Week 11 In-Depth — Integration, the Fundamental Theorem, and Why It All Works

Read the cheatsheet first. This note explains why each rule works, how the Fundamental Theorem links the two definitions of an integral, and walks several full worked examples.

1 — What integration actually is

There are two definitions of “integral” that coexist, and the Fundamental Theorem of Calculus says they happen to give the same answer for nice functions.

Definition A — antiderivative (“indefinite integral”)

If $F^{\prime }(x)=f(x)$, then $F$ is an antiderivative of $f$. We write

$$\int f(x)dx=F(x)+c.$$

The $+c$ is there because if $F$ is one antiderivative, so is $F+c$ for any constant $c$ — they all differentiate to the same $f$. So $\int fdx$ is a family of functions, not a single one.

This is the “undo differentiation” view. Every rule on the cheatsheet is just a derivative rule read backwards.

Definition B — Riemann sum (“definite integral”)

Take the interval $[a,b]$, chop it into $n$ thin strips of width $\Delta x=(b-a)/n$, pick a sample point $x_i^{\ast }$ in each strip, and form the sum

$$S_n=\sum _{i=1}^{n}f(x_i^{\ast })\Delta x.$$

Each term $f(x_i^{\ast })\Delta x$ is the (signed) area of a thin rectangle. Let $n\to \infty$ and the sum approaches a number we call

$${\int }_a^{b}f(x)dx=\underset{n\to \infty }{\lim }\sum _{i=1}^{n}f(x_i^{\ast })\Delta x.$$

This is the “signed area under the curve” view. It is a number, not a function.

The Fundamental Theorem of Calculus

The connection between A and B is the central result of single-variable calculus:

$${\int }_a^{b}f(x)dx=F(b)-F(a)$$

where $F$ is any antiderivative of $f$. This is what makes calculus useful — instead of summing infinitely many rectangles, you find an antiderivative and subtract.

Why does the $+c$ disappear? Because $(F(b)+c)-(F(a)+c)=F(b)-F(a)$. So even though the indefinite integral is a family, the definite integral picks out a single number.

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2 — Reading each rule as a backwards derivative

Every entry on the cheatsheet has a corresponding derivative rule:

Derivative	Antiderivative
$\frac{d}{dx}(kx)=k$	$\int kdx=kx+c$
$\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n$	$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$
$\frac{d}{dx}\ln |x|=\frac{1}{x}$	$\int \frac{1}{x}dx=\ln |x|+c$
$\frac{d}{dx}{e}^x=e^x$	$\int e^{x}dx=e^x+c$
$\frac{d}{dx}\sin (x)=\cos (x)$	$\int \cos (x)dx=\sin (x)+c$
$\frac{d}{dx}(-\cos (x))=\sin (x)$	$\int \sin (x)dx=-\cos (x)+c$
$\frac{d}{dx}\tan (x)={\sec }^2(x)$	$\int {\sec }^2(x)dx=\tan (x)+c$

If you ever forget a rule, differentiate your guess. If you get back the original integrand, the antiderivative is correct (up to $+c$).

The $\frac{1}{a}$ factor for $(ax+b)$ forms

The chain rule says

$$\frac{d}{dx}F(ax+b)=F^{\prime }(ax+b)\cdot a.$$

That extra $a$ on the right is the chain-rule kicker. Read it backwards:

$$\int F^{\prime }(ax+b)dx=\frac{1}{a}F(ax+b)+c.$$

You divide by $a$ to cancel the chain-rule factor. That is why every $(ax+b)$ rule on the cheatsheet has a $\frac{1}{a}$ in front.

This isn’t a special new rule — it’s the same chain rule you already know, run in reverse. Skipping the $\frac{1}{a}$ is the most common source of lost marks for this whole week.

The $1/x$ exception

The power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ blows up when $n=-1$ — you’d divide by zero. So $\int x^{-1}dx$ needs a different antiderivative entirely, and that antiderivative is $\ln |x|$. The absolute-value bars are required so the rule works for $x<0$ too.

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3 — Worked walk-throughs

3.1 An indefinite integral with everything in it

Evaluate $\int \left(5x^2-\cos \left(\frac{4x}{5}\right)\right)dx$.

Linearity splits the sum:

$$\int 5x^{2}dx-\int \cos \left(\frac{4x}{5}\right)dx.$$

First piece. Power rule with $n=2$ and constant 5 out front:

$$\int 5x^{2}dx=5\cdot \frac{x^3}{3}+c_1=\frac{5x^3}{3}+c_1.$$

Second piece. $\cos (ax)$ with $a=4/5$. Antiderivative is $\frac{1}{a}\sin (ax)=\frac{5}{4}\sin \left(\frac{4x}{5}\right)$.

So

$$\int \cos \left(\frac{4x}{5}\right)dx=\frac{5}{4}\sin \left(\frac{4x}{5}\right)+c_2.$$

Combine.

$$\int \left(5x^2-\cos \left(\frac{4x}{5}\right)\right)dx=\frac{5x^3}{3}-\frac{5}{4}\sin \left(\frac{4x}{5}\right)+c.$$

(Two constants collapse into one.)

3.2 Finding the antiderivative from $f^{\prime }(x)$

$f^{\prime }(x)=\frac{2}{3-7x}$. Find $f(x)$.

Use $\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+c$ with $a=-7$, $b=3$, and a constant $2$ pulled outside:

$$f(x)=2\cdot \frac{1}{-7}\ln |3-7x|+c=-\frac{2}{7}\ln |3-7x|+c.$$

3.3 Expand-then-integrate

$\int x(x+1{)}^{2}dx$.

There is no “product rule for integrals.” When two factors aren’t related by derivative/antiderivative, just expand:

$$x(x+1{)}^2=x(x^2+2x+1)=x^3+2x^2+x.$$

Integrate term by term:

$$\int (x^3+2x^2+x)dx=\frac{x^4}{4}+\frac{2x^3}{3}+\frac{x^2}{2}+c.$$

3.4 A definite integral end-to-end

${\int }_1^3(3x^2+x-2)dx$.

Antiderivative. $F(x)=x^3+\frac{x^2}{2}-2x$. (No $+c$ — it cancels.)

Top minus bottom.

$$\begin{array}{rl}F(3)& =27+\frac{9}{2}-6=25.5,\\ F(1)& =1+\frac{1}{2}-2=-0.5,\\ {\int }_1^3(3x^2+x-2)dx& =25.5-(-0.5)=26.\end{array}$$

3.5 Find-the-unknown-constant

Given ${\int }_1^4(3x^2+ax-5)dx=18$, find $a$.

Antiderivative. $F(x)=x^3+\frac{a{x}^2}{2}-5x$.

Top minus bottom.

$$\begin{array}{rl}F(4)& =64+8a-20=44+8a,\\ F(1)& =1+\frac{a}{2}-5=\frac{a}{2}-4,\\ F(4)-F(1)& =(44+8a)-\left(\frac{a}{2}-4\right)=48+\frac{15}{2}a.\end{array}$$

Set equal to 18:

$$48+\frac{15}{2}a=18⟹\frac{15}{2}a=-30⟹a=-4.$$

3.6 Geometric area vs. signed integral

Find the area under $y=x^2+2x-15$ between $x=0$ and $x=3$.

Step 1 — sketch. Factor: $y=(x+5)(x-3)$. Roots at $-5$ and $3$. The parabola opens up, so it is negative on $(-5,3)$. On $[0,3]$ it is entirely below the x-axis.

Step 2 — compute the signed integral first.

$${\int }_0^3(x^2+2x-15)dx={\left[\frac{x^3}{3}+x^2-15x\right]}_0^3=(9+9-45)-0=-27.$$

Step 3 — take the absolute value. Geometric area $=|-27|=27{\text{units}}^2$.

Don’t skip step 1. If the question said “find ${\int }_0^3\dots dx$”, the answer would be $-27$. If it says “find the area”, the answer is $+27$. Read the question.

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4 — The Fundamental Theorem, pictorially

If you sketch $f(x)$ and shade the region under the curve from $a$ to $b$:

• The Riemann-sum view says: “chop the region into thin rectangles, sum their (signed) areas, take a limit.”
• The antiderivative view says: “find $F$ such that $F^{\prime }(x)=f(x)$, evaluate $F(b)-F(a)$.”

The Fundamental Theorem says these two procedures give the same number. That is genuinely surprising — there’s no reason a priori that “find a function whose derivative is $f$” should have anything to do with “sum thin rectangles.” But it does.

This is what makes calculus so powerful. We almost never compute Riemann sums in practice — we always go through antiderivatives, because they’re enormously easier.

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5 — The mental model for “which rule do I use?”

When you stare at an integral, ask in this order:

1. Can I expand or simplify first? Distribute multiplications, split sums, cancel obvious terms. Most workshop integrals reduce to a sum of standard forms after expanding.
2. Is each term a standard rule? Power $x^n$, exponential $e^x$ or $e^{ax+b}$, trig, $1/x$ or $1/(ax+b)$.
3. If there’s a linear inner function $(ax+b)$, remember the $1/a$.
4. Differentiate the answer to check. If you don’t recover the original integrand, you missed a chain-rule factor or dropped a sign.

For this week’s workshop, no problem requires $u$-substitution or integration by parts. If you’re tempted to reach for something fancier, look again — there’s an expansion or a $(ax+b)$ form you can use instead.

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6 — Common mistakes, with the reasoning behind them

Mistake	Why it happens	Fix
Using the power rule on $1/x$	The formula $\frac{x^{n+1}}{n+1}$ is undefined at $n=-1$	Memorise $\int 1/xdx=\ln |x|+c$ as a separate rule
Forgetting the $+c$	Habit from definite integrals where it cancels	Write $+c$ as you write the antiderivative, not at the end
Dropping the $\frac{1}{a}$ on $(ax+b)$ forms	The chain-rule factor is invisible until you differentiate to check	Always differentiate your answer back. If a stray $a$ appears, you forgot $1/a$
$F(a)-F(b)$ instead of $F(b)-F(a)$	Inattention	Write “top − bottom” next to the antiderivative
Treating ${\int }_a^{b}f$ as area when $f$ dips below the axis	Conflating the two definitions	Sketch first when the question says “area”

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7 — A final exam-style worked example

Find $\int \left(6e^{2x}-\frac{3}{x}\right)dx$ and evaluate the same integrand between $x=1$ and $x=2$.

Indefinite.

$$\int 6e^{2x}dx=6\cdot \frac{1}{2}{e}^{2x}=3e^{2x}.$$ $$\int \frac{3}{x}dx=3\ln |x|.$$

So

$$\int \left(6e^{2x}-\frac{3}{x}\right)dx=3e^{2x}-3\ln |x|+c.$$

Definite on $[1,2]$. Let $F(x)=3e^{2x}-3\ln |x|$.

$$\begin{array}{rl}F(2)& =3e^4-3\ln 2\approx 163.79-2.08=161.71,\\ F(1)& =3e^2-3\ln 1=3e^2-0\approx 22.17,\\ {\int }_1^2\left(6e^{2x}-\frac{3}{x}\right)dx& =F(2)-F(1)\approx \mathrm{139.54.}\end{array}$$

That’s the entire workflow: rule lookup → linearity → antiderivative → (if definite) top minus bottom. Two-line setup, four-line solve.

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Where to look next

• The cheatsheet for revision + the quiz.
• The workshop PDFs (cited in frontmatter) for more practice.
• Your own additions in this folder — Lecture Atlas auto-discovers any new Markdown note.