| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Square root substitution: definite integral |
| Difficulty | Standard +0.3 This is a straightforward integration by substitution question where the substitution u = x + 2 is natural and leads directly to a polynomial integral. The question type 'show that' removes problem-solving difficulty since students know the target answer. The algebraic manipulation and evaluation at limits is routine for A-level, making this slightly easier than average. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Chooses suitable method for \(\int_0^2 2x\sqrt{x+2}\, dx\) using valid substitution \(u = \ldots\), changing terms to \(u\)'s and integrating with appropriate limits | M1 | Look for: terms and limits changed to \(u\)'s; attempted multiplication of terms raising at least one power of \(u\) by one; use of at least top correct limit |
| \(u = \sqrt{x+2} \Rightarrow \frac{dx}{du} = 2u\) oe or \(u = x+2 \Rightarrow \frac{dx}{du} = 1\) oe | B1 | For giving the substitution and stating correct \(\frac{dx}{du}\) |
| \(\int 2x\sqrt{x+2}\, dx = \int A(u^2 \pm 2)u^2\, du\) or \(= \int A(u \pm 2)\sqrt{u}\, du\) | M1 | Attempting to get all aspects of integral in terms of \(u\) |
| \(= Pu^5 \pm Qu^3\) or \(= Su^{\frac{5}{2}} \pm Tu^{\frac{3}{2}}\) | dM1 | Correct method of expanding and integrating each term. Dependent on previous M |
| \(= \frac{4}{5}u^5 - \frac{8}{3}u^3\) or \(= \frac{4}{5}u^{\frac{5}{2}} - \frac{8}{3}u^{\frac{3}{2}}\) | A1 | Correct answer in \(x\) or \(u\) |
| Uses limits 2 and \(\sqrt{2}\) (or limits 4 and 2) the correct way around | ddM1 | Dependent on previous M, correct limits used correctly. |
| \(= \frac{32}{15}(2 + \sqrt{2})\) | A1* | Proceeds correctly to given answer. At least one correct intermediate line must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Chooses by parts the correct way around, using limits | M1 | Formula applied correct way around; may condone incorrect attempts at integrating \(\sqrt{x+2}\) |
| \(\int(\sqrt{x+2})\, dx = \frac{2}{3}(x+2)^{\frac{3}{2}}\) | B1 | oe |
| \(\int 2x\sqrt{x+2}\, dx = Ax(x+2)^{\frac{3}{2}} - \int B(x+2)^{\frac{3}{2}}(dx)\) | M1 | Integration by parts the right way around |
| \(= Ax(x+2)^{\frac{3}{2}} - C(x+2)^{\frac{5}{2}}\) | dM1 | Integrating a second time |
| \(= \frac{4}{3}x(x+2)^{\frac{3}{2}} - \frac{8}{15}(x+2)^{\frac{5}{2}}\) | A1 | May be unsimplified |
| Uses limits 2 and 0 the correct way around | ddM1 | Dependent on previous M |
| \(= \frac{32}{15}(2+\sqrt{2})\) | A1* | Proceeds to given answer; at least one correct intermediate line must be seen |
# Question 13:
## Substitution Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Chooses suitable method for $\int_0^2 2x\sqrt{x+2}\, dx$ using valid substitution $u = \ldots$, changing terms to $u$'s and integrating with appropriate limits | M1 | Look for: terms and limits changed to $u$'s; attempted multiplication of terms raising at least one power of $u$ by one; use of at least top correct limit |
| $u = \sqrt{x+2} \Rightarrow \frac{dx}{du} = 2u$ oe **or** $u = x+2 \Rightarrow \frac{dx}{du} = 1$ oe | B1 | For giving the substitution and stating correct $\frac{dx}{du}$ |
| $\int 2x\sqrt{x+2}\, dx = \int A(u^2 \pm 2)u^2\, du$ **or** $= \int A(u \pm 2)\sqrt{u}\, du$ | M1 | Attempting to get all aspects of integral in terms of $u$ |
| $= Pu^5 \pm Qu^3$ **or** $= Su^{\frac{5}{2}} \pm Tu^{\frac{3}{2}}$ | dM1 | Correct method of expanding and integrating each term. Dependent on previous M |
| $= \frac{4}{5}u^5 - \frac{8}{3}u^3$ **or** $= \frac{4}{5}u^{\frac{5}{2}} - \frac{8}{3}u^{\frac{3}{2}}$ | A1 | Correct answer in $x$ or $u$ |
| Uses limits 2 and $\sqrt{2}$ (or limits 4 and 2) the correct way around | ddM1 | Dependent on previous M, correct limits used correctly. |
| $= \frac{32}{15}(2 + \sqrt{2})$ | A1* | Proceeds correctly to given answer. At least one correct intermediate line must be seen |
## Integration by Parts (Alternative):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Chooses by parts the correct way around, using limits | M1 | Formula applied correct way around; may condone incorrect attempts at integrating $\sqrt{x+2}$ |
| $\int(\sqrt{x+2})\, dx = \frac{2}{3}(x+2)^{\frac{3}{2}}$ | B1 | oe |
| $\int 2x\sqrt{x+2}\, dx = Ax(x+2)^{\frac{3}{2}} - \int B(x+2)^{\frac{3}{2}}(dx)$ | M1 | Integration by parts the right way around |
| $= Ax(x+2)^{\frac{3}{2}} - C(x+2)^{\frac{5}{2}}$ | dM1 | Integrating a second time |
| $= \frac{4}{3}x(x+2)^{\frac{3}{2}} - \frac{8}{15}(x+2)^{\frac{5}{2}}$ | A1 | May be unsimplified |
| Uses limits 2 and 0 the correct way around | ddM1 | Dependent on previous M |
| $= \frac{32}{15}(2+\sqrt{2})$ | A1* | Proceeds to given answer; at least one correct intermediate line must be seen |
---\begin{enumerate}
\item Show that
\end{enumerate}
$$\int _ { 0 } ^ { 2 } 2 x \sqrt { x + 2 } \mathrm {~d} x = \frac { 32 } { 15 } ( 2 + \sqrt { 2 } )$$
\hfill \mbox{\textit{Edexcel Paper 1 2018 Q13 [7]}}