2026-08-31T11:34:12.419271 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL The curve \(C\) has equation \(y = x^2 - 4x + 7\)...

The area of the finite region \(R\) bounded by \(C\) and \(l\) shown shaded is found by the standard "top curve minus bottom curve" principle. Between the two intersection points the line lies above the parabola, so the area is the area under the line minus the area under the curve over the same interval. That is why the question first makes you locate the intersections in part (b) and then evaluate the two definite integrals separately in parts (c) and (d): the limits of integration are the \(x\)-coordinates of the intersections, and nothing can be integrated until they are known.

(a) Complete the square and write down the minimum point [3]

Half of \(-4\) is \(-2\), so \((x - 2)^2 = x^2 - 4x + 4\), and four must be given back:

\[x^2 - 4x + 7 = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3 \qquad \textbf{[M1 A1]}\]

Matching against the requested form \((x + a)^2 + b\) gives \(a = -2\) and \(b = 3\). The [M1] is for the completing-the-square structure, the [A1] for both constants. The form has a plus sign inside the bracket, so \(a = -2\); quoting \(a = 2\) is the standard way to lose the accuracy mark here.

Since \((x - 2)^2 \ge 0\) with equality only at \(x = 2\), the least value of \(y\) is \(3\), so the minimum point of \(C\) is \((2,\, 3)\) [B1]. This is written down from the completed square, not obtained by differentiation.

(b) Coordinates of \(A\) and \(B\) [4]

At an intersection the two \(y\)-values agree, so equate the equations:

\[x^2 - 4x + 7 = x + 3 \qquad \textbf{[M1]}\] \[x^2 - 5x + 4 = 0 \qquad \textbf{[A1]}\] \[(x - 1)(x - 4) = 0 \quad\Rightarrow\quad x = 1 \text{ or } x = 4 \qquad \textbf{[M1]}\]

Substituting into the simpler equation, the line \(y = x + 3\), gives the points \(A(1,\, 4)\) and \(B(4,\, 7)\) [A1]. Always use the line for the \(y\)-values: it is one step instead of three and the arithmetic is safer. The final accuracy mark needs full coordinates, since the question asks for points; a pair of \(x\)-values alone is an incomplete answer.

(c) Show that \(\displaystyle\int_{1}^{4}\left(x^2 - 4x + 7\right)\mathrm{d}x = 12\) [4]

Integrate term by term, raising each index by one and dividing by the new index:

\[\int \left(x^2 - 4x + 7\right)\mathrm{d}x = \frac{x^3}{3} - 2x^2 + 7x \qquad \textbf{[M1 A1]}\]

The [M1] is for an attempt at integration, shown by at least one term correct; the [A1] is for all three terms. Because this is a definite integral, no constant of integration is required and none should be written; on an indefinite integral its omission would cost a mark, so the distinction is worth holding in mind. Note \(-4x\) integrates to \(-2x^2\), not \(-4x^2/2\) left unsimplified, and the constant \(7\) integrates to \(7x\).

Now substitute the upper limit and subtract the value at the lower limit:

\[= \left(\frac{64}{3} - 32 + 28\right) - \left(\frac{1}{3} - 2 + 7\right) \qquad \textbf{[M1]}\] \[= \left(\frac{64}{3} - 4\right) - \left(\frac{1}{3} + 5\right) = \frac{64}{3} - \frac{1}{3} - 9 = 21 - 9 = 12 \qquad \textbf{[A1]}\]

The [M1] is for correct substitution of both limits in the right order, and it survives an arithmetic slip; the [A1] is for the printed value \(12\). Since the answer is given, the working must be complete: writing \(12\) with no evaluation earns nothing. The frequent error is subtracting in the wrong order or losing the bracket around the lower-limit value, which flips the signs of its terms.

(d) Hence show that the area of \(R\) is \(\dfrac{9}{2}\) [6]

The region \(R\) lies between the line above and the curve below, over \(1 \le x \le 4\). Deal with the line first:

\[\int_{1}^{4}(x + 3)\,\mathrm{d}x = \left[\frac{x^2}{2} + 3x\right]_{1}^{4} \qquad \textbf{[M1 A1]}\] \[= (8 + 12) - \left(\frac{1}{2} + 3\right) = 20 - \frac{7}{2} = \frac{33}{2} \qquad \textbf{[M1 A1]}\]

The first [M1 A1] pair is for the integration, the second for the evaluation. The limits are the same \(1\) and \(4\) found in part (b), because that is the horizontal extent of \(R\).

Now the subtraction principle. The area of \(R\) is the area under the line minus the area under the curve, both taken between the same limits [M1], and part (c) has already supplied the second of these:

\[\text{Area of } R = \frac{33}{2} - 12 = \frac{33}{2} - \frac{24}{2} = \frac{9}{2} \qquad \textbf{[A1]}\]

The [M1] is for the correct principle, so it is earned by a candidate who subtracts the right way round even with a numerical slip; the [A1] is for the printed value. Subtracting the wrong way gives \(-\tfrac{9}{2}\), and an area cannot be negative, which is the built-in signal that the order has been reversed. The way to decide the order without guessing is to test a value between the limits: at \(x = 2\), the line gives \(y = 5\) and the curve gives \(y = 3\), so the line is on top.

An equivalent single-integral method scores the same marks and is less error-prone:

\[\int_{1}^{4}\left[(x + 3) - \left(x^2 - 4x + 7\right)\right]\mathrm{d}x = \int_{1}^{4}\left(-x^2 + 5x - 4\right)\mathrm{d}x = \left[-\frac{x^3}{3} + \frac{5x^2}{2} - 4x\right]_{1}^{4}\] \[= \left(-\frac{64}{3} + 40 - 16\right) - \left(-\frac{1}{3} + \frac{5}{2} - 4\right) = \frac{8}{3} - \left(-\frac{11}{6}\right) = \frac{16}{6} + \frac{11}{6} = \frac{27}{6} = \frac{9}{2}\]

which confirms the answer. Note that the integrand \(-x^2 + 5x - 4\) is the negative of the quadratic from part (b), which is no accident: it vanishes exactly at the intersection points, so it is positive throughout the interior of \(R\), another confirmation that the line is the upper boundary there.

Check: a rough magnitude test. The region is about \(3\) units wide and, at its widest point \(x = 2.5\), the vertical gap between line and curve is \(5.5 - 3.25 = 2.25\), so an area of \(4.5\) is entirely plausible for a lens-shaped region of those dimensions.