Edexcel C34 2016 June — Question 5 6 marks

Exam BoardEdexcel
ModuleC34 (Core Mathematics 3 & 4)
Year2016
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyStandard +0.3 This is a straightforward application of integration by parts with an exponential function. While it requires careful algebraic manipulation and evaluation at limits, it follows a standard template (polynomial times exponential) that students practice extensively. The main challenge is handling 2^x correctly and simplifying the final numerical answer to a fraction, but no novel insight is required.
Spec1.08i Integration by parts

5.Use integration by parts to find the exact value of $$\int _ { 0 } ^ { 2 } x 2 ^ { x } \mathrm {~d} x$$ Write your answer as a single simplified fraction.

Question 5 - Integration by parts:
AnswerMarks Guidance
WorkingMark Notes
\(\int x2^x\,dx = x\frac{2^x}{\ln 2}-\int\frac{2^x}{\ln 2}\,dx\)M1A1 M1: Integrates by parts the right way around to obtain form \(ax2^x-\int b2^x\,dx\). Allow \(a=1\) and/or \(b=1\). A1: \(x\frac{2^x}{\ln 2}-\int\frac{2^x}{\ln 2}\,dx\) (does not need to be on one line)
\(\int x2^x\,dx = x\frac{2^x}{\ln 2}-\frac{2^x}{(\ln 2)^2}\)dM1A1 dM1: Completes to obtain form \(\ldots -k2^x\). A1: \(x\frac{2^x}{\ln 2}-\frac{2^x}{(\ln 2)^2}\)
\(\left[x\frac{2^x}{\ln 2}-\frac{2^x}{(\ln 2)^2}\right]_0^2=\left(\frac{2\times2^2}{\ln 2}-\frac{2^2}{(\ln 2)^2}\right)-\left(\frac{0\times2^0}{\ln 2}-\frac{2^0}{(\ln 2)^2}\right)\)ddM1 Uses limits 0 and 2, subtracts right way round. \(F(0)\) may be implied by e.g. \(\frac{1}{(\ln 2)^2}\). Note: \(\left(\frac{2\times2^2}{\ln 2}-\frac{2^2}{(\ln 2)^2}\right)-(0)\) or just \(\left(\frac{2\times2^2}{\ln 2}-\frac{2^2}{(\ln 2)^2}\right)\) is ddM0
\(=\frac{8}{\ln 2}-\frac{4}{(\ln 2)^2}+\frac{1}{(\ln 2)^2}\)
\(=\frac{8\ln 2-3}{(\ln 2)^2}\)A1 Correct simplified fraction. Allow equivalent simplified forms e.g. \(\frac{\ln 256-3}{(\ln 2)^2}\), \(\frac{\ln 2^8-3}{(\ln 2)^2}\). Allow denominator as \((\ln 2)(\ln 2)\) and \(\ln^2 2\) but not \(\ln 2^2\)
Alternative by substitution (\(u=2^x\)):
AnswerMarks Guidance
WorkingMark Notes
\(u=2^x\Rightarrow\int x2^x\,dx=\int\frac{\ln u}{\ln 2}\cdot u\cdot\frac{1}{u\ln 2}\,du=\int\frac{\ln u}{(\ln 2)^2}\,du\)
\(\int\frac{\ln u}{(\ln 2)^2}\,du=\frac{1}{(\ln 2)^2}\left(u\ln u-\int du\right)\)M1A1 M1: Integrates by parts right way to obtain form \(au\ln u-\int b\,du\). Allow \(a=1\) and/or \(b=1\). A1: \(\frac{1}{(\ln 2)^2}\left(u\ln u-\int du\right)\)
\(\int\frac{\ln u}{(\ln 2)^2}\,du=\frac{1}{(\ln 2)^2}(u\ln u-u)\)dM1A1 dM1: Completes to form \(\ldots -ku\). A1: \(\frac{1}{(\ln 2)^2}(u\ln u - u)\)
\(\left[\frac{1}{(\ln 2)^2}(u\ln u-u)\right]_1^4=\frac{1}{(\ln 2)^2}(4\ln 4-4)-(\ln 1-1)\)M1 Uses limits 1 and 4, subtracts right way round
\(=\frac{4\ln 4-3}{(\ln 2)^2}\)A1 Correct simplified fraction. Allow equivalent forms e.g. \(\frac{\ln 256-3}{(\ln 2)^2}\), \(\frac{\ln 2^8-3}{(\ln 2)^2}\). Allow denominator as \((\ln 2)(\ln 2)\) and \(\ln^2 2\) but not \(\ln 2^2\)
# Question 5 - Integration by parts:

| Working | Mark | Notes |
|---------|------|-------|
| $\int x2^x\,dx = x\frac{2^x}{\ln 2}-\int\frac{2^x}{\ln 2}\,dx$ | M1A1 | M1: Integrates by parts the right way around to obtain form $ax2^x-\int b2^x\,dx$. Allow $a=1$ and/or $b=1$. A1: $x\frac{2^x}{\ln 2}-\int\frac{2^x}{\ln 2}\,dx$ (does not need to be on one line) |
| $\int x2^x\,dx = x\frac{2^x}{\ln 2}-\frac{2^x}{(\ln 2)^2}$ | dM1A1 | dM1: Completes to obtain form $\ldots -k2^x$. A1: $x\frac{2^x}{\ln 2}-\frac{2^x}{(\ln 2)^2}$ |
| $\left[x\frac{2^x}{\ln 2}-\frac{2^x}{(\ln 2)^2}\right]_0^2=\left(\frac{2\times2^2}{\ln 2}-\frac{2^2}{(\ln 2)^2}\right)-\left(\frac{0\times2^0}{\ln 2}-\frac{2^0}{(\ln 2)^2}\right)$ | ddM1 | Uses limits 0 and 2, subtracts right way round. $F(0)$ may be implied by e.g. $\frac{1}{(\ln 2)^2}$. Note: $\left(\frac{2\times2^2}{\ln 2}-\frac{2^2}{(\ln 2)^2}\right)-(0)$ or just $\left(\frac{2\times2^2}{\ln 2}-\frac{2^2}{(\ln 2)^2}\right)$ is ddM0 |
| $=\frac{8}{\ln 2}-\frac{4}{(\ln 2)^2}+\frac{1}{(\ln 2)^2}$ | | |
| $=\frac{8\ln 2-3}{(\ln 2)^2}$ | A1 | Correct simplified fraction. Allow equivalent simplified forms e.g. $\frac{\ln 256-3}{(\ln 2)^2}$, $\frac{\ln 2^8-3}{(\ln 2)^2}$. Allow denominator as $(\ln 2)(\ln 2)$ and $\ln^2 2$ but **not** $\ln 2^2$ |

## Alternative by substitution ($u=2^x$):

| Working | Mark | Notes |
|---------|------|-------|
| $u=2^x\Rightarrow\int x2^x\,dx=\int\frac{\ln u}{\ln 2}\cdot u\cdot\frac{1}{u\ln 2}\,du=\int\frac{\ln u}{(\ln 2)^2}\,du$ | | |
| $\int\frac{\ln u}{(\ln 2)^2}\,du=\frac{1}{(\ln 2)^2}\left(u\ln u-\int du\right)$ | M1A1 | M1: Integrates by parts right way to obtain form $au\ln u-\int b\,du$. Allow $a=1$ and/or $b=1$. A1: $\frac{1}{(\ln 2)^2}\left(u\ln u-\int du\right)$ |
| $\int\frac{\ln u}{(\ln 2)^2}\,du=\frac{1}{(\ln 2)^2}(u\ln u-u)$ | dM1A1 | dM1: Completes to form $\ldots -ku$. A1: $\frac{1}{(\ln 2)^2}(u\ln u - u)$ |
| $\left[\frac{1}{(\ln 2)^2}(u\ln u-u)\right]_1^4=\frac{1}{(\ln 2)^2}(4\ln 4-4)-(\ln 1-1)$ | M1 | Uses limits 1 and 4, subtracts right way round |
| $=\frac{4\ln 4-3}{(\ln 2)^2}$ | A1 | Correct simplified fraction. Allow equivalent forms e.g. $\frac{\ln 256-3}{(\ln 2)^2}$, $\frac{\ln 2^8-3}{(\ln 2)^2}$. Allow denominator as $(\ln 2)(\ln 2)$ and $\ln^2 2$ but **not** $\ln 2^2$ |
5.Use integration by parts to find the exact value of

$$\int _ { 0 } ^ { 2 } x 2 ^ { x } \mathrm {~d} x$$

Write your answer as a single simplified fraction.\\

\hfill \mbox{\textit{Edexcel C34 2016 Q5 [6]}}