Question 9:
Diagram 6 shows the shaded region bounded by the curves y = x3 – 6x2 + 12x – 6, y = g(x) and a straight line x = 2. The curves intersect at point A.
$$ \text { It is given that the area under the curve } y=g(x) \text { is represented }\left[-x^3+4 x^2-4 x\right]_1^2 \text {. } $$
Find
(a) the equation of g(x), [1 mark]
(b) the area of the shaded region. [4 marks]
Answer:
(a) $$ g(x)=-3 x^2+8 x-4 $$
(b) $$ \begin{aligned} &\text { Shaded area }\\ &\begin{aligned} & =\int_1^2\left(x^3-6 x^2+12 x-6\right) \mathrm{d} x-\left[-x^3+4 x^2-4 x\right]_1^2 \\ & =\left[\frac{x^4}{4}-\frac{6 x^3}{3}+\frac{12 x^2}{2}-6 x+x^3-4 x^2+4 x\right]_1^2 \\ & =\left[\frac{x^4}{4}-x^3+2 x^2-2 x\right]_1^2 \\ & =\left[\frac{2^4}{4}-2^3+2(2)^2-2(2)\right]-\left[\frac{1}{4}-1^3+2(1)^2-2(1)\right] \\ & =\left(\frac{16}{4}-8+8-4\right)-\left(\frac{1}{4}-1+2-2\right) \\ & =0-\left(-\frac{3}{4}\right) \\ & =\frac{3}{4} \text { unit }^2 \end{aligned} \end{aligned} $$
Diagram 6 shows the shaded region bounded by the curves y = x3 – 6x2 + 12x – 6, y = g(x) and a straight line x = 2. The curves intersect at point A.
$$ \text { It is given that the area under the curve } y=g(x) \text { is represented }\left[-x^3+4 x^2-4 x\right]_1^2 \text {. } $$
Find
(a) the equation of g(x), [1 mark]
(b) the area of the shaded region. [4 marks]
Answer:
(a) $$ g(x)=-3 x^2+8 x-4 $$
(b) $$ \begin{aligned} &\text { Shaded area }\\ &\begin{aligned} & =\int_1^2\left(x^3-6 x^2+12 x-6\right) \mathrm{d} x-\left[-x^3+4 x^2-4 x\right]_1^2 \\ & =\left[\frac{x^4}{4}-\frac{6 x^3}{3}+\frac{12 x^2}{2}-6 x+x^3-4 x^2+4 x\right]_1^2 \\ & =\left[\frac{x^4}{4}-x^3+2 x^2-2 x\right]_1^2 \\ & =\left[\frac{2^4}{4}-2^3+2(2)^2-2(2)\right]-\left[\frac{1}{4}-1^3+2(1)^2-2(1)\right] \\ & =\left(\frac{16}{4}-8+8-4\right)-\left(\frac{1}{4}-1+2-2\right) \\ & =0-\left(-\frac{3}{4}\right) \\ & =\frac{3}{4} \text { unit }^2 \end{aligned} \end{aligned} $$