P8.5aIntegration by substitution

• B1 for \(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\) or \(\mathrm{d}x = \mathrm{d}u\) oe
• M1 for substituting fully to obtain an integral of the form \(k\int (u - 3)u^{\frac{1}{2}}\,\mathrm{d}u\)
• A1 for a correct integral in \(u\), e.g. \(\int (u^{\frac{3}{2}} - 3u^{\frac{1}{2}})\,\mathrm{d}u\)
• dM1 for integrating \(u^{p} \to u^{p + 1}\) on at least one term
• A1 for \(\frac{2}{5}(x + 3)^{\frac{5}{2}} - 2(x + 3)^{\frac{3}{2}} + c\) oe, in terms of \(x\)

Worked solution: \(u = x + 3\), so \(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\), \(\mathrm{d}x = \mathrm{d}u\) and \(x = u - 3\).
\(\int x\sqrt{x + 3}\,\mathrm{d}x = \int (u - 3)u^{\frac{1}{2}}\,\mathrm{d}u = \int (u^{\frac{3}{2}} - 3u^{\frac{1}{2}})\,\mathrm{d}u\)
\(= \frac{2}{5}u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + c = \frac{2}{5}(x + 3)^{\frac{5}{2}} - 2(x + 3)^{\frac{3}{2}} + c\)

• B1 for stating that the limits were not changed: \(0\) and \(2\) are values of \(x\), but the integral is in \(u\), so the limits should be \(u = 1\) and \(u = 5\) (or the answer should be written in terms of \(x\) before using \(0\) and \(2\))

Worked solution: When \(x = 0\), \(u = 1\) and when \(x = 2\), \(u = 5\). The student integrated with respect to \(u\) but used the \(x\)-limits.

• B1 for limits \(1\) and \(5\)
• M1 for \(\frac{1}{2}\left[\frac{u^{4}}{4}\right]\) evaluated with their \(u\)-limits
• A1 for \(78\)

Worked solution: \(\frac{1}{2}\int_{1}^{5} u^{3}\,\mathrm{d}u = \frac{1}{2}\left[\frac{u^{4}}{4}\right]_{1}^{5} = \frac{1}{8}(625 - 1) = 78\)

• M1 for \(x = u^{2}\) and \(\mathrm{d}x = 2u\,\mathrm{d}u\) oe
• B1 for limits \(\frac{1}{\sqrt{3}}\) (or \(\frac{\sqrt{3}}{3}\)) and \(\sqrt{3}\)
• A1* for the given result with no errors seen (cso)

Worked solution: \(u = \sqrt{x}\), so \(x = u^{2}\) and \(\mathrm{d}x = 2u\,\mathrm{d}u\). When \(x = \frac{1}{3}\), \(u = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\); when \(x = 3\), \(u = \sqrt{3}\).
\(I = \int_{\frac{\sqrt{3}}{3}}^{\sqrt{3}} \frac{2u}{(1 + u^{2})u}\,\mathrm{d}u = 2\int_{\frac{\sqrt{3}}{3}}^{\sqrt{3}} \frac{1}{1 + u^{2}}\,\mathrm{d}u\)

• M1 for \(\mathrm{d}u = \sec^{2}\theta\,\mathrm{d}\theta\) and \(1 + \tan^{2}\theta = \sec^{2}\theta\), giving \(2\int 1\,\mathrm{d}\theta\)
• B1 for limits \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\)
• A1 for \(I = \frac{\pi}{3}\)

Worked solution: \(u = \tan\theta\), \(\mathrm{d}u = \sec^{2}\theta\,\mathrm{d}\theta\). When \(u = \frac{\sqrt{3}}{3}\), \(\theta = \frac{\pi}{6}\); when \(u = \sqrt{3}\), \(\theta = \frac{\pi}{3}\).
\(I = 2\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec^{2}\theta}{1 + \tan^{2}\theta}\,\mathrm{d}\theta = 2\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1\,\mathrm{d}\theta = 2\left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\pi}{3}\)