Standard +0.3 This is a straightforward substitution question where the substitution is given explicitly. Students must find du/dx, change limits, rewrite the integrand, integrate (which simplifies to a basic rational function), and verify the given answer. While it requires careful algebraic manipulation and multiple steps, it follows a standard template for C4 substitution questions with no conceptual surprises, making it slightly easier than average.
Attempt to diff to connect \(du\) and \(dx\) or find \(\frac{du}{dx}\) or \(\frac{dx}{du}\)
M1
no accuracy, not just \(du=dx\)
Correct e.g. \(\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}\) or \(dx=(2u-2)\,du\) AEF
*A1
Indefinite integral in terms of \(u = \int\frac{2u-2}{u}\,du\)
A1dep*
Provided of form \(\int\frac{au+b}{u}\,du\), change to \(\int a+\frac{b}{u}\,du\)
M1
Or by parts
Integrate to \(au + b\ln
u
\) or \(au + b\ln u\)
Use correct variable for limits after attempt at integral of \(f(u)\)
M1
i.e. use new values of \(u\) (usually) or orig values of \(x\) (if resubst)
Show as \(8-2\ln 4 - 6 + 2\ln 3\) (oe) \(= 2+2\ln\frac{3}{4}\) AG WWW
A1
Some 'numerical' working must be shown before giving final answer
Total: [7]
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to diff to connect $du$ and $dx$ or find $\frac{du}{dx}$ or $\frac{dx}{du}$ | M1 | no accuracy, not just $du=dx$ |
| Correct e.g. $\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$ or $dx=(2u-2)\,du$ AEF | *A1 | |
| Indefinite integral in terms of $u = \int\frac{2u-2}{u}\,du$ | A1dep* | |
| Provided of form $\int\frac{au+b}{u}\,du$, change to $\int a+\frac{b}{u}\,du$ | M1 | Or by parts |
| Integrate to $au + b\ln|u|$ or $au + b\ln u$ | A1 ft | |
| Use correct variable for limits after attempt at integral of $f(u)$ | M1 | i.e. use new values of $u$ (usually) or orig values of $x$ (if resubst) |
| Show as $8-2\ln 4 - 6 + 2\ln 3$ (oe) $= 2+2\ln\frac{3}{4}$ **AG WWW** | A1 | Some 'numerical' working must be shown before giving final answer |
| **Total: [7]** | | |
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