# Difference between revisions of "Math 22 The Area of a Region Bounded by Two Graphs"

## 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.
${\displaystyle {\text{Area}}=\int _{0}^{1}[f(x)-g(x)]dx=\int _{0}^{1}[x^{2}-x^{3}]dx=[{\frac {1}{3}}x^{3}-{\frac {1}{4}}x^{4}]{\Biggr |}_{0}^{1}={\frac {1}{3}}(1)^{3}-{\frac {1}{4}}(1)^{4}={\frac {1}{12}}}$