Answer

Solution

1. 1

   Factor: $x^2 - 7x + 10 = (x-2)(x-5)$. On $[0,2]$, both factors satisfy

2. 2

   $x-2 \le 0$ and $x-5 < 0$, so their product is $\ge 0$ on $[0,2]$ (with equality

3. 3

   only at $x=2$). Hence $|x^2 - 7x + 10| = x^2 - 7x + 10$ on $[0,2]$, and

4. 4
   $$\int_0^2 (x^2 - 7x + 10)\,dx = \left[\frac{x^3}{3} - \frac{7x^2}{2} + 10x\right]_0^2 = \frac{8}{3} - 14 + 20 = \frac{8}{3} + 6 = \frac{26}{3}.$$

Answer structure / marking notes

Content note