Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)/Solution 1

Since the two curves $y_{1} = f (x) = 4 - (x - 1)^{2}$ and $y_{2} = g (x) = x + 1$ intersects when

\begin{aligned} 0 & = f (x) - g (x) \\ = 4 - (x - 1)^{2} - (x + 1) \\ = 2 + x - x^{2} \\ = - (x - 2) (x + 1) \end{aligned}

Thus, the curves intersect at x=-1 and x=2. Moreover, from the sketch, we can see that f(x) > g(x) in the interval [-1, 2].

Therefore, the area A bounded by the curves is given by:

\begin{aligned} A & = \int_{- 1}^{2} f (x) - g (x) d x \\ = \int_{- 1}^{2} (2 + x - x^{2}) d x \\ = {[2 x + x^{2} / 2 - x^{3} / 3]}_{- 1}^{2} \\ = (4 + 2 - 8 / 3) - (- 2 + 1 / 2 + 1 / 3) \\ = 9 / 2 \end{aligned}