FP3 June 2014 Q3

EdexcelOld spec8 marksHyperbolic Functions Integration

3. Using calculus, find the exact value of

(a) \(\displaystyle\int_1^2 \frac{1}{\sqrt{(x^2 - 2x + 3)}}\,\mathrm{d}x\) (4)

(b) \(\displaystyle\int_0^1 \mathrm{e}^{2x}\sinh x\,\mathrm{d}x\) (4)

Mark scheme (a)

Scheme	Marks
\(x^2 - 2x + 3 = (x - 1)^2 + 2\) M1: Attempt to complete the square. Allow \((x - 1)^2 + k, k \neq 0\) A1: Correct expression	M1A1
\(\displaystyle\int \frac{1}{\sqrt{(x - 1)^2 + 2}}\,\mathrm{d}x = \alpha\,\mathrm{arsinh}(\mathrm{f}(x))\) Allow \(\alpha\ln\left(\mathrm{f}(x) + \sqrt{(\mathrm{f}(x))^2 + \beta}\right)\ (\beta > 0)\)	M1
\(\left[\mathrm{arsinh}\left(\dfrac{x - 1}{\sqrt{2}}\right)\right]_1^2 = \mathrm{arsinh}\dfrac{1}{\sqrt{2}}\) Any equivalent exact form. Allow \(\ln\left(\frac{1 + \sqrt{3}}{\sqrt{2}}\right)\) but no other terms e.g. arsinh(0)	A1
	(4)

Mark scheme (b)

Scheme	Marks
\(\mathrm{e}^{2x}\sinh x = \mathrm{e}^{2x}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) Substitutes the correct* exponential of \(\sinh x\)*	M1
\(\tfrac{1}{2}(\mathrm{e}^{3x} - \mathrm{e}^x)\) Correct expression with powers of e combined.	A1
\(\displaystyle\int_0^1 \tfrac{1}{2}(\mathrm{e}^{3x} - \mathrm{e}^x)\,\mathrm{d}x = \left[\tfrac{1}{2}\left(\tfrac{1}{3}\mathrm{e}^{3x} - \mathrm{e}^x\right)\right]_0^1\) \(= \tfrac{1}{2}\left(\tfrac{1}{3}\mathrm{e}^3 - \mathrm{e}^1\right) - \tfrac{1}{2}\left(\tfrac{1}{3}\mathrm{e}^0 - \mathrm{e}^0\right)\) \(\displaystyle\int \mathrm{e}^{px}\,\mathrm{d}x = q\mathrm{e}^{px}\) at least once and some correct use of the limits 0 and 1 and subtracts the right way round.	M1
\(= \left(\dfrac{\mathrm{e}^3}{6} - \dfrac{\mathrm{e}}{2} + \dfrac{1}{3}\right)\) Any exact equivalent (allow \(\mathrm{e}^1\)) but all like terms collected but isw following a correct answer.	A1
	(4)
	(8 marks)

Notes

(b) Integration by parts way 1

Scheme	Marks
\(I = \left[\tfrac{1}{2}\mathrm{e}^{2x}\sinh x\right]_0^1 - \displaystyle\int_0^1 \tfrac{1}{2}\mathrm{e}^{2x}\cosh x\,\mathrm{d}x = \left[\tfrac{1}{2}\mathrm{e}^{2x}\sinh x\right]_0^1 - \left[\tfrac{1}{4}\mathrm{e}^{2x}\cosh x\right]_0^1 + \tfrac{1}{4}I\) \(\tfrac{3}{4}I = \left[\tfrac{1}{2}\mathrm{e}^{2x}\sinh x\right]_0^1 - \left[\tfrac{1}{4}\mathrm{e}^{2x}\cosh x\right]_0^1\) M1: Parts twice in the correct direction A1: A correct expression for \(I\) or any constant multiple of \(I\)	M1A1
\(\displaystyle\int_0^1 \mathrm{e}^{2x}\sinh x\,\mathrm{d}x = \tfrac{4}{3}\left(\tfrac{1}{2}\mathrm{e}^2\sinh 1 - \tfrac{1}{4}\mathrm{e}^2\cosh 1 + \tfrac{1}{4}\right)\) M1: Correct use of limits having integrated by parts twice A1: Correct expression (oe)	M1A1 oe

Scheme

Marks

\(I = \left[\tfrac{1}{2}\mathrm{e}^{2x}\sinh x\right]_0^1 - \displaystyle\int_0^1 \tfrac{1}{2}\mathrm{e}^{2x}\cosh x\,\mathrm{d}x = \left[\tfrac{1}{2}\mathrm{e}^{2x}\sinh x\right]_0^1 - \left[\tfrac{1}{4}\mathrm{e}^{2x}\cosh x\right]_0^1 + \tfrac{1}{4}I\)
\(\tfrac{3}{4}I = \left[\tfrac{1}{2}\mathrm{e}^{2x}\sinh x\right]_0^1 - \left[\tfrac{1}{4}\mathrm{e}^{2x}\cosh x\right]_0^1\)
M1: Parts twice in the correct direction
A1: A correct expression for \(I\) or any constant multiple of \(I\)

M1A1

\(\displaystyle\int_0^1 \mathrm{e}^{2x}\sinh x\,\mathrm{d}x = \tfrac{4}{3}\left(\tfrac{1}{2}\mathrm{e}^2\sinh 1 - \tfrac{1}{4}\mathrm{e}^2\cosh 1 + \tfrac{1}{4}\right)\)
M1: Correct use of limits having integrated by parts twice
A1: Correct expression (oe)

M1A1 oe

(b) Integration by parts way 2

Scheme	Marks
\(I = \left[\mathrm{e}^{2x}\cosh x\right]_0^1 - \displaystyle\int_0^1 2\mathrm{e}^{2x}\cosh x\,\mathrm{d}x = \left[\mathrm{e}^{2x}\cosh x\right]_0^1 - \left[2\mathrm{e}^{2x}\sinh x\right]_0^1 + 4I\) \(-3I = \left[\mathrm{e}^{2x}\cosh x\right]_0^1 - \left[2\mathrm{e}^{2x}\sinh x\right]_0^1\) M1: Parts twice in the correct direction A1: A correct expression for \(I\) or any constant multiple of \(I\)
\(\displaystyle\int_0^1 \mathrm{e}^{2x}\sinh x\,\mathrm{d}x = -\frac{1}{3}\left(\mathrm{e}^2\cosh 1 - 1 - 2\mathrm{e}^2\sinh 1\right)\) M1: Correct use of limits having integrated by parts twice A1: Correct expression (oe)

Scheme

Marks

\(I = \left[\mathrm{e}^{2x}\cosh x\right]_0^1 - \displaystyle\int_0^1 2\mathrm{e}^{2x}\cosh x\,\mathrm{d}x = \left[\mathrm{e}^{2x}\cosh x\right]_0^1 - \left[2\mathrm{e}^{2x}\sinh x\right]_0^1 + 4I\)
\(-3I = \left[\mathrm{e}^{2x}\cosh x\right]_0^1 - \left[2\mathrm{e}^{2x}\sinh x\right]_0^1\)
M1: Parts twice in the correct direction
A1: A correct expression for \(I\) or any constant multiple of \(I\)

\(\displaystyle\int_0^1 \mathrm{e}^{2x}\sinh x\,\mathrm{d}x = -\frac{1}{3}\left(\mathrm{e}^2\cosh 1 - 1 - 2\mathrm{e}^2\sinh 1\right)\)
M1: Correct use of limits having integrated by parts twice
A1: Correct expression (oe)

Extra Notes: 3. (b) Parts once then exponentials

\(I = \left[\tfrac{1}{2}\mathrm{e}^{2x}\sinh x\right]_0^1 - \displaystyle\int_0^1 \tfrac{1}{2}\mathrm{e}^{2x}\cosh x\,\mathrm{d}x = \left[\tfrac{1}{2}\mathrm{e}^{2x}\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right]_0^1 - \int_0^1 \tfrac{1}{2}\mathrm{e}^{2x}\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\,\mathrm{d}x\)

M1 integrates by parts and writes \(\cosh x\) as exponentials
A1 Correct expression

\(= \left[\tfrac{1}{2}\mathrm{e}^{2x}\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right]_0^1 - \left[\frac{1}{12}\mathrm{e}^{3x} + \frac{1}{4}\mathrm{e}^x\right]_0^1 = \left[\tfrac{1}{2}\left(\tfrac{1}{3}\mathrm{e}^{3x} - \mathrm{e}^x\right)\right]_0^1 = \tfrac{1}{2}\left(\tfrac{1}{3}\mathrm{e}^3 - \mathrm{e}^1\right) - \tfrac{1}{2}\left(\tfrac{1}{3}\mathrm{e}^0 - \mathrm{e}^0\right)\)

M1 \(\displaystyle\int \mathrm{e}^{px}\,\mathrm{d}x = q\mathrm{e}^{px}\) at least once and correct use of the limits 0 and 1

\(= \left(\dfrac{\mathrm{e}^3}{6} - \dfrac{\mathrm{e}}{2} + \dfrac{1}{3}\right)\) A1
Any exact equivalent (allow \(\mathrm{e}^1\)) but all like terms collected but isw following a correct answer.

(corrected from the printed mark scheme: the second bracket is printed as \(\left[\frac{1}{12}\mathrm{e}^{2x} + \frac{1}{4}\mathrm{e}^x\right]_0^1\); integrating \(\tfrac{1}{4}(\mathrm{e}^{3x} + \mathrm{e}^x)\) gives \(\frac{1}{12}\mathrm{e}^{3x}\))