P9.3Numerical integration: the trapezium rule

• B1 for \(2.1011\)
• B1 for \(2.8829\)

Worked solution: When \(x = 2.4\): \(y = 2.4\ln 2.4 = 2.101125\ldots = 2.1011\).
When \(x = 2.8\): \(y = 2.8\ln 2.8 = 2.882934\ldots = 2.8829\).

• B1 for the strip width \(h = 0.4\) (or \(\frac{1}{2} \times 0.4\) seen)
• M1 for the correct bracket structure \(\left[y_0 + y_{5} + 2(y_1 + \cdots + y_{4})\right]\) with their values
• A1 for awrt \(6.713\)

Worked solution: There are 5 strips of width \(h = 0.4\):
\(\frac{1}{2} \times 0.4 \times \left[1.3863 + 5.5452 + 2(2.1011 + 2.8829 + 3.7221 + 4.6114)\right] = 6.7133 = 6.713\) (3 d.p.).

• B1 for \(h = 0.5\) and the ordinates \(0, 0.4055, 0.6931, 0.9163, 1.0986\)
• M1 for \(\frac{0.5}{2}\left[0 + 1.0986 + 2(0.4055 + 0.6931 + 0.9163)\right]\)
• A1 for 1.2821

Worked solution: \(h = 0.5\). Ordinates: \(\ln 1 = 0\), \(\ln 1.5 = 0.405465\), \(\ln 2 = 0.693147\), \(\ln 2.5 = 0.916291\), \(\ln 3 = 1.098612\).
Estimate \(= \frac{0.5}{2}\left[0 + 1.098612 + 2(2.014903)\right] = 0.25 \times 5.128418 = 1.2821\).

• M1 for \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{1}{x^2} \lt 0\), so the curve is concave (or a sketch showing the chords below the curve)
• A1 for the chords lie below the curve, so the trapezium rule gives an underestimate

Worked solution: \(y = \ln x\) has \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{1}{x^2} \lt 0\), so the curve is concave. The straight tops of the trapezia (chords) lie below the curve, so each trapezium's area is less than the area under the curve: the estimate is an underestimate.

• M1 for \(\int \ln x\,\mathrm{d}x = x\ln x - \int x \times \frac{1}{x}\,\mathrm{d}x = x\ln x - x\)
• A1 for \(3\ln 3 - 2\)
• A1 for awrt 1.06%

Worked solution: \(\int \ln x\,\mathrm{d}x = x\ln x - x + c\), so \(\int_{1}^{3} \ln x\,\mathrm{d}x = (3\ln 3 - 3) - (0 - 1) = 3\ln 3 - 2 = 1.29584\ldots\).
Percentage error \(= \dfrac{1.29584 - 1.28210}{1.29584} \times 100 = 1.06\%\).

• B1 for the 6 values of \(y\), \(4.0000, 4.7539, 5.4037, 5.9833, 6.5115, 7.0000\)
• B1 for the strip width \(h = 2.2\) (or \(\frac{1}{2} \times 2.2\) seen)
• M1 for the correct bracket structure \(\left[y_0 + y_{5} + 2(y_1 + \cdots + y_{4})\right]\) with their values
• A1 for awrt \(61.9\)

Worked solution: \(h = 2.2\). The values of \(y\) are \(4.0000, 4.7539, 5.4037, 5.9833, 6.5115, 7.0000\).
\(\frac{1}{2} \times 2.2 \times \left[4.0000 + 7.0000 + 2(4.7539 + 5.4037 + 5.9833 + 6.5115)\right] = 61.93528 = 61.9\) (3 s.f.).

• M1 for differentiating once: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{2\sqrt{3x + 1}}\)
• A1 for \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -\frac{9}{4(3x + 1)^{\frac{3}{2}}}\) oe
• M1 for considering the sign of \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) for \(5 \le x \le 16\)
• A1 for concluding concave, so an underestimate, with a reason (the tops of the trapezia lie below the curve)

Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{2\sqrt{3x + 1}}\) and \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -\frac{9}{4(3x + 1)^{\frac{3}{2}}}\), which is negative wherever it is defined.
So the curve is concave for \(5 \le x \le 16\), the straight tops of the trapezia lie below the curve, and the trapezium rule gives an underestimate.