Question
This question explores methods to determine the area bounded by an unknown curve.
The curve $y=f(x)$ is shown in the graph, for $0 \leqslant x \leqslant 4.4$.
The curve y=f(x) passes through the following points.
It is required to find the area bounded by the curve, the $x$-axis, the $y$-axis and the line $x=4.4$.
One possible model for the curve $y=f(x)$ is a cubic function.
A second possible model for the curve $y=f(x)$ is an exponential function, $y=p \mathrm{e}^{q x}$, where $p, q \in \mathbb{R}$.
▶️Answer/Explanation
a.i. Use the trapezoidal rule to find an estimate for the area.
a.ii.With reference to the shape of the graph, explain whether your answer to part (a)(i) will be an over-estimate or an underestimate of the area.
b.i.Use all the coordinates in the table to find the equation of the least squares cubic regression curve.
b.ii.Write down the coefficient of determination.
c.i. Write down an expression for the area enclosed by the cubic function, the $x$-axis, the $y$-axis and the line $x=4.4$.
c.ii.Find the value of this area.
d.i.Show that $\ln y=q x+\ln p$.
d.iiHence explain how a straight line graph could be drawn using the coordinates in the table.
d.iiiBy finding the equation of a suitable regression line, show that $p=1.83$ and $q=0.986$.
d.ivHence find the area enclosed by the exponential function, the $x$-axis, the $y$-axis and the line $x=4.4$.a.i. Area $=\frac{1.1}{2}(2+2(5+15+47)+148) \quad$ M1A1
Area $=156$ units $^2$
A1
[3 marks]
a.ii.The graph is concave up,
R1
so the trapezoidal rule will give an overestimate.
A1
[2 marks]
b.i. $f(x)=3.88 x^3-12.8 x^2+14.1 x+1.54 \quad$ M1A2
[3 marks]
b.ii. $R^2=0.999$
A1
[1 mark]
c.i. Area $=\int_0^{4.4}\left(3.88 x^3-12.8 x^2+14.1 x+1.54\right) d x \quad$ A1A1
[2 marks]
c.ii.Area $=145$ units $^2 \quad$ (Condone 143-145 units ${ }^2$, using rounded values.)
A2
[2 marks]
d.i. $\ln y=\ln \left(p \mathrm{e}^{q x}\right) \quad$ M1
$\ln y=\ln p+\ln \left(\mathrm{e}^{q x}\right) \quad \boldsymbol{A 1}$
$\ln y=q x+\ln p \quad$ AG
[2 marks]
d.iiPlot $\ln y$ against $p . \quad \boldsymbol{R} 1$
[1 mark]
d.iiiRegression line is $\ln y=0.986 x+0.602 \quad$ M1A1
So $q=$ gradient $=0.986 \quad \boldsymbol{R 1}$
$p=e^{0.602}=1.83 \quad$ M1A1
[5 marks]
M1A1