exact area. Try different functions and endpoint rules.

⬛ Interactive Riemann Sum Explorer
n = 6
Δx = —
Approx ≈ —
Exact = —
Error = —
Area as a Limit

The exact area under $f(x)$ from $a$ to $b$ is defined as the limit of the Riemann sum as $n \to \infty$:

$$\text{Area} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x$$

This limit exists whenever $f$ is continuous on $[a,b]$. The beautiful fact is: it does not matter which sample point $x_i^*$ we choose — left, right, or midpoint — the limit is always the same.

Example — Area Using the Limit

Find the area under $f(x) = 3x$ from $x = 0$ to $x = 2$ using the limit definition.

012x0246yf(x) = 3xArea = 6(exact)Δx = 2/nExact triangle (Area = 6)Rectangles (n = 4 shown)
Solution

Right endpoints: $x_i = \dfrac{2i}{n}$, width $\Delta x = \dfrac{2}{n}$

$$S_n = \sum_{i=1}^{n} 3\cdot\frac{2i}{n}\cdot\frac{2}{n} = \frac{12}{n^2}\sum_{i=1}^{n}i$$

$$= \frac{12}{n^2}\cdot\frac{n(n+1)}{2} = \frac{6(n+1)}{n}$$

$$\text{Area} = \lim_{n\to\infty}\frac{6(n+1)}{n} = 6$$

✓ Geometric check: $f(x)=3x$ forms a right triangle with base $2$ and height $f(2)=6$.

$$\text{Area} = \tfrac{1}{2}(2)(6) = \mathbf{6}$$

§ 3 — The Definite Integral

Defining the
Definite Integral

Definition — The Definite Integral

If $f$ is continuous on $[a,b]$, the definite integral of $f$ from $a$ to $b$ is:

$$\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\,\Delta x$$

The number $a$ is the lower limit and $b$ is the upper limit of integration. The result is a number, not a function.

Area as a Definite Integral

When $f(x) \geq 0$ on $[a,b]$, the definite integral equals the area under the curve:

$$\text{Area} = \int_a^b f(x)\,dx$$

When $f(x)$ takes negative values, the integral computes the signed area — area above the $x$-axis counts positive, area below counts negative.

📐 Signed Area Explorer — $\int_0^b \sin(x)\,dx$
Positive area: —
Negative area: —
Net integral: —
Important: $\displaystyle\int_a^b f(x)\,dx$ gives signed area. If you want total (unsigned) area when $f$ dips below the axis, you must split the integral at the zeros and add absolute values.
§ 4 — The Bridge Between Two Worlds

The Fundamental
Theorem of Calculus

Without the FTC, computing $\displaystyle\int_0^2 x^2\,dx$ requires evaluating a limit of a sum — tedious. The FTC provides a miraculous shortcut.

The Fundamental Theorem of Calculus

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ (i.e. $F'(x) = f(x)$), then:

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

We write $F(b) - F(a)$ as $\Big[F(x)\Big]_a^b$.

What this means: To compute a definite integral, find any antiderivative $F(x)$, then simply subtract: $F(b) - F(a)$. The $+C$ cancels out, so we can ignore it.

Using the FTC — Examples

Example 1 — Power Function

Compute $\displaystyle\int_0^2 x^2\,dx$.

Step 1: Find an antiderivative: $F(x) = \dfrac{x^3}{3}$

Step 2: Apply the FTC:

$$\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3}$$

This confirms our Riemann sum limit from earlier. The FTC computed it in two lines.

Example 2 — Mixed Terms

Compute $\displaystyle\int_1^3 (2x + e^x)\,dx$.

$$\int_1^3 (2x+e^x)\,dx = \Big[x^2 + e^x\Big]_1^3 = (9+e^3)-(1+e) = 8 + e^3 - e \approx 8 + 20.09 - 2.72 \approx 25.37$$

Example 3 — Rational Integrand

Compute $\displaystyle\int_1^e \frac{3}{x}\,dx$.

$$\int_1^e \frac{3}{x}\,dx = \Big[3\ln|x|\Big]_1^e = 3\ln e - 3\ln 1 = 3(1) - 3(0) = 3$$

Example 4 — Negative Exponent

Compute $\displaystyle\int_1^4 \frac{1}{\sqrt{x}}\,dx$.

$$\int_1^4 x^{-1/2}\,dx = \Big[2x^{1/2}\Big]_1^4 = 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2$$

§ 5 — Geometric Justification

Why Does the FTC Work?
A Visual Argument

Define the accumulation function $A(x) = \displaystyle\int_a^x f(t)\,dt$ — the area under $f$ from $a$ up to $x$.

🔗 Accumulation Function $A(x) = \int_0^x t^2\,dt$
Shaded area A(x) = —
F(x) = x³/3 = —
A'(x) = x² = —
The Key Geometric Insight

When $x$ increases by a tiny amount $h$, the extra area added is approximately a thin rectangle:

$$A(x+h) - A(x) \approx f(x) \cdot h$$

Dividing both sides by $h$ and taking $h \to 0$:

$$A'(x) = \lim_{h\to 0}\frac{A(x+h)-A(x)}{h} = f(x)$$

So $A(x)$ is an antiderivative of $f(x)$! Since $A(a) = 0$, and any antiderivative $F$ satisfies $F(x) = A(x) + C$, we get $\int_a^b f(x)\,dx = A(b) = F(b) - F(a)$.

The Profound Truth: The FTC says that accumulation (integration) and rate of change (differentiation) are inverse operations. The area function, when differentiated, gives back the original function. This is why $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$.
§ 6 — Properties of the Definite Integral

Integration Rules for
Definite Integrals

These rules let us manipulate definite integrals algebraically without recomputing from scratch.

Constant Multiple Rule

$$\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$$

Pull constants outside the integral.

Sum / Difference Rule

$$\int_a^b [f \pm g]\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$$

Split across addition or subtraction.

Same Limits Rule

$$\int_a^a f(x)\,dx = 0$$

No width means no area.

Reversed Limits Rule

$$\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$$

Switching limits negates the integral.

Subdivision Rule

$$\int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx$$

Split the interval at any interior point $c$.

Dominance Rule

$$\text{If } f(x)\geq g(x) \text{ on } [a,b]: \int_a^b f\,dx \geq \int_a^b g\,dx$$

Larger function, larger integral.

Examples Using the Rules

Example 1 — Constant Multiple + Sum

Given $\displaystyle\int_0^3 f(x)\,dx = 7$ and $\displaystyle\int_0^3 g(x)\,dx = 4$, find $\displaystyle\int_0^3 [5f(x) - 2g(x)]\,dx$.

$$\int_0^3 [5f(x)-2g(x)]\,dx = 5\int_0^3 f\,dx - 2\int_0^3 g\,dx = 5(7)-2(4) = 35-8 = 27$$

Example 2 — Subdivision Rule

Given $\displaystyle\int_0^5 f(x)\,dx = 12$ and $\displaystyle\int_0^3 f(x)\,dx = 7$, find $\displaystyle\int_3^5 f(x)\,dx$.

By the subdivision rule: $\displaystyle\int_0^5 f\,dx = \int_0^3 f\,dx + \int_3^5 f\,dx$

$$\int_3^5 f\,dx = 12 - 7 = 5$$

Example 3 — Reversed Limits

Compute $\displaystyle\int_4^1 \sqrt{x}\,dx$.

$$\int_4^1 \sqrt{x}\,dx = -\int_1^4 \sqrt{x}\,dx = -\Big[\frac{2}{3}x^{3/2}\Big]_1^4 = -\left(\frac{2}{3}(8) - \frac{2}{3}(1)\right) = -\frac{14}{3}$$

Example 4 — All Rules Combined

Compute $\displaystyle\int_{-1}^{2}(3x^2 - 4x + 1)\,dx$.

$$\Big[x^3 - 2x^2 + x\Big]_{-1}^{2} = (8-8+2)-(-1-2-1) = 2-(-4) = 6$$

§ 7 — Substitution in Definite Integrals

Using Substitution with
Definite Integrals

When a definite integral requires substitution, you have two approaches. The cleaner method is to change the limits of integration along with the variable.

Substitution Rule for Definite Integrals

If $u = g(x)$ and $g'$ is continuous on $[a,b]$, then:

$$\int_a^b f(g(x))\cdot g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$$

Key: when $x = a$, $u = g(a)$; when $x = b$, $u = g(b)$. Change both limits!

Example 1 — Change the Limits

Compute $\displaystyle\int_0^2 2x(x^2+1)^3\,dx$.

Let $u = x^2+1$, $du = 2x\,dx$.

New limits: When $x=0$: $u = 1$. When $x=2$: $u = 5$.

$$\int_1^5 u^3\,du = \left[\frac{u^4}{4}\right]_1^5 = \frac{625}{4} - \frac{1}{4} = \frac{624}{4} = 156$$

No back-substitution needed! The new limits handle everything.

Example 2 — Exponential with Substitution

Compute $\displaystyle\int_0^1 xe^{x^2}\,dx$.

Let $u = x^2$, $du = 2x\,dx$, so $x\,dx = \dfrac{du}{2}$.

New limits: $x=0 \Rightarrow u=0$; $x=1 \Rightarrow u=1$.

$$\int_0^1 e^u\cdot\frac{du}{2} = \frac{1}{2}\Big[e^u\Big]_0^1 = \frac{1}{2}(e-1) \approx \frac{1}{2}(1.718) \approx 0.859$$

Example 3 — Rational with Substitution

Compute $\displaystyle\int_0^3 \frac{x}{(x^2+1)^2}\,dx$.

Let $u = x^2+1$, $du = 2x\,dx$, so $x\,dx = \dfrac{du}{2}$.

New limits: $x=0 \Rightarrow u=1$; $x=3 \Rightarrow u=10$.

$$\int_1^{10} \frac{1}{u^2}\cdot\frac{du}{2} = \frac{1}{2}\left[-\frac{1}{u}\right]_1^{10} = \frac{1}{2}\left(-\frac{1}{10}+1\right) = \frac{1}{2}\cdot\frac{9}{10} = \frac{9}{20}$$

§ 8 — Net Change

The Net Change Theorem

Net Change Theorem

If $F'(x) = f(x)$, then the net change in $F$ from $x = a$ to $x = b$ is:

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

The integral of a rate of change gives the total (net) change in the quantity.

This is simply the FTC rewritten with emphasis on its interpretation. It tells us:

If $f(x)$ represents...Then $\int_a^b f(x)\,dx$ gives...
Velocity $v(t)$Net displacement (not total distance)
Marginal cost $C'(x)$Change in total cost from $x=a$ to $x=b$
Rate of population growth $P'(t)$Net change in population
Flow rate $R(t)$ (water, current)Total volume/charge accumulated
Rate of profit changeNet profit gained or lost

Example 1 — Displacement vs. Distance

A particle moves along a line with velocity $v(t) = t^2 - 4t + 3$ m/s for $0 \leq t \leq 4$.

Find (a) the net displacement and (b) the total distance traveled.

(a) Net displacement:

$$\int_0^4 (t^2-4t+3)\,dt = \left[\frac{t^3}{3}-2t^2+3t\right]_0^4 = \left(\frac{64}{3}-32+12\right)-0 = \frac{64}{3}-20 = \frac{4}{3} \approx 1.33 \text{ m}$$

(b) Total distance: First find where $v(t)=0$: $t^2-4t+3=(t-1)(t-3)=0$, so $t=1$ and $t=3$.

$v > 0$ on $[0,1]$, $v < 0$ on $[1,3]$, $v > 0$ on $[3,4]$.

$$d = \int_0^1(t^2-4t+3)\,dt + \left|\int_1^3(t^2-4t+3)\,dt\right| + \int_3^4(t^2-4t+3)\,dt$$

$$= \frac{4}{3} + \left|-\frac{4}{3}\right| + \frac{4}{3} = \frac{4}{3}+\frac{4}{3}+\frac{4}{3} = 4 \text{ m}$$

Example 2 — Marginal Cost

A factory's marginal cost is $C'(x) = 0.006x^2 - 0.6x + 30$ PKR/unit. Find the increase in cost when production goes from 50 to 100 units.

$$\int_{50}^{100}(0.006x^2-0.6x+30)\,dx = \Big[0.002x^3-0.3x^2+30x\Big]_{50}^{100}$$

$$= (2000-3000+3000)-(250-750+1500) = 2000-1000 = \text{PKR }1000$$

§ 9 — Applied Problems

Integration in Practice

🌍 Example 1 — Land Survey Along the Ravi River

A land surveyor in Lahore is measuring a plot between two boundary roads. The northern boundary follows the curve $f(x) = -x^2 + 8x$ and the southern boundary follows $g(x) = x^2 - 4x$, where $x$ is in hundreds of metres. Find the total area of the plot enclosed between the two curves.

Step 1 — Find where the boundaries meet (intersection points).

Set $f(x) = g(x)$:

$$-x^2 + 8x = x^2 - 4x \implies 0 = 2x^2 - 12x = 2x(x-6) \implies x = 0 \text{ and } x = 6$$

Step 2 — Determine which curve is on top.

Test $x = 3$ (midpoint): $f(3) = -9+24 = 15$, $g(3) = 9-12 = -3$. Since $f(3) > g(3)$, the northern boundary $f(x)$ lies above $g(x)$ throughout $[0,6]$.

Step 3 — Set up the area integral.

$$\text{Area} = \int_0^6 \Bigl[(-x^2+8x)-(x^2-4x)\Bigr]\,dx = \int_0^6 (-2x^2+12x)\,dx$$

Step 4 — Integrate and evaluate:

$$= \left[-\frac{2x^3}{3} + 6x^2\right]_0^6 = \left(-\frac{2(216)}{3} + 6(36)\right) - 0 = (-144 + 216) = 72$$

Step 5 — Convert to real units:

$$\text{Area} = 72 \times (100)^2 = 720{,}000 \text{ m}^2 = \boxed{72 \text{ hectares}}$$

Why this works: When a region is bounded above by $f(x)$ and below by $g(x)$, you subtract the lower from the upper before integrating. This automatically handles the fact that $g(x)$ dips below the $x$-axis — you never need to split the integral or worry about signs, because the difference $f(x)-g(x)$ is always positive on $[0,6]$.

⚡ Example 2 — Electricity Consumption at LUMS

LUMS's power consumption rate (in MW) during a 12-hour working day is modelled by

$$P(t) = 0.01t^3 - 0.18t^2 + 0.9t + 1.5, \qquad 0 \leq t \leq 12,$$

where $t$ is in hours. Find the total energy consumed (in MWh) over the full 12-hour period.

Step 1 — Set up the integral.

$$\text{Total energy} = \int_0^{12} P(t)\,dt = \int_0^{12}\!\left(0.01t^3 - 0.18t^2 + 0.9t + 1.5\right)dt$$

Step 2 — Find the antiderivative term by term:

$$F(t) = \frac{t^4}{400} - \frac{3t^3}{50} + \frac{9t^2}{20} + \frac{3t}{2}$$

Step 3 — Evaluate at the upper limit $t = 12$:

$$F(12) = \frac{20736}{400} - \frac{3\cdot 1728}{50} + \frac{9\cdot 144}{20} + 18 = 51.84 - 103.68 + 64.8 + 18 = 30.96$$

Step 4 — Apply the FTC:

$$E = F(12) - F(0) = 30.96 - 0 = \boxed{30.96 \text{ MWh}}$$

Sanity check: The average power over 12 hours is $\dfrac{30.96}{12} \approx 2.58$ MW. Since $P(0)=1.5$ MW and $P(12) = 3.66$ MW, an average of $\approx 2.58$ MW is very reasonable. ✓

📈 Example 3 — Revenue Accumulation at a Karachi Textile Firm

A Karachi textile firm's revenue rate (in million PKR/month) follows

$$R'(t) = \frac{3t}{\sqrt{2t^2 + 7}}, \qquad 0 \leq t \leq 9,$$

where $t$ is in months.

(a) Find the total revenue earned over the first 9 months.
(b) Find the revenue earned in the second 3-month period only (i.e. from $t=3$ to $t=6$).

Part (a) — Total revenue over 9 months.

Let $u = 2t^2 + 7$, $du = 4t\,dt$, so $t\,dt = \dfrac{du}{4}$.

Limits: $t=0 \Rightarrow u=7$; $t=9 \Rightarrow u=169$.

$$\int_0^9 \frac{3t}{\sqrt{2t^2+7}}\,dt = \frac{3}{4}\int_7^{169} u^{-1/2}\,du = \frac{3}{4}\Big[2u^{1/2}\Big]_7^{169} = \frac{3}{2}\left(13 - \sqrt{7}\right) \approx \boxed{15.53 \text{ million PKR}}$$

Part (b) — Revenue from $t = 3$ to $t = 6$ only.

Limits: $t=3 \Rightarrow u=25$; $t=6 \Rightarrow u=79$.

$$\int_3^6 \frac{3t}{\sqrt{2t^2+7}}\,dt = \frac{3}{2}\Big[\sqrt{u}\Big]_{25}^{79} = \frac{3}{2}\left(\sqrt{79} - 5\right) \approx \frac{3}{2}(3.888) \approx \boxed{5.83 \text{ million PKR}}$$

Key insight: You do not recompute the antiderivative for part (b) — the same substitution and antiderivative $\tfrac{3}{2}\sqrt{2t^2+7}$ works. Only the limits change.

🏃 Example 4 — LUMS Athletics Track Event

During a LUMS inter-faculty athletics event, a runner's speed (in km/min) is modelled by

$$v(t) = \frac{4t^3 - 2t}{t^4 - t^2 + 3}, \qquad 0 \leq t \leq 4,$$

where $t$ is in minutes.

(a) Find the total distance covered over the full 4 minutes.
(b) Find the distance covered in the first minute only, and express it as an exact logarithm.

Part (a) — Total distance over 4 minutes.

Let $u = t^4 - t^2 + 3$, then $\dfrac{du}{dt} = 4t^3 - 2t$ — the numerator is exactly $\dfrac{du}{dt}$. This is a $\int \dfrac{u'}{u}\,dt = \ln|u| + C$ pattern.

Limits: $t=0 \Rightarrow u=3$; $t=4 \Rightarrow u=243$.

$$\int_0^4 \frac{4t^3-2t}{t^4-t^2+3}\,dt = \int_3^{243}\frac{du}{u} = \Big[\ln u\Big]_3^{243} = \ln 243 - \ln 3 = \ln 81 = 4\ln 3 \approx \boxed{4.394 \text{ km}}$$

Part (b) — Distance in the first minute only.

Limits: $t=0 \Rightarrow u=3$, $t=1 \Rightarrow u = 1-1+3 = 3$.

$$\int_0^1 \frac{4t^3-2t}{t^4-t^2+3}\,dt = \Big[\ln u\Big]_3^{3} = \ln 3 - \ln 3 = \boxed{0 \text{ km}}$$

What just happened? When $t=0$ and $t=1$ both give $u=3$, the upper and lower limits of the transformed integral are identical — so by the same-limits rule $\int_a^a f\,du = 0$, the net distance is zero. This is a powerful reminder: always check your transformed limits carefully before concluding.
Coming up next — §5.4 Applying Definite Integration: Distribution of Wealth and Average Value — where we use integrals to measure economic inequality with the Lorenz curve and Gini coefficient.
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