| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show substitution transforms integral, then evaluate |
| Difficulty | Standard +0.8 This is a structured two-part integration by substitution question requiring multiple algebraic manipulations. Part (a) involves finding the differential, changing limits, and algebraic manipulation to reach the given form. Part (b) requires expanding (u-1)³/u, integrating term-by-term, and evaluating definite integrals including logarithms. While methodical, it demands careful algebra across multiple steps and is more demanding than standard single-substitution questions, placing it moderately above average difficulty. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = 1 + \sqrt{x} \Rightarrow x = (u-1)^2 \Rightarrow \frac{dx}{du} = 2(u-1)\) or \(u = 1+\sqrt{x} \Rightarrow \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\) | B1 | Correct expression for \(\frac{dx}{du}\) or \(\frac{du}{dx}\) (or \(u'\)) or \(dx\) in terms of \(du\) or \(du\) in terms of \(dx\) |
| \(\int \frac{x}{1+\sqrt{x}}dx = \int \frac{(u-1)^2}{u} \cdot 2(u-1)du\) or \(\int \frac{x}{1+\sqrt{x}}dx = \int \frac{x}{u} \times 2x^{\frac{1}{2}}du = \int \frac{2x^{\frac{3}{2}}}{u}du = \int \frac{2(u-1)^3}{u}du\) | M1 | Complete method using the given substitution; correct method for their \(\frac{dx}{du}\) or \(\frac{du}{dx}\) leading to integral in terms of \(u\) only |
| \(\int_0^{16} \frac{x}{1+\sqrt{x}}dx = \int_1^5 \frac{2(u-1)^3}{u}du\) | A1 | All correct with correct limits and no errors; "d\(u\)" must be present (at least once before final answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\int \frac{u^3 - 3u^2 + 3u - 1}{u}du = 2\int\left(u^2 - 3u + 3 - \frac{1}{u}\right)du = \ldots\) | M1 | Realises requirement to expand and divide by \(u\); makes progress obtaining at least 3 terms of form \(ku^3, ku^2, ku, k\ln u\) |
| \(= (2)\left[\frac{u^3}{3} - \frac{3u^2}{2} + 3u - \ln u\right]\) | A1 | Correct integration; "2" may still be outside integral or omitted, but if combined with integrand the integration must be correct |
| \(= 2\left[\frac{5^3}{3} - \frac{3(5)^2}{2} + 3(5) - \ln 5 - \left(\frac{1}{3} - \frac{3}{2} + 3 - \ln 1\right)\right]\) | dM1 | Completes by applying "changed" limits and subtracts correct way round; depends on first M mark |
| \(= \frac{104}{3} - 2\ln 5\) | A1 | Allow equivalents e.g. \(\frac{208}{6}\) |
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 1 + \sqrt{x} \Rightarrow x = (u-1)^2 \Rightarrow \frac{dx}{du} = 2(u-1)$ **or** $u = 1+\sqrt{x} \Rightarrow \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ | B1 | Correct expression for $\frac{dx}{du}$ or $\frac{du}{dx}$ (or $u'$) or $dx$ in terms of $du$ or $du$ in terms of $dx$ |
| $\int \frac{x}{1+\sqrt{x}}dx = \int \frac{(u-1)^2}{u} \cdot 2(u-1)du$ **or** $\int \frac{x}{1+\sqrt{x}}dx = \int \frac{x}{u} \times 2x^{\frac{1}{2}}du = \int \frac{2x^{\frac{3}{2}}}{u}du = \int \frac{2(u-1)^3}{u}du$ | M1 | Complete method using the given substitution; correct method for their $\frac{dx}{du}$ or $\frac{du}{dx}$ leading to integral in terms of $u$ only |
| $\int_0^{16} \frac{x}{1+\sqrt{x}}dx = \int_1^5 \frac{2(u-1)^3}{u}du$ | A1 | All correct with correct limits and no errors; "d$u$" must be present (at least once before final answer) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\int \frac{u^3 - 3u^2 + 3u - 1}{u}du = 2\int\left(u^2 - 3u + 3 - \frac{1}{u}\right)du = \ldots$ | M1 | Realises requirement to expand and divide by $u$; makes progress obtaining at least 3 terms of form $ku^3, ku^2, ku, k\ln u$ |
| $= (2)\left[\frac{u^3}{3} - \frac{3u^2}{2} + 3u - \ln u\right]$ | A1 | Correct integration; "2" may still be outside integral or omitted, but if combined with integrand the integration must be correct |
| $= 2\left[\frac{5^3}{3} - \frac{3(5)^2}{2} + 3(5) - \ln 5 - \left(\frac{1}{3} - \frac{3}{2} + 3 - \ln 1\right)\right]$ | dM1 | Completes by applying "changed" limits and subtracts correct way round; depends on first M mark |
| $= \frac{104}{3} - 2\ln 5$ | A1 | Allow equivalents e.g. $\frac{208}{6}$ |
---\begin{enumerate}
\item (a) Use the substitution $u = 1 + \sqrt { x }$ to show that
\end{enumerate}
$$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = \int _ { p } ^ { q } \frac { 2 ( u - 1 ) ^ { 3 } } { u } \mathrm {~d} u$$
where $p$ and $q$ are constants to be found.\\
(b) Hence show that
$$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = A - B \ln 5$$
where $A$ and $B$ are constants to be found.
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q12 [7]}}