- 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$
$\left. { = \left| {\frac{{ - 1}}{3} + \frac{1}{2}} \right.} \right| = \frac{1}{6}\,\, sq. \,unit$.
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$STATEMENT-1$ $:\left(\mathrm{p}^2-\mathrm{q}\right)\left(\mathrm{b}^2-\mathrm{ac}\right) \geq 0$ and
$STATEMENT$ $-2: \mathrm{b} \neq \mathrm{pa}$ or $\mathrm{c} \neq \mathrm{qa}$
$ + sin^{-1} \left\{ \,\frac{1}{{\sqrt {13} }}(2\cos x + 3\sin x)\,\,\,\right\} $ w.r.t. at $x = \frac{3}{4}$ is :
$\left(x^2+4\right)^2 d y+\left(2 x^3 y+8 x y-2\right) d x=0 \text {. If } y(0)=0 \text {, }$ then $y(2)$ is equal to