Question

Find the area of the closed figure bounded by the following curve
$y$ = $\frac{1}{3}{x}^2$ - 2$x$ + $4$, $y=10-x$. Make a drawing.

Answer 1

$\Rightarrow y=\frac{1}{3}{x}^2-2x+4$, $y=10-x$

$\Rightarrow 3y=x^2-6x+12$ , $y=10-x$

$\Rightarrow$ $3(y-1)=(x-3{)}^2$

$\Rightarrow (y-1)=\frac{1}{3}(x-3{)}^2$

Intersection between curves

putting $y=10-x$ in $3(y-1)=\frac{1}{3}(x-3{)}^2$

$\left(1\right)$ $y=4$ when $x=6$ and

$x=-3$ , $y=13$

${\int }_{-3}^6(10-x)dx+{\int }_{-3}^6$$\left[\frac{1}{3}\right(x-3{)}^2+1]dx$

$\Rightarrow$ $10\times 9-\frac{1}{2}(36-9)+\frac{1}{9}{\left[\right(x-3{)}^3]}_{-3}^6+[x{]}_{-3}^6$

$\Rightarrow$ $90-\frac{27}{2}+\frac{1}{9}[27+6^3]+9$

$\Rightarrow$ $99-\frac{27}{2}+27$

$\Rightarrow$ $112.5$