P8.4Integration as the limit of a sum

• B1 for writing \(L\) as \(\int_{2}^{3} (3x + 1)(x + 3)\,\mathrm{d}x\)
• M1 for integrating: \(x^{n} \to x^{n + 1}\) for at least one term (after multiplying out the brackets)
• dM1 for substituting the limits and subtracting
• A1* for \(47\) with no errors seen (cso)

Worked solution: \((3x + 1)(x + 3) = 3x^{2} + 10x + 3\), so
\(L = \int_{2}^{3} \left(3x^{2} + 10x + 3\right)\,\mathrm{d}x = \left[x^{3} + 5x^{2} + 3x\right]_{2}^{3} = \left(27 + 45 + 9\right) - \left(8 + 20 + 6\right) = 47\), as required.

• B1 for writing \(L\) as \(\int_{0}^{2} \frac{3}{3x + 2}\,\mathrm{d}x\)
• M1 for integrating to the form \(\lambda\ln(3x + 2)\)
• dM1 for substituting the limits and using a law of logarithms
• A1 for \(N = 4\)

Worked solution: \(L = \int_{0}^{2} \frac{3}{3x + 2}\,\mathrm{d}x = \left[\ln(3x + 2)\right]_{0}^{2} = \ln 8 - \ln 2 = \ln 4\), so \(N = 4\).

• B1 for \(\int_{1}^{a} \left(2 + \frac{2}{\sqrt{x}}\right)\,\mathrm{d}x\)
• M1 for integrating: \(x^{n} \to x^{n + 1}\) for at least one term, e.g. \(\frac{2}{\sqrt{x}} \to 4\sqrt{x}\)
• A1 for \(2a + 4\sqrt{a} - 6 = 10\) oe
• dM1 for solving their equation as a 3TQ in \(\sqrt{a}\), e.g. \(t^{2} + 2t - 8 = 0\) where \(t = \sqrt{a}\)
• A1 for \(a = 4\) only, rejecting \(\sqrt{a} = -4\) because \(\sqrt{a}\) cannot be negative

Worked solution: \(\int_{1}^{a} \left(2 + \frac{2}{\sqrt{x}}\right)\,\mathrm{d}x = \left[2x + 4\sqrt{x}\right]_{1}^{a} = 2a + 4\sqrt{a} - 6\).
So \(2a + 4\sqrt{a} - 6 = 10\), i.e. \(2a + 4\sqrt{a} - 16 = 0\).
With \(t = \sqrt{a}\): \(2t^{2} + 4t - 16 = 0 \Rightarrow t^{2} + 2t - 8 = 0 \Rightarrow (t - 2)(t + 4) = 0\), so \(t = 2\) or \(t = -4\).
\(\sqrt{a}\) cannot be negative, so \(\sqrt{a} = 2\) and \(a = 4\).