- A$6\,\, sq. \,unit$
- B$5/6\,\, sq. \,unit$
- ✓$1/6\,\, sq. \,unit$
- DNone of these
==> $y = {x^2} + x + c$
==> $y = {x^2} + x$, [ $\because$ $c = 0$ by putting $x = 1, y = 2$)
==> ${\left( {x + \frac{1}{2}} \right)^2} = y + \frac{1}{4}$,
which is a equation of parabola, whose vertices is, $V \left( {\frac{{ - 1}}{2},\,\frac{{ - 1}}{4}} \right)$
$\therefore $ Required area $ = \left. {\left| {\int_{ - 1}^0 {({x^2} + x)\;dx} } \right.} \right|$
$ = \left( {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} \right)_{ - 1}^0$
$\left. { = \left| {\frac{{ - 1}}{3} + \frac{1}{2}} \right.} \right| = \frac{1}{6}\,\, sq. \,unit$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$
$\frac{1}{2}$
$\frac{-1}{2}$
$2$
$2\text{e}^2$