2026-08-31T11:34:15.886856 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL The figure shows the curve \(C\) with equation \(...

Seven marks combining curve sketching with integration. The figure shows the curve \(C\) with equation \(y = 4x - x^2\) meeting the \(x\)-axis at the origin \(A\) and at \(B\), with the region \(R\) shown shaded between the curve and the \(x\)-axis. The three parts use three different techniques on the same curve: factorising to find intercepts, differentiating to find the maximum, and integrating to find an area. Recognising which tool answers which question is most of the skill here.

(a) Show that \(B\) is \((4, 0)\) [2 marks]

A curve meets the \(x\)-axis where \(y = 0\), so set the expression to zero and factorise rather than expand or use the formula: \[4x - x^2 = 0 \ \Rightarrow\ x(4 - x) = 0 \quad \textbf{M1}\] \[x = 0 \ \text{(the point } A) \quad \text{or} \quad x = 4, \ \text{so } B \ \text{is } (4,\, 0) \quad \textbf{A1}\] The result is printed, so the marks are for the working. The M1 is for setting \(y = 0\) and factorising; the A1 needs the root \(x = 4\) identified as \(B\) rather than merely listed. Do not divide by \(x\) to "simplify": that loses the root \(x = 0\), which the figure identifies as \(A\).

(b) The coordinates of the maximum point [2 marks]

At a maximum the gradient is zero, so differentiate and solve: \[\frac{\mathrm{d}y}{\mathrm{d}x} = 4 - 2x = 0 \ \Rightarrow\ x = 2 \quad \textbf{M1}\] \[y = 4(2) - 2^2 = 8 - 4 = 4, \ \text{so the maximum point is } (2,\, 4) \quad \textbf{A1}\] Both coordinates are needed. Because the coefficient of \(x^2\) is negative, the parabola opens downwards and this stationary point is a maximum, which agrees with the figure and means no second-derivative test is required here. Symmetry offers a check with no calculus at all: the vertex sits midway between the roots \(0\) and \(4\), so \(x = 2\).

(c) The exact area of \(R\) [3 marks]

The area between a curve and the \(x\)-axis is the definite integral of \(y\) between the two points where the curve meets that axis, and those limits are exactly the intercepts found in part (a): \[\text{Area} = \int_{0}^{4} \bigl(4x - x^2\bigr)\,\mathrm{d}x \quad \textbf{M1}\] Integrate by raising each index by one and dividing by the new index: \[= \left[2x^2 - \frac{x^3}{3}\right]_{0}^{4} \quad \textbf{A1}\] Evaluate at the upper limit and subtract the value at the lower limit, which is zero here: \[= \left(2(16) - \frac{64}{3}\right) - 0 = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3} \quad \textbf{A1}\] So the area of \(R\) is \(\dfrac{32}{3}\) square units. "Exact" means the fraction must be kept: \(10.7\) or \(10.67\) loses the final accuracy mark. The M1 is for a definite integral with the correct limits, the first A1 for correct integration, and the second for the evaluated exact value. Since this is a definite integral, no constant of integration is needed; it would cancel between the two limits.

Common wrong turns on this question

• Integrating with limits 0 and 2, which gives only the left half of \(R\). The region shown shaded runs from \(A\) all the way to \(B\).
• Differentiating instead of integrating in part (c), or integrating in part (b). Gradients come from differentiating; areas come from integrating.
• Forgetting to subtract the value at the lower limit. Here it happens to be zero, which conceals the omission and lets the habit survive into a question where it costs marks.
• Rounding to a decimal when the exact value is demanded.
• Adding a constant of integration to a definite integral.
• Dividing by \(x\) in part (a) and reporting only \(x = 4\) with no reference to \(A\).

How to check

An area can always be sanity-checked against a rectangle that contains it. The region \(R\) sits inside the rectangle of width 4 and height 4, of area 16, and it is a rounded arch filling rather more than half of that box, so a value of \(\dfrac{32}{3} = 10.67\) is the right order of magnitude. In fact for any parabolic arch the area is exactly two thirds of the enclosing rectangle, and \(\tfrac{2}{3} \times 16 = \dfrac{32}{3}\), which confirms the answer exactly. A further check on part (b): the maximum value 4 is the height of that rectangle, consistent with the vertex touching its top edge.