Challenging +1.2 This question requires understanding that the mid-ordinate rule uses the rectangle with height f((a+b)/2), and comparing this to the trapezium area formed by the tangent line. Students must recognize that for a concave function, the tangent lies above the curve, making the mid-ordinate an overestimate. While it requires geometric insight about tangent lines and numerical integration methods, it's a conceptual reasoning question without calculation, making it moderately above average difficulty for Further Maths.
6 The diagram below shows part of the graph of \(y = \mathrm { f } ( x )\)
The line \(T P Q\) is a tangent to the graph of \(y = \mathrm { f } ( x )\) at the point \(P \left( \frac { a + b } { 2 } , \mathrm { f } \left( \frac { a + b } { 2 } \right) \right)\)
The points \(S ( a , 0 )\) and \(T\) lie on the line \(x = a\)
The points \(Q\) and \(R ( b , 0 )\) lie on the line \(x = b\)
\includegraphics[max width=\textwidth, alt={}, center]{74b8239a-1f46-45e7-ad20-2dce7bf4baf6-05_748_696_669_671}
Sharon uses the mid-ordinate rule with one strip to estimate the value of the integral \(\int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { d } x\)
By considering the area of the trapezium QRST, state, giving reasons, whether you would expect Sharon's estimate to be an under-estimate or an over-estimate.
The area of the trapezium is greater than the integral
B1
Condone "the area of the trapezium is greater than the curve"
Area of trapezium \(= (b-a)y_{\frac{1}{2}}\); the area of the trapezium is the same as the area of the rectangle from use of the mid-ordinate rule, which equals Sharon's estimate
E1
Must explain that trapezium area equals Sharon's result via mid-ordinate rule
The trapezium includes the area represented by the integral, so Sharon's estimate is an over-estimate
R1
Must state over-estimate with reasoned argument
## Question 6:
The area of the trapezium is greater than the integral | B1 | Condone "the area of the trapezium is greater than the curve"
Area of trapezium $= (b-a)y_{\frac{1}{2}}$; the area of the trapezium is the same as the area of the rectangle from use of the mid-ordinate rule, which equals Sharon's estimate | E1 | Must explain that trapezium area equals Sharon's result via mid-ordinate rule
The trapezium includes the area represented by the integral, so Sharon's estimate is an over-estimate | R1 | Must state over-estimate with reasoned argument
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6 The diagram below shows part of the graph of $y = \mathrm { f } ( x )$
The line $T P Q$ is a tangent to the graph of $y = \mathrm { f } ( x )$ at the point $P \left( \frac { a + b } { 2 } , \mathrm { f } \left( \frac { a + b } { 2 } \right) \right)$\\
The points $S ( a , 0 )$ and $T$ lie on the line $x = a$\\
The points $Q$ and $R ( b , 0 )$ lie on the line $x = b$\\
\includegraphics[max width=\textwidth, alt={}, center]{74b8239a-1f46-45e7-ad20-2dce7bf4baf6-05_748_696_669_671}
Sharon uses the mid-ordinate rule with one strip to estimate the value of the integral $\int _ { a } ^ { b } \mathrm { f } ( x ) \mathrm { d } x$
By considering the area of the trapezium QRST, state, giving reasons, whether you would expect Sharon's estimate to be an under-estimate or an over-estimate.\\
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q6 [3]}}