# 7.1–7.2 Definite Integration

F.5 · M1 Mathematics · Definite Integrals · Lesson 5

From thin rectangles to a signed value: find a primitive, subtract the endpoints, and split where the formula or sign changes.

## Figure

For $y=x-1$, the two triangular areas are equal: signed contributions cancel, but the total area does not.

## Key points

- **Definite integral (定積分):** divide $[a,b]$ into $n$ equal widths $\Delta x=(b-a)/n$, choose $z_i$ in each sub-interval, add $f(z_i)\Delta x$, then take the limit as $n\to\infty$. For continuous $f$, the limit exists.
- **A value, not a family:** $\int_a^b f(x)\,dx$ is a number; $\int f(x)\,dx=F(x)+C$ is a family of primitive functions. Changing the dummy variable consistently leaves the value unchanged.
- **Signed area:** with $a<b$, regions above the axis contribute positively and those below negatively. Equal positive and negative areas can give a zero integral.
- **Primitive (原函數):** find $F$ with $F'(x)=f(x)$, then use $F(b)-F(a)$. Example: $\int_1^3 2x\,dx=[x^2]_1^3=9-1=8$.
- **Integration limits (積分上、下限):** upper-endpoint value minus lower-endpoint value. Reversing limits changes the sign; equal limits give zero.
- **No final $+C$:** $(F(b)+C)-(F(a)+C)=F(b)-F(a)$; the same constant cancels.
- **Linearity:** split sums or differences and take constants outside. For $3x^2-2x+1$, a primitive is $x^3-x^2+x$.
- **Split intervals:** $\int_a^b f=\int_a^c f+\int_c^b f$; use it to combine parts or find a missing part. Check continuity on the intervals involved.
- **Piecewise / absolute value:** split at formula breakpoints and relevant zeros. For $|x-1|$, use $1-x$ below 1 and $x-1$ above 1.
- **Check the domain:** the integrand must be defined throughout the interval before applying these continuous-integrand rules; endpoint substitution alone is not enough.

## Must know

| Situation | Rule / method |
|---|---|
| Evaluate a continuous integrand | $\int_a^b f(x)\,dx=F(b)-F(a)$, where $F'=f$ |
| Reverse / equal limits | $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$; $\int_a^a f(x)\,dx=0$ |
| Constant factor | $\int_a^b kf(x)\,dx=k\int_a^b f(x)\,dx$ |
| Sum / difference | $\int_a^b(f\pm g)\,dx=\int_a^b f\,dx\pm\int_a^b g\,dx$ |
| Split interval | $\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$ |
| Total area, $a<b$ | Split where $f$ changes sign; add the positive magnitudes, equivalently $\int_a^b |f(x)|\,dx$ |

These rules assume continuous integrands on the intervals involved. A primitive is found using the usual indefinite-integration rules; check it by differentiating before substituting the limits.

## Common mistakes

- ✗ $\int_1^3 2x\,dx=1-9=-8$. ✓ Upper value minus lower value: $[x^2]_1^3=9-1=8$.
- ✗ $\int_0^2 x\,dx=2+C$. ✓ The same $C$ cancels: $[x^2/2]_0^2=2$, a number.
- ✗ Reversing limits leaves the value unchanged. ✓ It changes the sign: if $\int_1^4 f=7$, then $\int_4^1 f=-7$.
- ✗ Zero integral means zero enclosed area. ✓ Above-axis and below-axis contributions can cancel; total area adds their magnitudes.
- ✗ $|x-1|=x-1$ throughout $[0,2]$. ✓ Split at $x=1$: use $1-x$ on $[0,1]$ and $x-1$ on $[1,2]$.

## Quick check

1. Evaluate $\int_0^2(3x^2-2x+1)\,dx$, showing endpoint substitution.
2. If $F'=f$, $F(1)=3$ and $F(4)=10$, find $\int_4^1 f(x)\,dx$.
3. Given $\int_0^3 f(x)\,dx=8$ and $\int_1^3 f(x)\,dx=5$, find $\int_0^1(2f(x)-1)\,dx$.
4. A continuous function is $f(x)=x$ for $0\le x\le1$ and $f(x)=2-x$ for $1<x\le2$. Find $\int_0^2 f(x)\,dx$.
5. For the graph $y=x-1$ on $[0,2]$, find its definite integral and total enclosed area. Explain the difference.

## Answers

1. $[x^3-x^2+x]_0^2=8-4+2=6$.
2. $F(1)-F(4)=3-10=-7$.
3. Missing part $=8-5=3$; $2(3)-[x]_0^1=5$.
4. $[x^2/2]_0^1+[2x-x^2/2]_1^2=1/2+1/2=1$.
5. Integral $=-1/2+1/2=0$; total area $=1/2+1/2=1$. Signed contributions cancel; area magnitudes add.

## Did you know?

Positive and negative signed areas can cancel, so a zero definite integral need not mean that the graph encloses no area.