Standard +0.8 This question requires students to set up two integrals (one for area under the curve, one for area to the left), recognize that 'area to left' means integrating x as a function of y (requiring rearrangement to x = y^(1/5)), and then prove a specific numerical relationship. The conceptual leap of integrating with respect to y and the algebraic manipulation to show the ratio equals exactly 5 elevates this above a routine area calculation.
3 In this question you must show detailed reasoning.
The diagram shows the curve with equation \(y = x ^ { 5 }\) and the square \(O A B C\) where the points \(A , B\) and \(C\) have coordinates \(( 1,0 ) , ( 1,1 )\) and \(( 0,1 )\) respectively.
The curve cuts the square into two parts.
\includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-04_658_780_1318_230}
Show that the relationship between the areas of the two parts of the square is
\(\frac { \text { Area to left of curve } } { \text { Area below curve } } = 5\).
Correct expression for area below curve OR area to left of curve
M1
Condone missing dx; must see correct limits \(\int_0^1\); reversed limits score M0 A0
Answer \(\frac{1}{6}\)
A1
Correct expression and answer
Finding other area as \(1 - \) first area
M1
Even if answer is negative
Fraction with both \(\frac{5}{6}\) and \(\frac{1}{6}\), answer 5
B1
Dependent on other 3 marks; words not required if intention clear; M0A0M1B0 possible
## Question 3:
| Correct expression for area below curve OR area to left of curve | M1 | Condone missing dx; must see correct limits $\int_0^1$; reversed limits score M0 A0 |
| Answer $\frac{1}{6}$ | A1 | Correct expression and answer |
| Finding other area as $1 - $ first area | M1 | Even if answer is negative |
| Fraction with both $\frac{5}{6}$ and $\frac{1}{6}$, answer 5 | B1 | Dependent on other 3 marks; words not required if intention clear; M0A0M1B0 possible |
3 In this question you must show detailed reasoning.
The diagram shows the curve with equation $y = x ^ { 5 }$ and the square $O A B C$ where the points $A , B$ and $C$ have coordinates $( 1,0 ) , ( 1,1 )$ and $( 0,1 )$ respectively.
The curve cuts the square into two parts.\\
\includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-04_658_780_1318_230}
Show that the relationship between the areas of the two parts of the square is\\
$\frac { \text { Area to left of curve } } { \text { Area below curve } } = 5$.
\hfill \mbox{\textit{OCR MEI Paper 3 2024 Q3 [4]}}