Question 5 - A-Level Maths

Edexcel C34 2017 October — Question 5 8 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2017
Session	October
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration by Substitution
Type	Algebraic manipulation before substitution
Difficulty	Moderate -0.8 Part (i) requires straightforward application of standard integration formulas (power rule with chain rule and exponential integration), while part (ii) involves recognizing a standard substitution u = x² + 5 and solving a simple logarithmic equation. Both parts are routine textbook exercises with no problem-solving insight required, making this easier than average.
Spec	1.08b Integrate x^n: where n != -1 and sums 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution

5. (i) Find $$\int \left( ( 3 x + 5 ) ^ { 9 } + \mathrm { e } ^ { 5 x } \right) \mathrm { d } x$$ (ii) Given that $b$ is a constant greater than 2 , and $$\int _ { 2 } ^ { b } \frac { x } { x ^ { 2 } + 5 } \mathrm {~d} x = \ln ( \sqrt { 6 } )$$ use integration to find the value of $b$.

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Question 5:

Part (i):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\int\left((3x+5)^9 + e^{5x}\right)dx = \frac{(3x+5)^{10}}{30} + \frac{e^{5x}}{5}(+c)$	M1A1, B1	M1: integral of form $C(3x+5)^{10}$ or $C(3x+5)^{9+1}$, no other powers of $(3x+5)$. A1: $\frac{(3x+5)^{10}}{30}$, no need for $+c$. B1: $e^{5x} \to \frac{e^{5x}}{5}$

Part (ii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\int \frac{x}{x^2+5}dx = \frac{1}{2}\ln(x^2+5)$	M1A1	M1: answer of form $C\ln k(x^2+5)$. A1: $\frac{1}{2}\ln k(x^2+5)$ or $\ln k(x^2+5)^{\frac{1}{2}}$ or \(\frac{1}{2}\ln\
\(\int_2^b \frac{x}{x^2+5}dx = \ln(\sqrt{6}) = \frac{1}{2}\ln\	b^2+5\	- \frac{1}{2}\ln\
$\ln\left(\frac{b^2+5}{9}\right) = \ln 6 \Rightarrow b = 7$	ddM1, A1	ddM1: removes logs to get equation in $b$. Dependent on both previous M marks. A1: $b=7$ only; $b=\pm7$ scores A0 unless $-7$ rejected

# Question 5:

## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\left((3x+5)^9 + e^{5x}\right)dx = \frac{(3x+5)^{10}}{30} + \frac{e^{5x}}{5}(+c)$ | M1A1, B1 | M1: integral of form $C(3x+5)^{10}$ or $C(3x+5)^{9+1}$, no other powers of $(3x+5)$. A1: $\frac{(3x+5)^{10}}{30}$, no need for $+c$. B1: $e^{5x} \to \frac{e^{5x}}{5}$ |

## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{x}{x^2+5}dx = \frac{1}{2}\ln(x^2+5)$ | M1A1 | M1: answer of form $C\ln k(x^2+5)$. A1: $\frac{1}{2}\ln k(x^2+5)$ or $\ln k(x^2+5)^{\frac{1}{2}}$ or $\frac{1}{2}\ln\|x^2+5\|$ |
| $\int_2^b \frac{x}{x^2+5}dx = \ln(\sqrt{6}) = \frac{1}{2}\ln\|b^2+5\| - \frac{1}{2}\ln\|2^2+5\| = \ln(\sqrt{6})$ | M1 | Substitutes $b$ and $2$, subtracts, sets equal to $\ln(\sqrt{6})$ |
| $\ln\left(\frac{b^2+5}{9}\right) = \ln 6 \Rightarrow b = 7$ | ddM1, A1 | ddM1: removes logs to get equation in $b$. Dependent on both previous M marks. A1: $b=7$ only; $b=\pm7$ scores A0 unless $-7$ rejected |

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5. (i) Find

$$\int \left( ( 3 x + 5 ) ^ { 9 } + \mathrm { e } ^ { 5 x } \right) \mathrm { d } x$$

(ii) Given that $b$ is a constant greater than 2 , and

$$\int _ { 2 } ^ { b } \frac { x } { x ^ { 2 } + 5 } \mathrm {~d} x = \ln ( \sqrt { 6 } )$$

use integration to find the value of $b$.\\

\hfill \mbox{\textit{Edexcel C34 2017 Q5 [8]}}

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