A2 October 2020 Paper 1 Q2

EdexcelCurrent spec7 marksIntegration

(a) Explain why \(\displaystyle\int_{1}^{\infty} \frac{1}{x(2x + 5)}\,\mathrm{d}x\) is an improper integral. (1)

(b) Prove that\[\int_{1}^{\infty} \frac{1}{x(2x + 5)}\,\mathrm{d}x = a\ln b\]where \(a\) and \(b\) are rational numbers to be determined. (6)

Mark scheme (a)

Scheme	Marks	AO
E.g. Because the interval being integrated over is unbounded Accept because the upper limit is infinity Accept because a limit is required to evaluate it	B1	2.4
	(1)

Notes

(a)
B1: For a suitable explanation with no contrary reasoning. Technically this should refer to the interval being unbounded, but this is unlikely to be seen. Accept “Because the upper limit is infinity”. Do not award if there are erroneous statements e.g. referring to as \(x = 0\) the integrand is not defined. Do not accept “because one of the limits is undefined” unless they state they mean \(\infty\). Do not accept “it is undefined when \(x = \infty\)” without reference to “it” being the upper limit.

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{1}{x(2x + 5)} = \dfrac{A}{x} + \dfrac{B}{2x + 5} \Rightarrow A = \ldots,\ B = \ldots\)	M1	3.1a
\(\dfrac{1}{x(2x + 5)} = \dfrac{1}{5x} - \dfrac{2}{5(2x + 5)}\)	A1	1.1b
\(\displaystyle\int \frac{1}{5x} - \frac{2}{5(2x + 5)}\,\mathrm{d}x = \frac{1}{5}\ln x - \frac{1}{5}\ln(2x + 5)\)	A1ft	1.1b
\(\dfrac{1}{5}\ln x - \dfrac{1}{5}\ln(2x + 5) = \dfrac{1}{5}\ln\dfrac{x}{(2x + 5)}\)	M1	2.1
\(\displaystyle\lim_{x \to \infty}\left\{\frac{1}{5}\ln\frac{x}{2x + 5}\right\} = \frac{1}{5}\ln\frac{1}{2}\)	B1	2.2a
\(\displaystyle\Rightarrow \int_{1}^{\infty} \frac{1}{x(2x + 5)}\,\mathrm{d}x = \frac{1}{5}\ln\frac{1}{2} - \frac{1}{5}\ln\frac{1}{7} = \frac{1}{5}\ln\frac{7}{2}\)	A1	1.1b
	(6)
	(7 marks)

Notes

(b)
M1: Selects the correct form for partial fractions and proceeds to find values for \(A\) and \(B\)
A1: Correct constants or partial fractions
A1ft: \(\displaystyle\int \frac{p}{x} + \frac{q}{2x + 5}\,\mathrm{d}x = p\ln x + \frac{q}{2}\ln(2x + 5)\) Note that \(\dfrac{1}{5}\ln 5x - \dfrac{1}{5}\ln(10x + 25)\) is correct.
M1: Combines logs correctly. May see \(-\dfrac{1}{5}\ln\left(\dfrac{2x + 5}{x}\right) = -\dfrac{1}{5}\ln\left(2 + \dfrac{5}{x}\right)\)
B1: Correct upper limit for \(x \to \infty\) by recognising the dominant terms. (Simply replacing \(x\) with \(\infty\) scores B0)
A1: Deduces the correct value for the improper integral in the correct form

(b) Way 2

Scheme	Marks	AO
\(\dfrac{1}{x(2x + 5)} = \dfrac{1}{2\left(x^2 + \dfrac{5}{2}x\right)} = \dfrac{1}{2} \times \dfrac{1}{\left(x + \dfrac{5}{4}\right)^2 - \dfrac{25}{16}}\)	M1 A1	3.1a 1.1b
\(\displaystyle\int \frac{1}{x(2x + 5)}\,\mathrm{d}x = \frac{1}{2} \times \frac{2}{5}\ln\left\|\frac{x + \frac{5}{4} - \frac{5}{4}}{x + \frac{5}{4} + \frac{5}{4}}\right\| = \frac{1}{5}\ln\left\|\frac{2x}{2x + 5}\right\|\)	M1 A1ft	2.1 1.1b
\(\displaystyle\lim_{x \to \infty}\left\{\frac{1}{5}\ln\frac{2x}{2x + 5}\right\} = \frac{1}{5}\ln\frac{2}{2} = 0\)	B1	2.2a
\(\displaystyle\Rightarrow \int_{1}^{\infty} \frac{1}{x(2x + 5)}\,\mathrm{d}x = 0 - \frac{1}{5}\ln\frac{2}{7} = \frac{1}{5}\ln\frac{7}{2}\)	A1	1.1b
	(6)

Note the method marks as MAMABA, and should be entered in this order on ePEN.
M1: Expands the denominator and completes the square.
A1: Correct expression
M1: For \(\dfrac{1}{(x + p)^2 - a^2} \to k\ln\left|\dfrac{x + p - a}{x + p + a}\right|\)
A1ft: \(\dfrac{1}{2}\dfrac{1}{(x + a)^2 - a^2} \to \dfrac{1}{2a}\ln\left|\dfrac{x}{x + 2a}\right|\) with their \(a\) (may be simplified as in scheme).
B1: Correct upper limit for \(x \to \infty\) by recognising the dominant terms. (Simply replacing \(x\) with \(\infty\) scores B0) Note in this method the upper limit evaluates to zero.
A1: Deduces the correct value for the improper integral in the correct form. Accept \(-\dfrac{1}{5}\ln\dfrac{2}{7}\)