C4 January 2007 Q8

EdexcelOld spec15 marksIntegration

\[I = \int_0^5 \mathrm{e}^{\surd(3x + 1)}\,\mathrm{d}x.\]

(a) Given that \(y = \mathrm{e}^{\surd(3x + 1)}\), complete the table with the values of \(y\) corresponding to \(x = 2\), 3 and 4.

\(x\)	0	1	2	3	4	5
\(y\)	\(\mathrm{e}^1\)	\(\mathrm{e}^2\)				\(\mathrm{e}^4\)

(2)

(b) Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the original integral \(I\), giving your answer to 4 significant figures. (3)

(c) Use the substitution \(t = \surd(3x + 1)\) to show that \(I\) may be expressed as \(\displaystyle\int_a^b kt\mathrm{e}^t\,\mathrm{d}t\), giving the values of \(a\), \(b\) and \(k\). (5)

(d) Use integration by parts to evaluate this integral, and hence find the value of \(I\) correct to 4 significant figures, showing all the steps in your working. (5)

Mark scheme (a)

\(x\)	0	1	2	3	4	5
\(y\)	\(\mathrm{e}^1\)	\(\mathrm{e}^2\)	\(\mathrm{e}^{\sqrt{7}}\)	\(\mathrm{e}^{\sqrt{10}}\)	\(\mathrm{e}^{\sqrt{13}}\)	\(\mathrm{e}^4\)
or \(y\)	2.71828…	7.38906…	14.09403…	23.62434…	36.80197…	54.59815…

Scheme	Marks
*At least* two correct	B1
All three correct	B1
	(2)

Notes

Either \(\mathrm{e}^{\sqrt{7}}\), \(\mathrm{e}^{\sqrt{10}}\) and \(\mathrm{e}^{\sqrt{13}}\) or awrt 14.1, 23.6 and 36.8 or e to the power awrt 2.65, 3.16, 3.61 (or mixture of decimals and e’s)

Mark scheme (b)

Scheme	Marks
\(I \approx \dfrac{1}{2} \times 1\ ; \times \underline{\left\{\mathrm{e}^1 + 2\left(\mathrm{e}^2 + \mathrm{e}^{\sqrt{7}} + \mathrm{e}^{\sqrt{10}} + \mathrm{e}^{\sqrt{13}}\right) + \mathrm{e}^4\right\}}\)	B1; M1ft
\(= \dfrac{1}{2} \times 221.1352227\ldots = 110.5676113\ldots = \underline{110.6}\) (4sf)	A1 cao
	(3)

Notes

B1 Outside brackets \(\frac{1}{2} \times 1\)

M1ft For structure of trapezium rule \(\{\ldots\ldots\ldots\}\);

A1 cao \(\underline{110.6}\)

Beware: In part (b) candidates can add up the individual trapezia:
(b) \(I \approx \tfrac{1}{2}.1\left(\underline{\mathrm{e}^1 + \mathrm{e}^2}\right) + \tfrac{1}{2}.1\left(\underline{\mathrm{e}^2 + \mathrm{e}^{\sqrt{7}}}\right) + \tfrac{1}{2}.1\left(\underline{\mathrm{e}^{\sqrt{7}} + \mathrm{e}^{\sqrt{10}}}\right) + \tfrac{1}{2}.1\left(\underline{\mathrm{e}^{\sqrt{10}} + \mathrm{e}^{\sqrt{13}}}\right) + \tfrac{1}{2}.1\left(\underline{\mathrm{e}^{\sqrt{13}} + \mathrm{e}^4}\right)\)

Mark scheme (c)

Scheme	Marks
\(t = (3x + 1)^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}x} = \tfrac{1}{2}.3.(3x + 1)^{-\frac{1}{2}}\) … or \(t^2 = 3x + 1 \Rightarrow \underline{2t\dfrac{\mathrm{d}t}{\mathrm{d}x} = 3}\)	M1 A1
so \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{3}{2.(3x + 1)^{\frac{1}{2}}} = \dfrac{3}{2t} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{2t}{3}\) \(\therefore I = \displaystyle\int \mathrm{e}^{\sqrt{(3x + 1)}}\,\mathrm{d}x = \int \mathrm{e}^t\dfrac{\mathrm{d}x}{\mathrm{d}t}.\mathrm{d}t = \int \mathrm{e}^t.\frac{2t}{3}.\mathrm{d}t\)	dM1
\(\therefore I = \underline{\displaystyle\int \tfrac{2}{3}t\mathrm{e}^t\,\mathrm{d}t}\)	A1
change limits: when \(x = 0\), \(t = 1\) & when \(x = 5\), \(t = 4\)	B1
Hence \(I = \displaystyle\int_1^4 \tfrac{2}{3}t\mathrm{e}^t\,\mathrm{d}t\); where \(a = 1,\ b = 4,\ k = \tfrac{2}{3}\)
	(5)

Notes

M1 \(A(3x + 1)^{-\frac{1}{2}}\) or \(t\frac{\mathrm{d}t}{\mathrm{d}x} = A\)

A1 \(\tfrac{3}{2}(3x + 1)^{-\frac{1}{2}}\) or \(\underline{2t\frac{\mathrm{d}t}{\mathrm{d}x} = 3}\)

dM1 Candidate obtains either \(\frac{\mathrm{d}t}{\mathrm{d}x}\) or \(\frac{\mathrm{d}x}{\mathrm{d}t}\) in terms of \(t\) … … and moves on to substitute this into \(I\) to convert an integral wrt \(x\) to an integral wrt \(t\).

A1 \(\underline{\int \frac{2}{3}t\mathrm{e}^t}\)

B1 changes limits \(x \to t\) so that \(0 \to 1\) and \(5 \to 4\)

Mark scheme (d)

Scheme	Marks
\(\left\{\begin{aligned} u &= t \Rightarrow \tfrac{\mathrm{d}u}{\mathrm{d}t} = 1 \\ \tfrac{\mathrm{d}v}{\mathrm{d}t} &= \mathrm{e}^t \Rightarrow v = \mathrm{e}^t \end{aligned}\right\}\)
\(k\displaystyle\int t\mathrm{e}^t\,\mathrm{d}t = k\left(t\mathrm{e}^t - \int \mathrm{e}^t.1\,\mathrm{d}t\right)\)	M1 A1
\(= k\left(\underline{t\mathrm{e}^t - \mathrm{e}^t}\right) + c\)	A1
\(\therefore \displaystyle\int_1^4 \tfrac{2}{3}t\mathrm{e}^t\,\mathrm{d}t = \frac{2}{3}\left\{\left(4\mathrm{e}^4 - \mathrm{e}^4\right) - \left(\mathrm{e}^1 - \mathrm{e}^1\right)\right\}\)	dM1 oe
\(= \tfrac{2}{3}(3\mathrm{e}^4) = \underline{2\mathrm{e}^4} = 109.1963\ldots\)	A1
	(5)
	(15 marks)

Notes

Let \(k\) be any constant for the first three marks of this part.

M1 Use of ‘integration by parts’ formula in the correct direction.

A1 Correct expression with a constant factor \(k\).

A1 Correct integration with/without a constant factor \(k\)

dM1 oe Substitutes their changed limits into the integrand and subtracts oe.

A1 either \(2\mathrm{e}^4\) or awrt 109.2

Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark. ddM1 denotes a method mark which is dependent upon the award of the previous two method marks.