Standard +0.8 This is a Further Maths integration question requiring completion of the square for the quadratic under the square root, then applying the substitution u = x - 2 to transform it into a standard arcsin integral form. While it involves multiple steps (algebraic manipulation, substitution, recognizing the standard form, and evaluating limits), these are well-practiced techniques in FP2. The question is moderately challenging but follows a recognizable pattern for this topic.
6 By using the substitution \(u = x - 2\), or otherwise, find the exact value of
$$\int _ { - 1 } ^ { 5 } \frac { \mathrm {~d} x } { \sqrt { 32 + 4 x - x ^ { 2 } } }$$
$u = x - 2$ | B1 | clearly seen
$du = dx$ or $\frac{du}{dx} = 1$ |
$32 + 4x - x^2 = 36 - u^2$ | B1 | if $32 + 4x - x^2$ is written as $36 - (x-2)^2$, give B2
$\int \frac{du}{\sqrt{36 - u^2}} = \sin^{-1} \frac{u}{6}$ | M1 | allow if dx is used instead of du
limits $-3$ and $3$ | A1
or substitute back to give $\sin^{-1} \frac{x-2}{6}$ |
$I = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ | A1 | 5 marks for question
6 By using the substitution $u = x - 2$, or otherwise, find the exact value of
$$\int _ { - 1 } ^ { 5 } \frac { \mathrm {~d} x } { \sqrt { 32 + 4 x - x ^ { 2 } } }$$
\hfill \mbox{\textit{AQA FP2 2008 Q6 [5]}}