KCSE 2025 Maths P2 Q24 — Area Under a Curve by Integration

KCSE 2025 Form 4 Calculus

Published

The Question

“The curve y = x squared minus 4x plus 3 has y-intercept at y = 3 and x-intercepts at x = 1 and x = 3. The region bounded by the curve, the line y = 3 and the line x = 3 is shaded. (a)(i) Evaluate the integral of (x squared minus 4x plus 3) with respect to x from 0 to 1. (a)(ii) Calculate the area of the shaded region above the x-axis. (b) Calculate the area of the shaded region below the x-axis. (c) Calculate the area of the entire shaded region.”

1

Evaluate the definite integral (a)(i)

Integrate the curve term by term: raise each power by one and divide by the new power. Then substitute the upper limit 1 and the lower limit 0 into the result and subtract. Putting in 0 gives nothing, so the value is just what you get at x = 1.

∫01(x2−4x+3) dx=[x33−2x2+3x]01\int_{0}^{1}(x^{2} - 4x + 3)\,dx = \left[\frac{x^{3}}{3} - 2x^{2} + 3x\right]_{0}^{1}
=(13−2+3)−0=43= \left(\tfrac{1}{3} - 2 + 3\right) - 0 = \frac{4}{3}
2

Area above the x-axis (a)(ii)

The shaded region above the axis is the big rectangle stretching from x = 0 to x = 3 with height 3 (the line y = 3), but with the sliver under the curve between x = 0 and x = 1 removed. That sliver is exactly the integral you just found, so subtract it from the rectangle's area.

Rectangle=3×3=9\text{Rectangle} = 3 \times 3 = 9
Area above=9−43=27−43=233≈7.67 sq units\text{Area above} = 9 - \frac{4}{3} = \frac{27 - 4}{3} = \frac{23}{3} \approx 7.67 \text{ sq units}
3

Area below the x-axis (b)

Between x = 1 and x = 3 the curve dips below the x-axis, so integrating it there gives a negative value. Area is always positive, so evaluate the definite integral over these limits and then take its magnitude.

∫13(x2−4x+3) dx=[x33−2x2+3x]13\int_{1}^{3}(x^{2} - 4x + 3)\,dx = \left[\frac{x^{3}}{3} - 2x^{2} + 3x\right]_{1}^{3}
=(9−18+9)−(13−2+3)=0−43=−43= (9 - 18 + 9) - \left(\tfrac{1}{3} - 2 + 3\right) = 0 - \frac{4}{3} = -\frac{4}{3}
Area below=∣−43∣=43≈1.33 sq units\text{Area below} = \left|-\tfrac{4}{3}\right| = \frac{4}{3} \approx 1.33 \text{ sq units}
4

Total shaded area (c)

The whole shaded region is just the piece above the axis plus the piece below the axis. Add the two areas, using a common denominator, to get the final answer as a single exact value.

Total=233+43=273=9 sq units\text{Total} = \frac{23}{3} + \frac{4}{3} = \frac{27}{3} = 9 \text{ sq units}

Final Result

The integral from 0 to 1 evaluates to 4/3. The shaded area above the x-axis is 23/3, about 7.67 square units. The shaded area below the x-axis is 4/3, about 1.33 square units. The entire shaded region has area exactly 9 square units.

Why this method works

Definite integration measures signed area between a curve and the x-axis, so where the curve is above the axis the integral is positive and where it dips below it is negative. That is why the region above the axis is found by taking the enclosing rectangle and subtracting the curve's integral, while the region below must have its negative integral turned positive by taking the magnitude. Adding the two positive pieces gives the true total area, because physical area cannot cancel out even though signed integrals can.

✓

The two pieces recombine neatly: 23/3 plus 4/3 is 27/3, which is exactly 9, matching the total shaded area.