Area Under a Curve That Crosses the Axis
A student is asked to find the area of the region enclosed between the curve y = x² − 4x + 3 and the x-axis for 0 ≤ x ≤ 4. She evaluates ∫₀⁴ (x² − 4x + 3) dx, obtains 4/3, and submits this as the area. The examiner marks the answer wrong and awards method marks only. The same question paper has two further parts: find the positive value of c for which ∫₀^c (x² − 4x + 3) dx = 0, and find the area of the region bounded by the curve and the line y = x − 1. The student's sketch shows the curve crossing the x-axis twice but she does not use the crossings in her working, and she says she has 'no idea' how to start the last part.
1. Show the student's integral computation, explain why 4/3 is not the area, and obtain the correct area for 0 ≤ x ≤ 4.
Read the worked answer
The student's computation is arithmetically right:
$$\int_0^4 (x^2-4x+3),dx = \left[\frac{x^3}{3}-2x^2+3x\right]_0^4 = \frac{64}{3}-32+12 = \frac{4}{3}.$$
Why it is not the area: the definite integral gives the net signed area, counting parts below the axis as negative. Factorising,
$$x^2-4x+3 = (x-1)(x-3),$$
so the curve is positive on $[0,1)$, negative on $(1,3)$ and positive on $(3,4]$. The above-axis and below-axis pieces cancel rather than add.
Correct area: split at the roots. With $F(x) = \frac{x^3}{3}-2x^2+3x$:
- $F(0)=0$, $F(1)=\frac{1}{3}-2+3=\frac{4}{3}$, $F(3)=9-18+9=0$, $F(4)=\frac{4}{3}$.
$$A = \left|F(1)-F(0)\right| + \left|F(3)-F(1)\right| + \left|F(4)-F(3)\right| = \frac{4}{3}+\frac{4}{3}+\frac{4}{3} = 4.$$
The correct area is 4 square units, exactly three times the net integral - a coincidence here only because the three pieces happen to be equal.
2. Solve ∫₀^c (x² − 4x + 3) dx = 0 for positive c, and interpret the result geometrically.
Read the worked answer
Using the antiderivative above:
$$F(c) = \frac{c^3}{3}-2c^2+3c = \frac{c}{3}\left(c^2-6c+9\right) = \frac{c}{3}(c-3)^2.$$
Setting $F(c)=0$ gives $c = 0$ or $(c-3)^2 = 0$, i.e. $c = 3$ (a repeated root). Both roots are legitimate; the positive solution asked for is
$$c = 3.$$
Interpretation: the signed area from 0 to 3 is zero because the positive area between $x=0$ and $x=1$ is exactly cancelled by the negative area between $x=1$ and $x=3$. Each equals $4/3$. So a zero integral does not mean there is no area - it means the two regions are equal in magnitude and opposite in sign. This is the same trap as in part one: the student should always ask where the integrand changes sign before interpreting an integral as an area.
3. Find the area of the region bounded by the curve y = x² − 4x + 3 and the line y = x − 1.
Read the worked answer
Step 1 - intersections. Set
$$x^2-4x+3 = x-1 \Rightarrow x^2-5x+4 = 0 \Rightarrow (x-1)(x-4)=0,$$
so the two curves meet at $x=1$ and $x=4$.
Step 2 - which is on top. On $(1,4)$ take $x=2$: curve gives $4-8+3=-1$, line gives $1$. The line is above the curve. Equivalently,
$$(x-1) - (x^2-4x+3) = -x^2+5x-4 = (x-1)(4-x) > 0 \text{ on } (1,4).$$
Step 3 - integrate the difference.
$$A = \int_1^4 \left(-x^2+5x-4\right)dx = \left[-\frac{x^3}{3}+\frac{5x^2}{2}-4x\right]_1^4.$$
At $x=4$: $-\frac{64}{3}+40-16 = -21.33+24 = 2.667 = \frac{8}{3}$.
At $x=1$: $-\frac{1}{3}+\frac{5}{2}-4 = -\frac{1}{3}-\frac{3}{2} = -\frac{11}{6}$.
$$A = \frac{8}{3}+\frac{11}{6} = \frac{16+11}{6} = \frac{27}{6} = 4.5.$$
The required area is 4.5 square units.
Method note for the student: 'bounded by a curve and a line' always means (i) find intersections, (ii) decide which function is greater on the interval, (iii) integrate the difference with the greater function first. No sign-splitting is needed here because the difference does not change sign between the intersection points.