| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show definite integral equals specific value (algebraic/exponential substitution) |
| Difficulty | Moderate -0.3 This is a straightforward integration by substitution question with a simple linear expression under the square root. The substitution u = 3x - 2 is obvious, the integration of u^(-1/2) is standard, and evaluating definite integral limits is routine. It's slightly easier than average because it's a 'show that' question (answer given) with only one step of integration and simple arithmetic. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
Question 1:
$\int_1^2 (3x-2)^{1/2} \, dx$
M1: $\left[\frac{2}{3}(3x-2)^{1/2}\right]$ with $k = \frac{2}{3}$
M1dep: Substituting limits (dependent on 1st M1)
A1: Correct substitution of upper and lower limits
OR
M1: $u = 3x - 2$, $\frac{du}{dx} = 3$
A1: $\frac{1}{3} \int_1^4 u^{1/2} \, du$
M1dep: $\frac{2}{3}u^{1/2} \times \frac{1}{3}$ or equivalent (dependent on 1st M1)
A1: Correct simplification to $\frac{2}{9}u^{3/2}$
M1dep: Substituting correct limits (dependent on 1st M1)
A1: Answer $= \frac{2}{3}$
NB: AG (Answer Given)
OR
M1: $w^2 = 3x - 2$, $\frac{dw}{dx} = \frac{3}{2w}$
A1: $\frac{2}{3} \int_1^2 w^2 \, dw$
M1dep: Upper minus lower; limits 1 to 4 for $u$ or 1 to 2 for $w$ (dependent on 1st M1)
A1: Substituting back correctly for $x$ and using limits 1 to 2
[5 marks total]1 Show that $\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 3 x - 2 } } \mathrm {~d} x = \frac { 2 } { 3 }$.
\hfill \mbox{\textit{OCR MEI C3 Q1 [5]}}