Question

The area bounded by the curves $y=(x-3{)}^2$ and $y=x$ (in sq. units to nearest integer):

• A. $28$
• B. $32$
• C. $4$
• D. $8$

Answer 1

The correct option is D $8$

$po\mathrm{int}of\mathrm{int}er\sec tionofgivencurves$
${(x-3)}^2=x$
$\Rightarrow x^2+9-6x-x=0$
$\Rightarrow x^2-7x+9=0$
$x=\frac{7\pm \sqrt{49-36}}{2}$
$=5.303,1.697$
$\mathrm{Re}quiredArea=\underset{a}{\overset{b}{\int }}\left(uppercurve\right)-\left(lowercurve\right)dx$
$A=\underset{1.697}{\overset{5.303}{\int }}[\left(x\right)-{(x-3)}^2]dx$
$=\underset{1.697}{\overset{5.303}{\int }}[\left(x\right)-(x^2+9-6x)]dx$
$=\underset{1.697}{\overset{5.303}{\int }}[-x^2-9+7x]dx$
$={[\frac{-x^3}{3}+\frac{7x^2}{2}-9x]}_{1.697}^{5.303}$
$=[(\frac{-{\left(5.303\right)}^3}{3}+\frac{7{\left(5.303\right)}^2}{2}-9\left(5.303\right))-(\frac{-{\left(1.697\right)}^3}{3}+\frac{7{\left(1.697\right)}^2}{2}-9\left(1.697\right))]$
$=7.811$