Question

Area bounded by curve $y^2=x$ and x = 4 is divided into 4 equal parts by the lines x = a and y = b then.

• A. Area of each part = $\frac{8}{3}$
• B. $b=0$
• C. $a=\sqrt{2}$
  
• D. $a=(16{)}^{\frac{1}{3}}$

Answer 1

The correct option is D $a=(16{)}^{\frac{1}{3}}$



Total area = $2{\int }_0^4\sqrt{x}dx=\frac{32}{3}$
Area of each part= $\frac{8}{3}$
$A_3=A_4\Rightarrow {\int }_a^4(\sqrt{x}-b)dx={\int }_a^4(b+\sqrt{x})dx=\frac{8}{3}\Rightarrow b=0$
${\int }_a^4\sqrt{x}dx=\frac{8}{3}\Rightarrow a^3=16$