WAEC 2011 · Paper 2 · Q9

(a)
Use the trapezium rule with ordinates at $x = 1, 2, 3, 4$ and $5$ to calculate, correct to two decimal places, an approximate value for $\displaystyle\int_1^5 (2x + 8x^{-2})\,dx$ .
(b)
Given that $f: x \to x^2 + 1$ and $g: x \to \dfrac{3x + 1}{x - 1}$ , $x \ne 1$ , find: (i) $g\circ f$ ; (ii) $g\circ f(2)$ .

Worked solution (try it first)

(a)

Four strips from 1 to 5, so $h = 1$ .
Work out $y = 2x + \dfrac{8}{x^2}$ at each ordinate.
$x = 1$ : $10$ .
$x = 2$ : $6$ .
$x = 3$ : $6 + 0.8889 = 6.8889$ .
$x = 4$ : $8.5$ .
$x = 5$ : $10.32$ .
Trapezium rule: $\frac12h[(\text{first} + \text{last}) + 2(\text{the rest})]$ .
$\frac12[(10 + 10.32) + 2(6 + 6.8889 + 8.5)] = \frac12[20.32 + 42.7778]$
$= 31.5489$ .
So the integral is about $31.55$ .

(b)(i)

$f$ acts first: $g \circ f(x) = g(x^2 + 1)$
$= \dfrac{3(x^2 + 1) + 1}{(x^2 + 1) - 1}$ .
Simplify the top and bottom: $\dfrac{3x^2 + 4}{x^2}$ .
Divide each term by $x^2$ : $g \circ f(x) = 3 + \dfrac{4}{x^2}$ .

(ii)

$g \circ f(2) = 3 + \dfrac44 = 4$ .
Check: $f(2) = 5$ and $g(5) = \dfrac{16}{4} = 4$ ✓.

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