Solution — 2025 Question 2 Graphs| Integration & Applications

(a)

(b)

$\begin{matrix} x + y & = 6 \\ y & = 6 - x & (1) \end{matrix}$

[making y the subject for the line equation]

Substituting $(1)$ into the equation of the curve,

$\begin{matrix} 6 - x = x^{2} + x + 3 \\ x^{2} + 2 x - 3 = 0 \\ (x + 3) (x - 1) = 0 \end{matrix}$

[setting up quadratic equation in x]

$\begin{matrix} x = - 3 or x = 1 \end{matrix}$

[x coordinates of the points of intersection]

Substituting back to $(1),$

$\begin{matrix} y & = 6 - (- 3) & or & y & = 6 - 1 \\ = 9 & = 5 \end{matrix}$

[y coordinates of the points of intersection]

Hence the coordinates of the points are

$\begin{matrix} (- 3, 9) ∎ and (1, 5) ∎ \end{matrix}$

[coordinates of the points of intersection]

(c)

Graph of area between curve and line

$\begin{matrix} Area between C and line \\ = Area of trapezium - \int_{- 3}^{1} (x^{2} + x + 3) d x \\ = \frac{1}{2} (1 - (- 3)) (9 + 5) - \int_{- 3}^{1} (x^{2} + x + 3) d x \\ = 28 - {[\frac{x^{3}}{3} + \frac{x^{2}}{2} + 3 x]}_{- 3}^{1} \\ = 28 - (\frac{1^{3}}{3} + \frac{1^{2}}{2} + 3 (1) - (\frac{{(- 3)}^{3}}{3} + \frac{{(- 3)}^{2}}{2} + 3 (- 3))) \\ = \frac{32}{3} ∎ \end{matrix}$

[integration to find area]