OCR MEI Paper 3 2019 June — Question 11 6 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeSquare root substitution: indefinite integral
DifficultyStandard +0.3 This is a straightforward substitution question where the substitution is explicitly given. Students must find du/dx, rearrange to express x and dx in terms of u, substitute into the integral, and integrate a polynomial. While it requires multiple steps and careful algebraic manipulation, it's a standard textbook exercise with no novel insight required, making it slightly easier than average.
Spec1.08h Integration by substitution
11 By using the substitution \(u = 1 + \sqrt { x }\), find \(\int \frac { x } { 1 + \sqrt { x } } \mathrm {~d} x\). Answer all the questions.
Question 11:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Let \(u = 1 + \sqrt{x}\), \(\Rightarrow dx = 2(u-1)\,du\) or \(dx = \frac{du}{\frac{1}{2}x^{-\frac{1}{2}}}\) oe or \(2\sqrt{x}\,du\)M1 1.1
\(\int \frac{x}{1+\sqrt{x}}\,dx = \int \frac{2(u-1)^3}{u}\,du\) oeA1 1.1
\(\int \frac{2(u^3 - 3u^2 + 3u - 1)}{u}\,du\)M1 1.1
\(\int 2\left(u^2 - 3u + 3 - \frac{1}{u}\right)du\)M1 3.1a
\(\frac{2u^3}{3} - 3u^2 + 6u - 2\lnu + c\)
\(\frac{2(1+\sqrt{x})^3}{3} - 3(1+\sqrt{x})^2 + 6(1+\sqrt{x}) - 2\ln\left1+\sqrt{x}\right + c\)
[6] 
## Question 11:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $u = 1 + \sqrt{x}$, $\Rightarrow dx = 2(u-1)\,du$ or $dx = \frac{du}{\frac{1}{2}x^{-\frac{1}{2}}}$ oe or $2\sqrt{x}\,du$ | M1 | 1.1 |
| $\int \frac{x}{1+\sqrt{x}}\,dx = \int \frac{2(u-1)^3}{u}\,du$ oe | A1 | 1.1 | In terms of $u$ |
| $\int \frac{2(u^3 - 3u^2 + 3u - 1)}{u}\,du$ | M1 | 1.1 | For expanding cubic (allow one error) |
| $\int 2\left(u^2 - 3u + 3 - \frac{1}{u}\right)du$ | M1 | 3.1a | Dividing by $u$ |
| $\frac{2u^3}{3} - 3u^2 + 6u - 2\ln|u| + c$ | M1 | 1.1 | 3 of *their* terms correctly integrated to include $\ln u$; Constant term may be missing at this stage; Condone omission of modulus (since $1+\sqrt{x}$, and hence $u$, must be positive) |
| $\frac{2(1+\sqrt{x})^3}{3} - 3(1+\sqrt{x})^2 + 6(1+\sqrt{x}) - 2\ln\left|1+\sqrt{x}\right| + c$ | A1 | 2.5 | Correct answer in terms of $x$ with constant term; oe |
| **[6]** | | |

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11 By using the substitution $u = 1 + \sqrt { x }$, find $\int \frac { x } { 1 + \sqrt { x } } \mathrm {~d} x$.

Answer all the questions.

\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q11 [6]}}
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