Science:Math Exam Resources/Courses/MATH103/April 2010/Question 02/Solution 1

To find the limits of the integral that calculates the area, we need to find the intersection points of the two functions $f$ and $g$ and so we set the functions equal.

$f(x)\ =g(x)$

That means

$\ x^{3}-4x=-x^{2}+2x$

and leads to

$\ x(x+3)(x-2)=0$

So, we find that the functions intersect three times at $x=-3,\ x=0,\ x=2$. However, we have a constraint that $x\geq {0}$ and so we will only be concerned with the interval [0,2]. We need to integrate the function $f-g$ or $g-f$ (depending on which is the upper and which is the lower curve).

Now we check which function is the upper function. One way to do this is to simply plug in any point in the interval and compare the function values. On [0,2] we choose x = 1 and get f(1) = -3, g(1)=1. Hence g > f on [0,2].

Therefore, the area A is given by

$A=\int _{0}^{2}(g(x)-f(x))dx.$

${\begin{aligned}\ A&=\int _{0}^{2}(g(x)-f(x))dx\\&=\int _{0}^{2}(-x^{2}+2x-x^{3}+4x)dx\\&=\left[-{\frac {x^{3}}{3}}+6{\frac {x^{2}}{2}}-{\frac {x^{4}}{4}}\right]_{0}^{2}\\&=-{\frac {2^{3}}{3}}+6{\frac {2^{2}}{2}}-{\frac {2^{4}}{4}}-0\\&={\frac {16}{3}}\end{aligned}}$

Therefore the area of A is 16/3.