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 \(I _ { 0 }\).