CAIE
FP1
2009
June
Q7
8 marks
Challenging +1.2
7 Let
$$I _ { n } = \int _ { 0 } ^ { 1 } t ^ { n } \mathrm { e } ^ { - t } \mathrm {~d} t$$
where \(n \geqslant 0\). Show that, for all \(n \geqslant 1\),
$$I _ { n } = n I _ { n - 1 } - \mathrm { e } ^ { - 1 }$$
Hence prove by induction that, for all positive integers \(n\),
$$I _ { n } < n ! .$$
CAIE
FP1
2010
June
Q5
9 marks
Challenging +1.2
5 Let
$$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } x ( \ln x ) ^ { n } \mathrm {~d} x$$
where \(n \geqslant 1\). Show that
$$I _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { 2 } - \frac { 1 } { 2 } ( n + 1 ) I _ { n }$$
Hence prove by induction that, for all positive integers \(n , I _ { n }\) is of the form \(A _ { n } \mathrm { e } ^ { 2 } + B _ { n }\), where \(A _ { n }\) and \(B _ { n }\) are rational numbers.
CAIE
FP1
2012
June
Q11 EITHER
Challenging +1.8
Show that
$$\int _ { 0 } ^ { \pi } \mathrm { e } ^ { x } \sin x \mathrm {~d} x = \frac { 1 + \mathrm { e } ^ { \pi } } { 2 }$$
Given that
$$I _ { n } = \int _ { 0 } ^ { \pi } \mathrm { e } ^ { x } \sin ^ { n } x \mathrm {~d} x$$
show that, for \(n \geqslant 2\),
$$I _ { n } = n ( n - 1 ) \int _ { 0 } ^ { \pi } \mathrm { e } ^ { x } \cos ^ { 2 } x \sin ^ { n - 2 } x \mathrm {~d} x - n I _ { n }$$
and deduce that
$$\left( n ^ { 2 } + 1 \right) I _ { n } = n ( n - 1 ) I _ { n - 2 } .$$
A curve has equation \(y = \mathrm { e } ^ { x } \sin ^ { 5 } x\). Find, in an exact form, the mean value of \(y\) over the interval \(0 \leqslant x \leqslant \pi\).
CAIE
FP1
2006
November
Q2
5 marks
Challenging +1.2
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 \(I _ { 0 }\).
CAIE
FP1
2012
November
Q5
8 marks
Challenging +1.2
5 Let \(I _ { n }\) denote \(\int _ { 0 } ^ { \infty } x ^ { n } \mathrm { e } ^ { - 2 x } \mathrm {~d} x\). Show that \(I _ { n } = \frac { 1 } { 2 } n I _ { n - 1 }\), for \(n \geqslant 1\).
Prove by mathematical induction that, for all positive integers \(n , I _ { n } = \frac { n ! } { 2 ^ { n + 1 } }\).
CAIE
FP1
2016
November
Q9
11 marks
Challenging +1.2
9 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x\).
Given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x\), prove that, for \(n > 1\),
$$I _ { n } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 }$$
By first using the substitution \(x = \cos ^ { - 1 } u\), find the value of
$$\int _ { 0 } ^ { 1 } \left( \cos ^ { - 1 } u \right) ^ { 3 } \mathrm {~d} u$$
giving your answer in an exact form.