91Ó°ÊÓ

The two given functions are $f\left(x\right)=4-x^2,g\left(x\right)=1-2x$and the given interval $[-4,4].$

The area $A$ enclosed by these curves on the given interval $[0,\mathrm{Ï€}]$is

The graphs of functions $f\left(x\right)=4-x^{2}andg\left(x\right)=1-2x$are shown in Figure-1 here.

The curves intersect at points $A(-1,3)andF(3,-5).$

The area enclosed by the curves of functions $f\left(x\right)=4-x^{2}andg\left(x\right)=1-2x$is

$AreaAGEFCA=AreaBAGECB+AreaCDFC-AreaBACB-AreaEFDE$ ....(1)

Now, $AreaAGEFCA=AreaBAGECB+AreaCDFC-AreaBACB-AreaEFDE$

$$\begin{gathered} AreaAGEFCA=\left|{∫}_{-1}^{2}f\left(x\right)dx\right|+\left|{∫}_{-1}^{\frac{1}{2}}g\left(x\right)dx\right|-\left|{∫}_{\frac{1}{2}}^{3}g\left(x\right)dx\right|-\left|{∫}_2^{3}f\left(x\right)dx\right| \\ AreaAGEFCA=\left|{∫}_{-1}^2(4-x^2)dx\right|+\left|{∫}_{-1}^{\frac{1}{2}}(1-2x)dx\right|-\left|{∫}_{\frac{1}{2}}^3(1-2x)dx\right|-\left|{∫}_2^3(4-x^2)dx\right| \end{gathered}$$

Now, using graphing calculator

$$\begin{gathered} \left|{∫}_{-1}^2(4-x^2)dx\right|=\left|9\right|=9,\left|{∫}_2^3(4-x^2)dx\right|=\left|-2.33\right|=2.33, \\ \left|{∫}_{-1}^{\frac{1}{2}}(1-2x)dx\right|=\left|2.25\right|=2.25,\left|{∫}_{\frac{1}{2}}^3(1-2x)dx\right|=\left|-6.25\right|=6.25 \end{gathered}$$

Thus,

localid="1661249443514" $$\begin{gathered} AreaAGEFCA=\left|{∫}_{-1}^2(4-x^2)dx\right|+\left|{∫}_{-1}^{\frac{1}{2}}(1-2x)dx\right|-\left|{∫}_{\frac{1}{2}}^3(1-2x)dx\right|-\left|{∫}_2^3(4-x^2)dx\right| \\ =9.00+2.25-2.33-6.25 \\ =2.67 \end{gathered}$$