Since the two curves y 1 = f ( x ) = 4 − ( x − 1 ) 2 {\displaystyle y_{1}=f(x)=4-(x-1)^{2}} and y 2 = g ( x ) = x + 1 {\displaystyle y_{2}=g(x)=x+1} intersects when
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:
and 
intersects when
![{\displaystyle {\begin{aligned}A&=\int _{-1}^{2}f(x)-g(x)dx\\&=\int _{-1}^{2}(2+x-x^{2})dx\\&=\left[2x+x^{2}/2-x^{3}/3\right]_{-1}^{2}\\&=\left(4+2-8/3\right)-\left(-2+1/2+1/3\right)\\&=9/2\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9ab54dd6d6d2d4c4405d19a2496c8a9a656d2b36)
.png)