Edexcel IAL Maths FP3.4.5 Reduction Formulae

[Maximum number: 9]

I_n=\int_0^k x^n(k-x)^{\frac12}\,dx \quad n\geqslant 0

where k is a positive constant.

(a)

Show that

I_n=\frac{2kn}{3+2n}I_{n-1}\quad n\ge1

[ 5 ]

u=x^n,\quad \mathrm{d}v=(k-x)^{1/2}\,\mathrm{d}x,\quad v=-\frac23(k-x)^{3/2}

I_n=\left[-\frac23x^n(k-x)^{3/2}\right]_0^k+\frac{2n}{3}\int_0^k x^{n-1}(k-x)^{3/2}\,\mathrm{d}x

=\frac{2n}{3}\int_0^k x^{n-1}(k-x)(k-x)^{1/2}\,\mathrm{d}x

=\frac{2n}{3}\left(kI_{n-1}-I_n\right)

\left(1+\frac{2n}{3}\right)I_n=\frac{2kn}{3}I_{n-1}

I_n=\frac{2kn}{3+2n}I_{n-1}

M1/A1: integration by parts in the correct direction with correct expression.
dM1/ddM1: rewrites (k-x)^(3/2) and expresses the result in I_n and I_(n-1).
A1*: reaches the printed reduction formula.

(b)

Given that

\int_0^k x^2(k-x)^{1/2}\,\mathrm{d}x=\frac{9\sqrt3}{280}

use the result in part (a) to determine the exact value of k.

[ 4 ]

I_2=\frac{4k}{7}I_1,\quad I_1=\frac{2k}{5}I_0

I_0=\int_0^k(k-x)^{1/2}\,\mathrm{d}x=\left[-\frac23(k-x)^{3/2}\right]_0^k=\frac23k^{3/2}

I_2=\frac{4k}{7}\cdot\frac{2k}{5}\cdot\frac23k^{3/2}=\frac{16}{105}k^{7/2}

\frac{16}{105}k^{7/2}=\frac{9\sqrt3}{280}

k^{7/2}=\frac{27\sqrt3}{128}=\left(\frac34\right)^{7/2}

k=\frac34

M1: uses the reduction formula to express I_2 in terms of I_0.
M1: evaluates I_0.
ddM1: solves an equation of the form a k^(7/2)=9sqrt(3)/280.
A1: exact value k=3/4.

Edexcel A-Level Mathematics A2 Fp3 4 5 Reduction Formulae Questions

Exam points

Edexcel A-Level Mathematics A2 Fp3 4 5 Reduction Formulae Questions question 1

Question (a)

Question (b)