WAEC 2010 · Paper 2 · Q11

(a)
Find the area enclosed by the $x$ -axis and the curve $y = 3x^2 + 2x - 1$ .
(b)
If $2x^2 - x - 3 = P(x + Q)^2 + R$ , where $P$ , $Q$ and $R$ are constants, find the: (i) values of $P$ , $Q$ and $R$ ; (ii) minimum value of $2x^2 - x - 3$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Find where the curve meets the $x$ -axis: $3x^2 + 2x - 1 = 0$ , i.e. $(3x - 1)(x + 1) = 0$ , so $x = -1$ or $x = \frac13$ .
Integrate between these limits: $\displaystyle\int_{-1}^{1/3}(3x^2 + 2x - 1)\,dx = \big[x^3 + x^2 - x\big]_{-1}^{1/3}$ .
Upper limit: $\frac1{27} + \frac19 - \frac13 = -\frac5{27}$ .
Lower limit: $-1 + 1 + 1 = 1$ .
Subtract: $-\frac5{27} - 1 = -\frac{32}{27}$ .
The region is below the $x$ -axis, so the area is $\dfrac{32}{27}$ square units.

Take out 2 from the $x$ terms: $2x^2 - x - 3 = 2\left(x^2 - \frac12x\right) - 3$ .
Complete the square: $2\left[\left(x - \frac14\right)^2 - \frac1{16}\right] - 3 = 2\left(x - \frac14\right)^2 - \frac{25}{8}$ .
Compare: $P = 2$ , $Q = -\frac14$ and $R = -\frac{25}{8}$ .

The square is never negative, so the least value is when $x = \frac14$ : minimum $= -\dfrac{25}{8}$ .