A2 June 2023 Q8

EdexcelCurrent spec7 marksFurther Calculus

\[I_n = \int_0^2 (x - 2)^n \mathrm{e}^{4x}\,\mathrm{d}x \qquad n \geqslant 0\]

(a) Prove that for \(n \geqslant 1\)\[I_n = -a^{n-2} - \frac{n}{4}I_{n-1}\]where \(a\) is a constant to be determined. (4)

(b) Hence determine the exact value of\[\int_0^2 (x - 2)^2 \mathrm{e}^{4x}\,\mathrm{d}x\] (3)

Mark scheme (a)

Scheme	Marks	AO
\[I_n = \int_0^2 (x - 2)^n \mathrm{e}^{4x}\,\mathrm{d}x = \Big[(x - 2)^n \times A\mathrm{e}^{4x}\Big]_0^2 - \int_0^2 n(x - 2)^{n-1} \times A\mathrm{e}^{4x}\,\mathrm{d}x\]\[\left\{= \left[(x - 2)^n \times \frac{1}{4}\mathrm{e}^{4x}\right]_0^2 - \int_0^2 n(x - 2)^{n-1} \times \frac{1}{4}\mathrm{e}^{4x}\,\mathrm{d}x\right\}\]	M1	1.1b
\[= \big(0 - A(-2)^n\big) - A\int_0^2 (x - 2)^{n-1}\mathrm{e}^{4x}\,\mathrm{d}x\]\[\left\{= \left(0 - \frac{1}{4}(-2)^n\right) - \frac{n}{4}\int_0^2 (x - 2)^{n-1}\mathrm{e}^{4x}\,\mathrm{d}x\right\}\]	M1	1.1b
\[= -\frac{1}{(-2)^2}(-2)^n - \frac{n}{4}I_{n-1}\]	M1	3.1a
\[I_n = -(-2)^{n-2} - \frac{n}{4}I_{n-1}\]	A1	2.1
	(4)

Notes

M1: Applies integration by parts to achieve the correct form

M1: Substitutes in the limits 0 and 2 and simplifies integral to match the form of \(I_n\)

M1: Writes 4 as \((-2)^2\) in the first terms (or correctly combines powers) and replaces integral by \(I_{n-1}\)

In the alternative award this M for a full process to rearrange to the form \(I_{n+1}\) as well as the above constraints.

A1: Achieves the correct answer following correct working, cso. In the alternative a replacement of \(n\) by \(n - 1\) must also occur. isw

Alternative (a)

Scheme	Marks	AO
\[I_n = \int_0^2 (x - 2)^n \mathrm{e}^{4x}\,\mathrm{d}x = \left[\frac{(x - 2)^{n+1}}{n + 1} \times \mathrm{e}^{4x}\right]_0^2 - \int_0^2 \frac{(x - 2)^{n+1}}{n + 1} \times A\mathrm{e}^{4x}\,\mathrm{d}x\]\[\left\{= \left[\frac{(x - 2)^{n+1}}{n + 1} \times \mathrm{e}^{4x}\right]_0^2 - \int_0^2 \frac{(x - 2)^{n+1}}{n + 1} \times 4\mathrm{e}^{4x}\,\mathrm{d}x\right\}\]	M1	1.1b
\[= \left(0 - \frac{(-2)^{n+1}}{n + 1}\right) - \frac{A}{n + 1}\int_0^2 (x - 2)^{n+1}\mathrm{e}^{4x}\,\mathrm{d}x\]\[\left\{= \left(0 - \frac{(-2)^{n+1}}{n + 1}\right) - \frac{4}{n + 1}\int_0^2 (x - 2)^{n+1}\mathrm{e}^{4x}\,\mathrm{d}x\right\}\]	M1	1.1b
\[\Rightarrow (n + 1)I_n = -(-2)^{n+1} - 4I_{n+1} \Rightarrow I_{n+1} = -\frac{1}{(-2)^2}(-2)^{n+1} - \frac{n + 1}{4}I_n\]	M1	3.1a
\[\Rightarrow I_n = -(-2)^{n-2} - \frac{n}{4}I_{n-1}\]	A1	2.1
	(4)

Mark scheme (b)

Scheme	Marks	AO
\[I_2 = \int_0^2 (x - 2)^2 \mathrm{e}^{4x}\,\mathrm{d}x = -(-2)^0 - \frac{2}{4}I_1 = -1 - \frac{1}{2}\left(-(-2)^{-1} - \frac{1}{4}I_0\right)\]	M1	2.1
\[= -1 - \frac{1}{4} + \frac{1}{8}\left(\frac{1}{4}\mathrm{e}^8 - \frac{1}{4}\right)\]	M1	1.1b
\[= \frac{1}{32}\mathrm{e}^8 - \frac{41}{32}\]	A1	1.1b
	(3)
	(7 marks)

Notes

M1: For a full process of reducing the integral to an expression in \(I_0\)

M1: Evaluates \(I_0\) and substitutes into the expression.

A1: cso \(\dfrac{1}{32}\mathrm{e}^8 - \dfrac{41}{32}\) or simplified equivalent.

Note: Candidates might use

M1: \(I_2 = \int_0^2 (x - 2)^2 \mathrm{e}^{4x}\,\mathrm{d}x = -(-2)^0 - \dfrac{2}{4}I_1 = -1 - \dfrac{1}{2}I_1\)

M1: \(I_1 = \int_0^2 (x - 2)\mathrm{e}^{4x}\,\mathrm{d}x = -\dfrac{1}{16}\mathrm{e}^8 + \dfrac{9}{16}\) therefore \(I_2 = -1 - \dfrac{1}{2}\left(-\dfrac{1}{16}\mathrm{e}^8 + \dfrac{9}{16}\right)\)

A1: cso \(\dfrac{1}{32}\mathrm{e}^8 - \dfrac{41}{32}\) or simplified equivalent. isw