Area Between a Curve and the x-axis

The trap in this question is not the integration, it is knowing that a region below the x-axis produces a negative integral, and that area is the modulus of it. Getting $-\frac43$ and calling it the answer is the difference between full marks and a lost accuracy mark.

The curve $y = x^2 - 4x + 3$ meets the x-axis at two points. Find the area of the region enclosed between the curve and the x-axis. [6]

The working

Step 1, find the limits (where the curve meets the x-axis). Factorise: $x^2 - 4x + 3 = (x - 1)(x - 3) = 0 \;\Rightarrow\; x = 1 \text{ and } x = 3 \quad \text{(M1, A1)}$

Between $x = 1$ and $x = 3$ the parabola dips below the axis (it opens upward with roots at 1 and 3), so expect a negative integral.

Step 2, integrate: $\int (x^2 - 4x + 3)\,dx = \frac{x^3}{3} - 2x^2 + 3x \quad \text{(M1)}$

Step 3, apply the limits $1$ to $3$: $\left[\frac{x^3}{3} - 2x^2 + 3x\right]_1^3$

At $x = 3$: $\dfrac{27}{3} - 2(9) + 9 = 9 - 18 + 9 = 0$

At $x = 1$: $\dfrac{1}{3} - 2 + 3 = \dfrac{1}{3} + 1 = \dfrac{4}{3}$

$\int_1^3 = 0 - \frac{4}{3} = -\frac{4}{3} \quad \text{(A1)}$

Step 4, report the area as the modulus: $\text{Area} = \left|-\frac{4}{3}\right| = \frac{4}{3}\ \text{square units} \quad \text{(A1)}$

Where the marks are won and lost

• The limits are marks. They come from solving $y = 0$, not from the question text, which deliberately does not give them.
• The modulus is the final accuracy mark. Leaving the answer as $-\frac43$ concedes it.
• If a region straddles the axis (part above, part below), split the integral at the root and add the two moduli separately. A single integral across the crossing lets the halves cancel.

Common mistakes

• Reporting $-\frac43$ as the area.
• Integrating with the wrong limits, or guessing limits instead of solving $y = 0$.
• For a region that crosses the axis, integrating straight through and understating the area.

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