How Adding Up Tiny Strips Gives an Exact Area

An indefinite integral is a family of functions; a definite integral is a single number. It adds up infinitely many thin pieces — of area, distance or work — between two fixed limits, and the Fundamental Theorem of Calculus lets us compute it using antiderivatives.

This part covers the definition and the Fundamental Theorem, evaluation, substitution with changed limits, and the properties of definite integrals.

**The definite integral ${\int }_a^{b}f(x)dx$ is the signed area between $y=f(x)$ and the $x$-axis from $x=a$ to $x=b$, and the Fundamental Theorem of Calculus says it equals $F(b)-F(a)$ for any antiderivative $F$ of $f$.

The notation:**

- $a$ is the lower limit and $b$ the upper limit
- $f(x)$ is continuous on $[a,b]$
- The result is a number, not a function

Area as a limit of sums. Divide $[a,b]$ into $n$ strips of width $h=\frac{b-a}{n}$; as $n\to \infty$, the total area of the rectangles approaches the definite integral.

First Fundamental Theorem. The area function $A(x)={\int }_a^{x}f(t)dt$ satisfies $A^{\prime }(x)=f(x)$.

Second Fundamental Theorem. If $F^{\prime }(x)=f(x)$ on $[a,b]$, then

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

Worked example. ${\int }_0^{4}xdx$ is the area under $y=x$ from $0$ to $4$, a triangle with base $4$ and height $4$:

$${\int }_0^{4}xdx={\left[\frac{x^2}{2}\right]}_0^4=8-0=8=\frac{1}{2}\times 4\times 4$$

An everyday example. Adding up a car's speed over many tiny time intervals gives the total distance travelled — a definite integral of speed.

The substance. **No constant $C$ is needed** — it cancels in $F(b)-F(a)$.

**Find any antiderivative $F$ of the integrand, substitute the upper limit and then the lower limit, and subtract to get $F(b)-F(a)$.

Worked example 1.**

$${\int }_1^3(3x^2+2)dx=[x^3+2x{]}_1^3=(27+6)-(1+2)=30$$

Worked example 2.

$${\int }_0^{\pi /2}\cos xdx=[\sin x{]}_0^{\pi /2}=1-0=1$$

Worked example 3.

$${\int }_1^e\frac{1}{x}dx=[\log x{]}_1^e=1-0=1$$

Worked example 4 — signed area.

$${\int }_0^{2\pi }\sin xdx=[-\cos x{]}_0^{2\pi }=-1+1=0$$

The area above the axis from $0$ to $\pi$ cancels the equal area below it from $\pi$ to $2\pi$.

An everyday example. Subtracting the start-of-month water meter reading from the end-of-month reading gives the water used — exactly the $F(b)-F(a)$ step.

The substance. A zero definite integral does not mean zero area — regions below the axis count as negative.

**When substituting $t=g(x)$ in a definite integral, convert $dx$ into $dt$ and convert both limits into values of $t$, so the integral can be finished in $t$ without substituting back.

The method:**

- Choose $t=g(x)$ and find $dt=g^{\prime }(x)dx$
- New limits: $g(a)$ and $g(b)$
- Evaluate the new integral between these limits

Worked example 1. ${\int }_0^{1}2x(x^2+1{)}^{3}dx$. Put $t=x^2+1$, so $dt=2xdx$; $x=0$ gives $t=1$ and $x=1$ gives $t=2$.

$${\int }_1^2{t}^{3}dt={\left[\frac{t^4}{4}\right]}_1^2=\frac{16}{4}-\frac{1}{4}=\frac{15}{4}$$

Worked example 2. ${\int }_0^{\pi /2}{\sin }^{2}x\cos xdx$. Put $t=\sin x$, so $dt=\cos xdx$; the limits become $0$ and $1$.

$${\int }_0^1{t}^{2}dt={\left[\frac{t^3}{3}\right]}_0^1=\frac{1}{3}$$

An everyday example. Converting a route map from kilometres to miles means changing both the start and end markers, not just the scale — as substitution changes both limits.

The substance. **Using the old $x$-limits with the new variable $t$ gives a wrong answer** — either change the limits or substitute back, never neither.

**Definite integrals change sign when the limits are swapped, can be split at an interior point, vanish over symmetric limits for odd functions, double over half the range for even functions, and stay unchanged when $x$ is replaced by $a+b-x$.

The properties:

- Reversal** — ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$
- Additive — ${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$
- Complementary limits — ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$
- Even and odd — ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$ if $f$ is even, and $0$ if $f$ is odd

Worked example 1 — an odd function. $x^3\cos x$ is odd, so ${\int }_{-1}^1{x}^3\cos xdx=0$.

Worked example 2 — splitting a modulus.

$${\int }_{-1}^2|x|dx={\int }_{-1}^0(-x)dx+{\int }_0^{2}xdx=\frac{1}{2}+2=\frac{5}{2}$$

Worked example 3 — complementary limits. Let $I={\int }_0^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx$. Replacing $x$ by $\frac{\pi }{2}-x$ gives $I={\int }_0^{\pi /2}\frac{\cos x}{\cos x+\sin x}dx$. Adding the two forms,

$$2I={\int }_0^{\pi /2}1dx=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$$

An everyday example. Reading a symmetric rangoli pattern from either end shows the same design — the idea behind replacing $x$ by $a+b-x$.

The substance. Check whether the integrand is odd or even first — it can turn a long calculation into a one-line answer.