2010 H2 Mathematics Paper 1 Question 6

Definite Integrals: Areas and Volumes

Answers

{\beta = 0.347, \; \gamma = 1.532}

{0.781 \textrm{ units}^2}

{\frac{9}{4} \textrm{ units}^2}

{- 1 < k < 3}

Using a GC,

\begin{align*} \beta &= 0.34729 \\ &= 0.347 \textrm{ (3 dp)} \; \blacksquare \\ \gamma &= 1.5321 \\ &= 1.532 \textrm{ (3 dp)} \; \blacksquare \\ \end{align*}

\begin{align*} & \textrm{Area of region} \\ & = - \int_{0.34729}^{1.5321} \left( x^3 - 3 x + 1 \right) \; \mathrm{d}x \\ & = 0.781 \textrm{ units}^2 \textrm{ (3 sf)} \; \blacksquare \end{align*}

At intersection between curve and the line

{y=1,}

\begin{gather*} x^3 - 3 x + 1 = 1 \\ x^3 - 3x = 0 \\ x(x^2 - 3) = 0 \\ x=0 \textrm{ or } x = \pm \sqrt{3} \end{gather*}

\begin{align*} & \textrm{Area of shaded region} \\ & = \int_{-\sqrt{3}}^0 \left( x^3 - 3 x + 1 - 1 \right) \; \mathrm{d}x \\ & = \left[ \frac{1}{3} x^3 - 3 x \right]_{-\sqrt{3}}^0 \\ & = 0 - \frac{1}{4}(9) + \frac{3}{2}(3) \\ &= \frac{9}{4} \textrm{ units}^2 \; \blacksquare \end{align*}

Using a GC, the turning points of the curve are

{\left( - 1, 3 \right)}

and

{\left( 1, - 1 \right)}

For the equation to have three real distinct roots, the line

{y=k}

must cut the curve at three distinct points

From the graph,

- 1 < k < 3