# The Definite Integral and Area Under a Curve

> Area from rectangles and trapeziums, the trapezoidal rule, the fundamental theorem of calculus and signed area.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/definite-integrals-and-area/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The bushwalk distance
Lucas's fitness watch records his speed every minute during a bushwalk, but it lost signal and forgot the total distance. All he has is a graph of speed against time.

At a constant speed, distance is speed times time: the area of a rectangle under the graph. But his speed kept changing, so the graph is a wiggly curve. The distance is still the **area under the curve**. Integration is the tool that finds that area exactly.

**In short:** A definite integral $\int_a^b f(x)\,dx$ gives the area under a curve between $x = a$ and $x = b$, found as $F(b) - F(a)$ from a primitive $F$. Area below the x-axis comes out negative, so split the integral at the roots to find the total area.

## Starting point: Years 9 and 10

### Area formulas

$$\text{rectangle: } A = lw \qquad \text{triangle: } A = \frac{1}{2}bh \qquad \text{trapezium: } A = \frac{h}{2}(a + b)$$

### Area under a velocity graph is distance

If a car travels at 20 m/s for 10 s, the velocity-time graph is a horizontal line, and the area under it is $20 \times 10 = 200$ m, the distance travelled.

### Worked example: A car speeding up
A car goes from 0 m/s to 30 m/s steadily over 12 s. The graph is a straight line and the area under it is a triangle: $\dfrac{1}{2} \times 12 \times 30 = 180$ m.

### Estimating area under a curve

For a curve, split the area into thin strips and add them up. Using rectangles:

*Diagram: Four rectangles under the curve y equals one half x squared plus 1 from x equals 0 to 4*
*Four strips of width 1. Each rectangle uses the height at its left edge.*

The rectangles here use the height at the left of each strip: $1 + 1.5 + 3 + 5.5 = 11$. The true area is a little more, because the curve rises above each rectangle. **Thinner strips give a better estimate.**

### Key idea: Area as a sum of strips
Area under a curve $\approx$ sum of (height $\times$ width) of many thin strips. As the strips get thinner, the sum gets closer to the exact area.

### Watch out: Units of area
On a velocity-time graph the area has units of m/s $\times$ s $=$ m. Multiply the units on the two axes.

### Calculator tip: Check with the integral key
If your calculator has a $\int \square \, dx$ key, use it to check a definite integral. It gives the signed value: parts below the $x$-axis count as negative. For an area, split at the $x$-intercepts and add the sizes. Use radian mode for trig functions.

**Mind map**

```text
Area under a curve
  Shapes
    Rectangle, triangle, trapezium
  Estimate
    Thin strips
    Trapezoidal rule
  Exact
    Definite integral
    F(b) - F(a)
  Sign
    Above axis: positive
    Below axis: negative
    Split at roots for area
  Meaning
    Velocity area: distance
    Rate area: total change
```

### Year 11: Upper and lower sums
Use rectangles touching the curve at the **top** of each strip for an upper sum, and at the **bottom** for a lower sum. For an increasing curve, right-hand heights give the upper sum and left-hand heights give the lower sum. The true area is between them.

For the curve above, the right-hand heights are $1.5 + 3 + 5.5 + 9 = 19$. So $11 < \text{area} < 19$. Halving the strip width squeezes the two sums closer together.

### Year 12: The trapezoidal rule
Using trapeziums instead of rectangles is much more accurate. With strips of width $h$ and heights $y_0, y_1, \dots, y_n$:

$$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0 + 2(y_1 + y_2 + \dots + y_{n-1}) + y_n\right]$$

For the curve above with $h = 1$: $\dfrac{1}{2}[1 + 2(1.5 + 3 + 5.5) + 9] = 15$.

**Over or under?** For a curve that is **concave up**, the straight tops of the trapeziums sit above the curve, so the rule **overestimates**. For a curve that is **concave down**, it **underestimates**.

### Year 12: The definite integral and the fundamental theorem
The exact area under $y = f(x)$ from $x = a$ to $x = b$ (with $f \geq 0$) is the **definite integral**:

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

The **fundamental theorem of calculus** links this area to anti-derivatives. If $F$ is any primitive of $f$:

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

**Example.** The curve above is $y = \dfrac{1}{2}x^2 + 1$.

$$\int_0^4 \left(\frac{1}{2}x^2 + 1\right)dx = \left[\frac{x^3}{6} + x\right]_0^4 = \frac{64}{6} + 4 = 14\frac{2}{3}$$

The trapezoidal estimate of 15 was close.

**Signed area.** Area below the x-axis counts as **negative** in a definite integral. $\displaystyle\int_0^{2\pi}\sin x\,dx = 0$ because the part above cancels the part below. To find the actual **area**, split at the x-intercepts and add the absolute values:

$$\text{Area} = \int_0^{\pi}\sin x\,dx + \left|\int_{\pi}^{2\pi}\sin x\,dx\right| = 2 + 2 = 4$$

**Useful properties.** $\displaystyle\int_a^a f(x)\,dx = 0$, $\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$, and an odd function integrated from $-a$ to $a$ gives 0.

**More properties.** Integrals split over sums and over intervals: $\displaystyle\int_a^b \big(f(x) + g(x)\big)dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx$, $\displaystyle\int_a^b kf(x)\,dx = k\int_a^b f(x)\,dx$, and $\displaystyle\int_a^c f = \int_a^b f + \int_b^c f$. Shifting the variable shifts the limits: $\displaystyle\int_1^4 f(x + 2)\,dx = \int_3^6 f(x)\,dx$.

**Use geometry when you can.** $\displaystyle\int_0^4 |x - 2|\,dx$ is two triangles, each of area 2, so it equals 4. $\displaystyle\int_0^3 \sqrt{9 - x^2}\,dx$ is a quarter of a circle of radius 3, so it equals $\dfrac{9\pi}{4}$.

**Unknown limits or constants.** Evaluate as usual, then solve. $\displaystyle\int_1^k \frac{1}{x}\,dx = \ln k = 3$ gives $k = e^3$. $\displaystyle\int_0^a 2x\,dx = a^2 = 9$ gives $a = 3$ (for $a > 0$).

**Area against the $y$-axis.** For the region between $y = \ln x$, the $y$-axis and the lines $y = 0$ and $y = 1$, write $x = e^y$ and integrate with respect to $y$: $\displaystyle\int_0^1 e^y\,dy = e - 1$.

## Quick summary

- The area under a velocity-time graph is the distance.
- Estimate area with thin rectangles or trapeziums. Thinner strips are more accurate.
- Trapezoidal rule: $\dfrac{h}{2}\left[y_0 + 2(\text{middle heights}) + y_n\right]$.
- Fundamental theorem: $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$.
- Area below the axis gives a negative integral. For total area, split at the roots.

## Quiz

Try these questions from the chapter quiz.

1. What is the area of a trapezium with parallel sides 4 and 10 and height 6?
   - A. 60
   - B. 42
   - C. 84
   - D. 20
   - Answer: B. 42. $\dfrac{6}{2}(4 + 10) = 42$.

2. A cyclist rides at 8 m/s for 30 s. How far does she travel?
   - A. 38 m
   - B. 240 m
   - C. 3.75 m
   - D. 120 m
   - Answer: B. 240 m. Area of the rectangle: $8 \times 30$.

3. A car speeds up steadily from 0 to 20 m/s in 10 s. How far does it go?
   - A. 200 m
   - B. 100 m
   - C. 20 m
   - D. 2 m
   - Answer: B. 100 m. Triangle: $\dfrac{1}{2} \times 10 \times 20$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/definite-integrals-and-area/#quiz