Math 22 The Area of a Region Bounded by Two Graphs

Area of a Region Bounded by Two Graphs

 If $f$ and $g$ are continuous on $[a,b]$ and $g(x)\leq f(x)$ for all $x$ in $[a,b]$, 
 then the area of the region bounded by the graphs of $f,g,x=a,x=b$ given by
 
 $A=\int _{a}^{b}[f(x)-g(x)]dx$

Exercises

1) Find the area of the region bounded by the graph of $f(x)=x^{2}$ and the graph of $g(x)=x^{3}$.

Solution:
Find the bound of the region by setting $f(x)=g(x)$, so $x^{2}=x^{3}$, hence $x^{3}-x^{2}=0$, then $x^{2}(x-1)=0$, therefore $x=0$ and $x=1$
Check which function is the top function by choosing one number in between the bound and plug in:
Pick $x={\frac {1}{2}}$, so $f({\frac {1}{2}})={\frac {1}{4}}$, and $g({\frac {1}{2}})={\frac {1}{8}}$. Therefore, $f(x)$ will be the top function.
${\text{Area}}=\int _{0}^{1}[f(x)-g(x)]dx=\int _{0}^{1}[x^{2}-x^{3}]dx=[{\frac {1}{3}}x^{3}-{\frac {1}{4}}x^{4}]{\Biggr |}_{0}^{1}={\frac {1}{3}}(1)^{3}-{\frac {1}{4}}(1)^{4}={\frac {1}{12}}$

Consumer Surplus and Producer Surplus

 Given the demand function is $d(x)$ and the supply function is $s(x)$. 
 Let $(x_{0},p_{0})$ be the solution of $p(x)=g(x)$.
 
 Then, the Consumer Surplus is $CS=\int _{0}^{x_{0}}[d(x)-p_{0}]dx$
 
 and the Producer Surplus is $PS=\int _{0}^{x_{0}}[p_{0}-s(x)]dx$

Exercises

2) Find the consumer and producer surpluses by using the demand $p=50-0.5x$ and supply functions $p=0.125x$.

Solution:
Find the solution (equilibrium point): $50-0.5x=0.125x$, so $0.625x=50$, so $x=80$, then $p(80)=0.125(80)=50-0.5(80)=10$. Therefor, $x_{0}=80$ and $p_{0}=10$
So $CS=\int _{0}^{80}[50-0.5x-10]dx=\int _{0}^{80}[40-0.5x]dx=[40x-{\frac {1}{4}}x^{2}]{\Biggr |}_{0}^{80}=40(80)-{\frac {1}{4}}(80)^{2}=1600$
and $PS=\int _{0}^{80}[10-0.125x]dx=[10x-{\frac {0.125}{2}}x^{2}]{\Biggr |}_{0}^{80}=10(80)-{\frac {0.125}{2}}(80^{2})=400$

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