Area Between a Curve and a Line

The area between two graphs is the integral of their difference, $\int(\text{upper} - \text{lower})\,dx$, taken between their intersection points. Finding the limits from the intersection, and subtracting in the right order, are the whole skill.

Find the area of the region enclosed between the curve $y = x^2$ and the line $y = 2x$. [6]

The working

Step 1, find the intersection points (the limits) by setting curve $=$ line: $x^2 = 2x \;\Rightarrow\; x^2 - 2x = 0 \;\Rightarrow\; x(x - 2) = 0 \;\Rightarrow\; x = 0 \text{ and } x = 2 \quad \text{(M1, A1)}$

Step 2, decide which is on top. Between $0$ and $2$, test $x = 1$: line $y = 2$, curve $y = 1$, so the line is above the curve.

Step 3, integrate (upper $-$ lower) from $0$ to $2$: $\text{Area} = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 \quad \text{(M1, M1)}$

Step 4, evaluate: $= \left(4 - \frac{8}{3}\right) - 0 = \frac{12 - 8}{3} = \frac{4}{3} \text{ square units} \quad \text{(A1, A1)}$

Where the marks are won and lost

• The limits come from the intersection ($x = 0, 2$), not from the question text. Solving curve $=$ line is the first method mark.
• Integrate (line $-$ curve) because the line is on top here. Reversing the order gives $-\frac43$; the area is the positive value.
• Integrating the difference in one go handles the region automatically, no need to compute two separate areas and subtract.

Common mistakes

• Subtracting in the wrong order (curve $-$ line) and getting a negative.
• Integrating each graph separately and mishandling the subtraction.
• Using wrong limits instead of the intersection points.

Full method: Definite Integrals & Area notes. Topic home: Calculus pillar.