| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Indefinite integral with linear substitution |
| Difficulty | Moderate -0.8 This is a straightforward application of the reverse chain rule for integration, requiring only pattern recognition and knowledge of standard integration formulas. Part (i) uses the power rule with a linear substitution, and part (ii) recognizes the logarithmic form. Both are routine C3 exercises with minimal steps and no problem-solving required beyond direct application of learned techniques. |
| Spec | 1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain integral of form \(k(4-3x)^8\) | M1 | Any non-zero constant \(k\); using substitution to obtain \(ku^8\) earns M1 |
| Obtain \(-\frac{1}{24}(4-3x)^8\) | A1 | Or unsimplified equiv; must be in terms of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain integral of form \(k\ln(4-3x)\) | M1 | Any non-zero constant \(k\); allow M1 if brackets missing; using substitution to obtain \(k\ln u\) earns M1; \(\log(4-3x)\) with base e not specified is M1A0 |
| Obtain \(-\frac{1}{3}\ln(4-3x)\) | A1 | Now with either brackets or modulus signs; must be in terms of \(x\); note that \(-\frac{1}{3}\ln(x-\frac{4}{3})\) and \(-\frac{1}{3}\ln(\frac{4}{3}-x)\) are correct alternatives |
| Include \(+c\) or \(+k\) at least once | B1 | Anywhere in solution to question 1; this mark available even if no other marks earned |
## Question 1:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain integral of form $k(4-3x)^8$ | M1 | Any non-zero constant $k$; using substitution to obtain $ku^8$ earns M1 |
| Obtain $-\frac{1}{24}(4-3x)^8$ | A1 | Or unsimplified equiv; must be in terms of $x$ |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain integral of form $k\ln(4-3x)$ | M1 | Any non-zero constant $k$; allow M1 if brackets missing; using substitution to obtain $k\ln u$ earns M1; $\log(4-3x)$ with base e not specified is M1A0 |
| Obtain $-\frac{1}{3}\ln(4-3x)$ | A1 | Now with either brackets or modulus signs; must be in terms of $x$; note that $-\frac{1}{3}\ln(x-\frac{4}{3})$ and $-\frac{1}{3}\ln(\frac{4}{3}-x)$ are correct alternatives |
| Include $+c$ or $+k$ at least once | B1 | Anywhere in solution to question 1; this mark available even if no other marks earned |
---1 Find\\
(i) $\quad \int ( 4 - 3 x ) ^ { 7 } \mathrm {~d} x$,\\
(ii) $\quad \int ( 4 - 3 x ) ^ { - 1 } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR C3 2013 Q1 [5]}}