Let A_1 be the bounded area enclosed by the curves y=x^2+2,x+y=8 and y-axis that lies in t…

Question

Let ${A}_{1}$ be the bounded area enclosed by the curves $y={x}^{2}+2,x+y=8$ and $y$-axis that lies in the first quadrant. Let ${A}_{2}$ be the bounded area enclosed by the curves $y={x}^{2}+2,{y}^{2}=x,x=2$, and $y$-axis that lies in the first quadrant. Then ${A}_{1}-{A}_{2}$ is equal to

Accepted Answer

Area ${A}_{1}$ enclosed by curves

$y={x}^{2}+2,x+y=8,y$-axis in first quadrant as shown in graph.

Solve $y={x}^{2}+2$ and $x+y=8$ to find point of intersection.

Substituting $y=8-x$ in $y={x}^{2}+2$

$\begin{gathered} 8-x={x}^{2}+2 \ {x}^{2}+x-6=0 \ {x}^{2}+3x-2x-6=0 \ x(x+3)-2(x+3)=0 \ (x+3)(x-2)=0 \ x=-3,2 \end{gathered}$

$x=2,y=8-2=6\Rightarrow (2,6)$

$\begin{gathered} {A}_{1}={\int }_{2}^{0}[(8-x)-({x}^{2}+2)]dx \ ={\int }_{2}^{0}(-{x}^{2}-x+6)dx \ ={[-\frac{{x}^{3}}{3}-\frac{{x}^{2}}{2}+6x]}_{2}^{0} \ =[-\frac{8}{3}-\frac{4}{2}+12-{0-0-0}] \ =\frac{-16-12+72}{6}=\frac{72-28}{6}=\frac{44}{6} \ {A}_{1}=\frac{22}{3} \end{gathered}$

Now, ${A}_{2}$ is area enclosed by curves

$y={x}^{2}+2,{y}^{2}=x,x=2$

Solve $y={x}^{2}+2$ and $y=\sqrt{x}$

$y={y}^{4}+2$

${y}^{4}-y+2=0$

$D=(-1{)}^{2}-4	imes 1	imes 2=1-8=-7$

discriminant (D) is negative. This means curves do not intersect.

$\begin{gathered} {A}_{2}x=2,y={x}^{2}+2y=\sqrt{x} \ {A}_{2}={\int }_{2}^{0}({x}^{2}+2-\sqrt{x})dx \ ={[\frac{{x}^{3}}{3}+2x-\frac{2{x}^{3/2}}{3}]}_{2}^{0} \ =[\frac{8}{3}+2	imes 2-\frac{2	imes 2\sqrt{2}}{3}-{0+0-0}] \ =[\frac{8}{3}+4-\frac{4\sqrt{2}}{3}]=[\frac{20}{3}-\frac{4\sqrt{2}}{3}] \ {A}_{1}-{A}_{2}=\frac{22}{3}-(\frac{20}{3}-\frac{4\sqrt{2}}{3}) \ =\frac{22}{3}-\frac{20}{3}+\frac{4\sqrt{2}}{3}=\frac{2}{3}+\frac{4\sqrt{2}}{3} \ =\frac{2}{3}(2\sqrt{2}+1) \end{gathered}$

Solutions · 1