Question

The part of the area bounded by the curve $x^2(y-1)=6$ between the ordinates $x=2$ and $x=3$ is 2 sq. units, if the ordinate $x=a$ divides the area into equal parts then $a=$

• A. $\sqrt{6}$
• B. $\sqrt{8}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A $\sqrt{6}$

$x^2(y-1)=6$ $x=2$ and
$x=3$ is 2 sq. units.a divides the area into 2 equal parts
$y=\frac{6}{x^2}+1$
$\stackrel{a}{\underset{2}{\int }}(\frac{6}{x^2}+1)dx=1$
$\stackrel{a}{\underset{2}{\int }}\frac{6}{x^2}dx+\stackrel{a}{\underset{2}{\int }}1dx=1$
$-6/x{|}_2^a+x{|}_2^a=1$
$=-[6/a-3]+[a-2]=1$
$=-6/a+3+a-2=1$
$+6/a=+a$
$a=\sqrt{6}$