Chapter 06

Medium

Integrals — Definition and FTC

00 · Symbol Glossary

$\Delta x$Delta x — subinterval width

A small change in $x$ : the width of one rectangular strip used to approximate an area. Capital Greek delta $\Delta$ denotes "change in."

$x_i^*$x star sub i — sample point

A chosen point inside the $i$ -th subinterval, used to determine the height $f(x_i^*)$ of that subinterval's approximating rectangle.

$\sum_{i=1}^{n} f(x_i^*)\Delta x$Riemann sum

The total area of $n$ approximating rectangles: sum, over each strip, of (height $\times$ width). Named after Bernhard Riemann. As $n\to\infty$ and each $\Delta x\to0$ , this sum approaches the exact area.

$\int_a^b f(x)\,dx$definite integral from a to b

The exact limiting value of the Riemann sum as the number of rectangles goes to infinity. Read "the integral from $a$ to $b$ of $f(x)$ d-x." $a$ and $b$ are the limits of integration.

$dx$d x — differential

Formally, the limit of $\Delta x$ as it shrinks to an infinitesimal width. It marks which variable the integration is performed with respect to.

$F(x)$F of x — antiderivative

A function whose derivative is $f(x)$ : $F'(x) = f(x)$ . Capital letters are used by convention to denote an antiderivative of the corresponding lowercase function.

$\int f(x)\,dx$indefinite integral

The family of all antiderivatives of $f$ , written without limits of integration. Always includes an arbitrary constant $+C$ , since any two antiderivatives of the same function differ only by a constant.

$C$constant of integration

An arbitrary real-number constant, appended to every indefinite integral, because $\dfrac{d}{dx}[F(x)+C]=F'(x)=f(x)$ for any constant $C$ — differentiation destroys constant terms, so integration cannot recover which one was there.

$\big[F(x)\big]_a^b$evaluation bracket

Notation meaning $F(b)-F(a)$ : evaluate $F$ at the upper limit, subtract $F$ evaluated at the lower limit.

01 · The Area Problem and Riemann Sums

Finding the area under a curve $y=f(x)$ between $x=a$ and $x=b$ cannot generally be done with a simple geometric formula — only rectangles, triangles, and circles have exact area formulas. The fix: approximate the region with many thin rectangles, then let the number of rectangles grow without bound.

Definition — Riemann Sum

Divide $[a,b]$ into $n$ subintervals of equal width $\Delta x = \dfrac{b-a}{n}$ . In each subinterval, pick a sample point $x_i^*$ and form a rectangle of height $f(x_i^*)$ and width $\Delta x$ . The Riemann sum is

\sum_{i=1}^{n} f(x_i^*)\,\Delta x

Step-by-step — Approximate the area under $f(x)=x^2$ on $[0,2]$ with 4 rectangles

Compute the subinterval width:

\Delta x = \dfrac{b-a}{n} = \dfrac{2-0}{4} = 0.5

List the right-endpoint sample points: using the right edge of each strip,

x_1^*=0.5

x_2^*=1.0

x_3^*=1.5

x_4^*=2.0

Compute the height at each sample point:

f(0.5)=0.25

f(1.0)=1

f(1.5)=2.25

f(2.0)=4

Multiply each height by the width and sum:

(0.25+1+2.25+4)\times0.5 = 7.5\times0.5=3.75

Compare to the exact area (computed in Section 05): the exact area under

x^2

[0,2]

\frac83\approx2.667

. The right-endpoint estimate

3.75

overshoots because

f(x)=x^2

is increasing, so right endpoints always sit above the curve's average height on each strip — more rectangles would shrink this gap.

Definition — Definite Integral

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

provided the limit exists (true for any continuous $f$ ). This is the exact, signed area between the curve and the $x$ -axis.

Signed area

Where $f(x)<0$ , the corresponding rectangles have negative height, contributing negatively to the sum. $\int_a^b f(x)\,dx$ measures area above the axis minus area below the axis — not total unsigned area.

02 · Properties of the Definite Integral

Definition — Basic Integral Properties

\int_a^b \big[f(x)+g(x)\big]dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx \qquad \int_a^b c\,f(x)\,dx = c\int_a^b f(x)\,dx

\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx \qquad \int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx \qquad \int_a^a f(x)\,dx = 0

03 · Antiderivatives and Indefinite Integrals

Instead of computing area directly from Riemann sums every time, calculus provides a shortcut built on reversing differentiation.

Definition — Antiderivative

$F(x)$ is an antiderivative of $f(x)$ if $F'(x) = f(x)$ .

Example — Finding an antiderivative by reversing the power rule

For $f(x)=x^2$ , recall from Chapter 02 that $\dfrac{d}{dx}\left[\dfrac{x^3}{3}\right] = \dfrac{1}{3}\cdot3x^2 = x^2$ — the $3$ in the denominator was chosen precisely to cancel the $3$ that the power rule brings down. So $F(x)=\dfrac{x^3}{3}$ is one antiderivative of $x^2$ .

❌ Assuming the antiderivative is unique

Both $F_1(x) = \dfrac{x^3}{3}$ and $F_2(x) = \dfrac{x^3}{3}+7$ satisfy $F'(x)=x^2$ , since differentiating a constant gives $0$ regardless of its value (Chapter 02).

Why it breaks: believing "the" antiderivative is a single specific function ignores that differentiation is not one-to-one — every constant shift produces the same derivative.

Consequence: any antiderivative of $f$ must be reported as a whole family, $F(x)+C$ , unless additional information (like a known value of $F$ at one point) pins down the specific constant.

Definition — Indefinite Integral and Power Rule for Integration

\int f(x)\,dx = F(x) + C \qquad \text{where } F'(x)=f(x)

For powers of $x$ (the reverse of Chapter 02's power rule), valid whenever $n\neq-1$ :

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C

Step-by-step — Compute $\int (3x^2 - 4x + 5)\,dx$

Integrate term by term (linearity, mirroring Chapter 02's sum rule): handle each of the three terms separately.

$3x^2$ : power rule for integration with

n=2

\dfrac{x^3}{3}

; multiply by the constant

3

3\cdot\dfrac{x^3}{3}=x^3

$-4x$ :

n=1

\dfrac{x^2}{2}

; multiply by

-4

-4\cdot\dfrac{x^2}{2}=-2x^2

$5$ : treat as

5x^0

n=0

\dfrac{x^1}{1}=x

; multiply by

5

5x

Assemble, adding one constant of integration for the whole expression:

x^3 - 2x^2 + 5x + C

❌ The power rule for integration fails at n = -1

Attempting $\int x^{-1}\,dx$ using the formula: $\dfrac{x^{-1+1}}{-1+1} = \dfrac{x^0}{0}$ — division by zero.

Why it breaks: the formula $\int x^n\,dx = \frac{x^{n+1}}{n+1}+C$ explicitly requires $n\neq-1$ ; at $n=-1$ the denominator vanishes and the formula is undefined.

Consequence: $\int x^{-1}\,dx = \int \frac{1}{x}\,dx$ requires a different antiderivative entirely (the natural logarithm, $\ln\lvert x\rvert + C$ ) — outside the scope of the power rule, and a standing exception to remember whenever $n=-1$ appears.

04 · The Fundamental Theorem of Calculus, Part 1

Riemann sums (Section 01) define the integral; antiderivatives (Section 03) reverse the derivative. The Fundamental Theorem of Calculus proves these two ideas are the same operation viewed from opposite directions.

Definition — FTC Part 1

If $f$ is continuous on $[a,b]$ , and $G(x) = \displaystyle\int_a^x f(t)\,dt$ for $x\in[a,b]$ , then $G$ is differentiable and

G'(x) = f(x)

The derivative of an integral (as a function of its upper limit) is simply the original integrand back again. $t$ is used as the integration variable here purely to avoid reusing $x$ , which already names the upper limit.

Example — Differentiating an integral with a variable upper limit

Let $G(x) = \displaystyle\int_2^x (t^2+1)\,dt$ . By FTC Part 1, without computing the integral at all: $G'(x) = x^2+1$ — substitute $x$ for $t$ in the original integrand.

05 · The Fundamental Theorem of Calculus, Part 2

Part 1 connects integration and differentiation conceptually. Part 2 turns that connection into the everyday computational tool for evaluating definite integrals — no Riemann sum limits required.

Definition — FTC Part 2

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

\int_a^b f(x)\,dx = F(b) - F(a) = \big[F(x)\big]_a^b

Step-by-step — Compute the exact area under $f(x)=x^2$ on $[0,2]$ (revisiting Section 01)

Find an antiderivative: from Section 03,

F(x) = \dfrac{x^3}{3}

satisfies

F'(x)=x^2

Apply FTC Part 2:

\displaystyle\int_0^2 x^2\,dx = F(2)-F(0)

Evaluate $F$ at the upper limit:

F(2) = \dfrac{2^3}{3}=\dfrac{8}{3}

Evaluate $F$ at the lower limit:

F(0) = \dfrac{0^3}{3}=0

Subtract:

\dfrac{8}{3}-0=\dfrac{8}{3}\approx2.667

. This matches the value the Section 01 Riemann sum was converging toward, and confirms the

4

-rectangle estimate of

3.75

was an overshoot, as predicted there.

Any antiderivative works, and the +C always cancels

FTC Part 2 says "any" antiderivative $F$ , not "the" antiderivative $F+C$ . Using $F(x)+C$ instead: $\big[F(x)+C\big]_0^2 = \big(F(2)+C\big)-\big(F(0)+C\big) = F(2)-F(0)$ — the $C$ cancels regardless of its value, which is why the constant of integration is never written when evaluating a definite integral.

❌ Reversing the order of subtraction

On the same integral, computing $F(0)-F(2) = 0 - \dfrac83 = -\dfrac83$ instead of $F(2)-F(0)$ .

Why it breaks: FTC Part 2 fixes the order as (value at the upper limit) minus (value at the lower limit) — reversing it flips the sign, corresponding to the $\int_a^b f\,dx = -\int_b^a f\,dx$ property from Section 02, but applied by accident rather than intentionally.

Consequence: since $f(x)=x^2\geq0$ on $[0,2]$ , the true signed area must be non-negative; a negative result like $-\frac83$ is an immediate signal that the subtraction order was inverted.

06 · Exercises

EXERCISE 6.1

Compute the subinterval width, list the right-endpoint sample points, evaluate the function at each, then sum height times width.

$f(x)=x+1$ on $[0,4]$ , $n=4$ . $\Delta x = 4/4=1$ . Right endpoints: $1,2,3,4$ . Heights: $f(1)=2,f(2)=3,f(3)=4,f(4)=5$ . Sum: $(2+3+4+5)\times1=14$ .

Approximate $\int_0^4 (x+1)\,dx$ using a right-endpoint Riemann sum with $4$ rectangles.

EXERCISE 6.2

Find an antiderivative using the power rule for integration, then apply FTC Part 2 and compare to Exercise 6.1's approximation.

$F(x)=\dfrac{x^2}{2}+x$ . $\int_0^4(x+1)\,dx = F(4)-F(0) = (8+4)-(0+0)=12$ . Comparing to Exercise 6.1's estimate of $14$ : the right-endpoint sum overshoots because $f(x)=x+1$ is increasing, same pattern as Section 01.

Compute the exact value of $\int_0^4(x+1)\,dx$ using FTC Part 2, and compare to the Riemann sum estimate from Exercise 6.1.

EXERCISE 6.3

Integrate term by term using the power rule for integration; don't forget the constant of integration.

$\int(6x^2-4x+3)\,dx$ . Term by term: $6x^2\to6\cdot\frac{x^3}{3}=2x^3$ ; $-4x\to-4\cdot\frac{x^2}{2}=-2x^2$ ; $3\to3x$ . Result: $2x^3-2x^2+3x+C$ .

Find $\int(6x^2-4x+3)\,dx$ .

EXERCISE 6.4

Apply FTC Part 1 directly — the derivative of a definite integral with variable upper limit is just the integrand evaluated there.

$G(x)=\displaystyle\int_1^x \sqrt{t^3+1}\,dt$ . By FTC Part 1: $G'(x) = \sqrt{x^3+1}$ — substitute $x$ for $t$ , no integration needed.

Given $G(x)=\displaystyle\int_1^x \sqrt{t^3+1}\,dt$ , find $G'(x)$ using FTC Part 1.

EXERCISE 6.5

Find an antiderivative first, then evaluate at the upper limit minus the lower limit, being careful with the order.

$\int_1^3 (2x-x^2)\,dx$ . Antiderivative: $F(x)=x^2-\dfrac{x^3}{3}$ . $F(3)=9-9=0$ . $F(1)=1-\dfrac13=\dfrac23$ . $F(3)-F(1)=0-\dfrac23=-\dfrac23$ . The negative result is valid here (unlike Section 05's fail case) because $2x-x^2<0$ for $x>2$ , so the region on $(2,3)$ contributes negative signed area that outweighs the positive part on $(1,2)$ .

Evaluate $\int_1^3(2x-x^2)\,dx$ . The result is negative — explain why that is a valid, not an error, given the sign of the integrand.

EXERCISE 6.6

Total accumulated quantity from a rate function is exactly what a definite integral computes — integrate the rate function over the given time interval.

Cash flow rate $r(t) = 500 - 20t$ (dollars/day). Total accumulated cash over $[0,10]$ : $\displaystyle\int_0^{10}(500-20t)\,dt$ . Antiderivative: $F(t)=500t-10t^2$ . $F(10)=5000-1000=4000$ . $F(0)=0$ . Total $=4000-0=4000$ dollars.

A trading desk's cash flow rate is $r(t)=500-20t$ dollars per day, for $t$ in $[0,10]$ . Find the total accumulated cash over those 10 days using a definite integral.

07 · Chapter Summary

Concept	Formula / Rule
Riemann sum	$\sum_{i=1}^n f(x_i^*)\Delta x$ ; approximates area with rectangles
Definite integral	$\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum f(x_i^*)\Delta x$ ; signed area
Antiderivative	$F'(x)=f(x)$
Indefinite integral	$\int f(x)\,dx = F(x)+C$
Power rule for integration	$\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C$ , $n\neq-1$
FTC Part 1	$\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$
FTC Part 2	$\int_a^b f(x)\,dx = F(b)-F(a)$

Next: Chapter 07 — Techniques of Integration, which builds a toolkit for antiderivatives that the power rule alone cannot reach.