Challenging +1.2 This is a guided reduction formula question where the method (differentiation of a given expression) is explicitly provided. The product rule application is straightforward, and the subsequent integration by parts follows mechanically. Finding I_6 requires only routine substitution into the derived formula twice. While reduction formulae are a Further Maths topic, the heavy scaffolding and standard technique make this easier than average for FM students.
2 The integral \(I _ { n }\), where \(n\) is a non-negative integer, is defined by
$$I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { - x ^ { 3 } } \mathrm {~d} x$$
By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x ^ { n + 1 } \mathrm { e } ^ { - x ^ { 3 } } \right)\) or otherwise, show that
$$3 I _ { n + 3 } = ( n + 1 ) I _ { n } - \mathrm { e } ^ { - 1 }$$
Hence find \(I _ { 6 }\) in terms of e and \(I _ { 0 }\).
2 The integral $I _ { n }$, where $n$ is a non-negative integer, is defined by
$$I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { - x ^ { 3 } } \mathrm {~d} x$$
By considering $\frac { \mathrm { d } } { \mathrm { d } x } \left( x ^ { n + 1 } \mathrm { e } ^ { - x ^ { 3 } } \right)$ or otherwise, show that
$$3 I _ { n + 3 } = ( n + 1 ) I _ { n } - \mathrm { e } ^ { - 1 }$$
Hence find $I _ { 6 }$ in terms of e and $I _ { 0 }$.
\hfill \mbox{\textit{CAIE FP1 2006 Q2 [5]}}