2014 H2 Mathematics Paper 1 Question 7

Definite Integrals: Areas and Volumes

Answers

{\alpha = 1.885}

{\beta = \sqrt{3}}

\int_\beta^\alpha f(x) \; \mathrm{d}x =  -0.597

{\frac{54}{35} \sqrt{3} \textrm{ units}^2}

The six roots of

{f(x)=0}

consists of two real roots,

{x=\alpha}

and

{x=-\alpha}

and four complex roots which made up of two complex conjugate pairs of the form

{x=a+b\mathrm{i}, }

{x=a-b\mathrm{i}, }

{x=-a-b\mathrm{i}, }

and

{x=-a+b\mathrm{i} }

where

{a,b \in \mathbb{R}}

Using a GC,

\begin{align*} \alpha &= 1.8852 \\ &= 1.885 \textrm{ (3 dp)} \; \blacksquare \\ \end{align*}

For

{\beta,}

\begin{gather*} x^6 - 3 x^4 - 7 = -7 \\ x^6 - 3x^4 = 0 \\ x^4 (x^ - 3 ) = 0 \\ x = 0 \textrm{ or } x = \pm \sqrt{3} \end{gather*}

Since

{\beta > 0,}

\beta = \sqrt{3} \; \blacksquare

\int_\beta^\alpha f(x) \; \mathrm{d}x = -0.597 \textrm{ (3 dp)} \; \blacksquare

\begin{align*} & \textrm{Area of region} \\ & = \int_0^{\sqrt{3}} -7 - (x^6 - 3x^4 - 7) \; \mathrm{d}x \\ & = \int_0^{\sqrt{3}} ( - x^6 + 3x^4 ) \; \mathrm{d}x \\ & = \left[ -\frac{1}{7} x^7 + \frac{3}{5} x^5 \right]_0^{\sqrt{3}} \\ & = -\frac{1}{7} \left( \sqrt{3} \right)^7 + \frac{3}{5} \left( \sqrt{3} \right)^5 \\ & = - \frac{1}{7} (27 \sqrt{3}) + \frac{3}{5} (9 \sqrt{3}) \\ & = \frac{54}{35} \sqrt{3} \textrm{ units}^2 \; \blacksquare \end{align*}

\begin{align*} \textrm{RHS} &= f(-x) \\ &= (-x)^6 - 3(-x)^4 - 7 \\ &= x^6 - 3 x^4 - 7 \\ & = f(x) \\ &= \textrm{LHS} \; \blacksquare \end{align*}

From the graph, the positive real root to the equation

{f(x)=0}

{x = \alpha.}

Since

{f(x)=f(-x), }

{x=-\alpha}

is also a root.

The remaining four roots of the equation are complex.
Let

{x = a+b\mathrm{i}}

be a complex root of

{f(x)=0}

where

{a,b \in \mathbb{R}}

Since

{f(x)=f(-x), }

{x = -a - b\mathrm{i}}

is also a root.

Since all the coefficients are real, by the conjugate root theorem, the complex conjugates

{(a+b\mathrm{i})^* = a-b\mathrm{i}}

and

{(-a-b\mathrm{i})^* = -a+b\mathrm{i}}

are also roots

Hence the six roots of

{f(x)=0}

consists of two real roots,

{x=\alpha}

and

{x=-\alpha}

and four complex roots which made up of two complex conjugate pairs of the form

{x=a+b\mathrm{i}, }

{x=a-b\mathrm{i}, }

{x=-a-b\mathrm{i}, }

and

{x=-a+b\mathrm{i}. }