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

MATH103 April 2010

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[hide] Question 02
Calculate the area of the bounded region enclosed by the curves $f(x)=x^{3}-4x\quad {\text{ and }}\quad g(x)=-x^{2}+2x$ for x ≥ 0.

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[show] Hint
Find the intersection points of the two functions.

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[show] Solution
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.