Areas
Integration lets us measure the space under a curve. Like calculating the total rainfall from a rate graph, the definite integral accumulates area between a function and the axis, revealing totals that simple geometry cannot. The fundamental theorem of calculus is the key that unlocks the calculation.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Before we start, what do you already know about measuring the space under a curve? How would you estimate the area under $y = x^2$ between $x = 0$ and $x = 2$?
There are only two moves in this entire lesson. Lock them into muscle memory and the rest is just calculation.
Move 1 — Integrate. Find the antiderivative $F(x)$ of $f(x)$. This is the reverse of differentiation: increase the power by one and divide.
Move 2 — Evaluate at limits and subtract. Compute $F(b) - F(a)$. This gives the signed area between the curve and the $x$-axis from $x = a$ to $x = b$.
Key facts
- The fundamental theorem: $\int_a^b f(x)\,dx = F(b) - F(a)$
- Area above the axis is positive; area below is negative (signed area)
- Area between curves: $\int_a^b (\text{top} - \text{bottom})\,dx$
Concepts
- Why the definite integral gives signed area
- How to handle regions that cross the $x$-axis
- Why we need to identify the “top” curve when finding area between two curves
Skills
- Evaluate definite integrals using the fundamental theorem
- Find the area between a curve and the $x$-axis
- Find the area enclosed between two curves
The definite integral $\int_a^b f(x)\,dx$ measures the signed area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$. The fundamental theorem of calculus converts this into a calculation with antiderivatives:
Areas above the $x$-axis contribute positively. Areas below the $x$-axis contribute negatively. To find total area, split at $x$-intercepts and take absolute values.
Green (above axis): positive signed area. Red (below axis): negative signed area. For total area, take the absolute value of each piece.
Area between two curves
When two curves intersect, the enclosed area is found by integrating the difference between the top and bottom curves over the interval between their intersection points:
Fundamental theorem: $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ where $F'(x) = f(x)$; Signed area: above axis $= +$; below axis $= -$
Pause — copy the Fundamental Theorem $\int_a^b f(x)\,dx = F(b) - F(a)$ and the signed-area rule (above $x$-axis = positive, below = negative) into your book.
Quick check: True or false — $\displaystyle\int_0^2 (x^2 - 1)\,dx$ gives the total (unsigned) area between $y = x^2 - 1$ and the $x$-axis on $[0, 2]$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$.
Find the total area bounded by $y = x^2 - 4$, the $x$-axis, $x = 0$ and $x = 3$.
Find the area enclosed by $y = x^2$ and $y = x$.
Quick check: What is $\displaystyle\int_0^2 3x^2\,dx$?
Common errors · the 3 traps that cost marks
Odd one out: Three of the following describe the definite integral correctly. Which one is WRONG?
Quick-fire practice · 5 problems
Evaluate $\displaystyle\int_0^2 (3x^2 + 2x)\,dx$.
Find the area bounded by $y = x^2$, the $x$-axis, $x = 1$ and $x = 3$.
Find the total area between $y = x^2 - 1$ and the $x$-axis from $x = 0$ to $x = 2$.
Find the area enclosed by $y = x^2$ and $y = 2x$.
Find the area between $y = x^3$ and $y = x$ for $x \ge 0$.
Fill the blanks: drag each token into the matching blank.
The fundamental theorem says: find the ___ $F(x)$, then ___ $F(a)$ from $F(b)$. For total area when the curve dips below the axis, take the ___ value of each piece. For area between two curves, first find their ___ points.
Match each integral to its correct value.
- $\displaystyle\int_0^1 x^2\,dx$
- $\displaystyle\int_1^3 2x\,dx$
- $\displaystyle\int_0^2 (x+1)\,dx$
- $\displaystyle\int_0^3 1\,dx$
- $3$
- $4$
- $8$
- $\tfrac{1}{3}$
Earlier you were asked about measuring space under curves. The definite integral does exactly this. The fundamental theorem lets us calculate it using antiderivatives, avoiding the need for approximation methods. The key rule to remember: the integral gives signed area, so you must handle below-axis regions with care.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\displaystyle\int_1^4 (2x - 3)\,dx$. Show all working. 2 MARKS
Q2. Find the total area bounded by $y = x^2 - 4x + 3$, the $x$-axis, and the lines $x = 0$ and $x = 4$. Show all working. 4 MARKS
Q3. Find the area enclosed by the curves $y = x^2$ and $y = 2 - x^2$. Show all working. 4 MARKS
📖 Comprehensive answers (click to reveal)
Drill 1: $\left[x^3 + x^2\right]_0^2 = 12$.
Drill 2: $\left[\frac{x^3}{3}\right]_1^3 = 9 - \frac{1}{3} = \frac{26}{3}$ sq units.
Drill 3: Intercept at $x = 1$. $\left|\int_0^1(x^2-1)\,dx\right| + \int_1^2(x^2-1)\,dx = \frac{2}{3} + \frac{4}{3} = 2$ sq units.
Drill 4: Intersect at $x = 0, 2$. $\int_0^2(2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$ sq units.
Drill 5: $x^3 = x$ at $x = 0, 1$. On $[0,1]$, $x \ge x^3$. Area $= \int_0^1(x-x^3)\,dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ sq units.
Q1 (2 marks): $\left[x^2 - 3x\right]_1^4 = (16-12) - (1-3) = 4 - (-2) = 6$ [2].
Q2 (4 marks): $y = (x-1)(x-3)$, intercepts at $x = 1, 3$ [0.5]. $\int_0^1 = \frac{4}{3}$; $\left|\int_1^3\right| = \frac{4}{3}$; $\int_3^4 = \frac{4}{3}$ [2.5]. Total area $= \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ sq units [1].
Q3 (4 marks): $x^2 = 2 - x^2 \Rightarrow x = \pm 1$ [1]. Top curve: $y = 2 - x^2$ [0.5]. Area $= \int_{-1}^{1}(2 - 2x^2)\,dx = \left[2x - \frac{2x^3}{3}\right]_{-1}^{1} = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ sq units [2.5].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering integration and area questions. Lighter alternative to the boss.
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