Question

Area bounded by curves $x=\sqrt{y-1}$
and y=x+1 is -

• A. $\frac{1}{3}sq\cdot unit$
• B. $\frac{8}{3}sq\cdot unit$
• C. $\frac{1}{6}sq\cdot unit$
• D. $5sq.unit$

Answer 1

The correct option is C $\frac{1}{6}sq\cdot unit$

$\text{given:}{eq}^n\text{of curves are:}$

$x=\sqrt{y-1}\to y=x^2+1$

$y=x+1$

$\text{So, the area enclosed :-}$

$A={\int }_0^1(x+1)dx-{\int }_0^1(x^2+1)dx$

$={[\frac{x^2}{2}+x]}_0^1-{(\frac{x^3}{3}+x)}^1$

$=((\frac{1}{2}+1)-0)-((\frac{1}{3}+1)-0)$

$=(\frac{3}{2}-\frac{4}{3})\text{sq. Urits}$

$\text{Ans.}A=\frac{1}{6}sq.\text{units}$