Use the trapezoidal rule to find an estimate for the area.

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.

Use all the coordinates in the table to find the equation of the least squares cubic regression curve.

Write down the coefficient of determination.

Write down an expression for the area enclosed by the cubic function, the $x$-axis, the $y$-axis and the line $x=4.4$.

Find the value of this area.

Show that $\text{ln}y=qx+\text{ln}p$.

Hence explain how a straight line graph could be drawn using the coordinates in the table.

By finding the equation of a suitable regression line, show that $p=1.83$ and $q=0.986$.

Hence find the area enclosed by the exponential function, the $x$-axis, the $y$-axis and the line $x=4.4$.

Area $=\frac{1.1}{2}\left(2+2\left(5+15+47\right)+148\right)$         M1A1

Area = 156 units²          A1

[3 marks]

The graph is concave up,         R1

so the trapezoidal rule will give an overestimate.         A1

[2 marks]

$f\left(x\right)=3.88x^3-12.8x^2+14.1x+1.54$         M1A2

[3 marks]

$R^2=0.999$        A1

[1 mark]

Area $=\underset{0}{\overset{4.4}{\int }}\left(3.88x^3-12.8x^2+14.1x+1.54\right)dx$        A1A1

[2 marks]

Area = 145 units²    (Condone 143–145 units², using rounded values.)      A2

[2 marks]

$\text{ln}y=\text{ln}\left(p{\text{e}}^{qx}\right)$      M1

$\text{ln}y=\text{ln}p+\text{ln}\left({\text{e}}^{qx}\right)$      A1

$\text{ln}y=qx+\text{ln}p$      AG

[2 marks]

Plot $\text{ln}y$ against $p$.      R1

[1 mark]

Regression line is $\text{ln}y=0.986x+0.602$       M1A1

So $q=$ gradient = 0.986    R1

$p=e^{0.602}=1.83$       M1A1

[5 marks]

Area $=\underset{0}{\overset{4.4}{\int }}1.83e^{0.986x}dx=140$ units²     M1A1

[2 marks]