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 $PS=\int _{0}^{x_{0}}[p_{0}-s(x)]dx$

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