WAEC 2017 · Paper 2 · Q11

(a)
Find the values of $x$ at the points of intersection of the curves $y = (2x - 5)^3$ and $y = \dfrac{1}{(2x - 5)^3}$ .
Separate values with commas, e.g. 3, −2
(b)
Use the trapezium rule with ordinates at $x = 1, 2, 3, 4$ and $5$ to calculate, correct to two decimal places, an approximate value of $\displaystyle\int_1^5 \left(x + \frac{8}{x^2}\right)dx$ .

Try it on a graph

Move n to see how the trapezium estimate approaches the exact value 18.4.

Worked solution (try it first)

(a)

The curves meet where $(2x - 5)^3 = \dfrac{1}{(2x - 5)^3}$ .
Multiply both sides by $(2x - 5)^3$ : $(2x - 5)^6 = 1$ .
So $2x - 5 = 1$ or $2x - 5 = -1$ , which gives $x = 3$ or $x = 2$ .

(b)

Four strips from 1 to 5, so $h = 1$ .
Work out $y = x + \dfrac{8}{x^2}$ at each ordinate.
$x = 1$ : $9$ .
$x = 2$ : $4$ .
$x = 3$ : $3.8889$ .
$x = 4$ : $4.5$ .
$x = 5$ : $5.32$ .
$\frac12[(9 + 5.32) + 2(4 + 3.8889 + 4.5)] = \frac12[14.32 + 24.7778]$
$= 19.5489$ , about $19.55$ .

Report a problem with this question