# Math 22 The Area of a Region Bounded by Two Graphs

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## Area of a Region Bounded by Two Graphs

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

${\displaystyle A=\int _{a}^{b}[f(x)-g(x)]dx}$


Exercises

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

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

## Consumer Surplus and Producer Surplus

 Given the demand function is ${\displaystyle d(x)}$ and the supply function is ${\displaystyle s(x)}$.
Let ${\displaystyle (x_{0},p_{0})}$ be the solution of ${\displaystyle p(x)=g(x)}$.

Then, the Consumer Surplus is ${\displaystyle CS=\int _{0}^{x_{0}}[d(x)-p_{0}]dx}$

and the Producer Surplus is ${\displaystyle PS=\int _{0}^{x_{0}}[p_{0}-s(x)]dx}$


Exercises

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

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