MCQ
Let $y$ be the function which passes through $(1, 2)$ having slope $(2x + 1)$. The area bounded between the curve and $x -$ axis is
- A$6\,\, sq. \,unit$
- B$5/6\,\, sq. \,unit$
- ✓$1/6\,\, sq. \,unit$
- DNone of these
✓
Answer
Correct option: C.
$1/6\,\, sq. \,unit$
c
(c) $\frac{{dy}}{{dx}} = 2x + 1$
(c) $\frac{{dy}}{{dx}} = 2x + 1$
==> $y = {x^2} + x + c$
==> $y = {x^2} + x$, [ $\because$ $c = 0$ 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$
$\left. { = \left| {\frac{{ - 1}}{3} + \frac{1}{2}} \right.} \right| = \frac{1}{6}\,\, sq. \,unit$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = $
→If $\left[\begin{array}{cc}3 c+6 & a-d \\ a+d & 2-3 b\end{array}\right]=\left[\begin{array}{cc}12 & 2 \\ -8 & -4\end{array}\right]$ then find $a b-c d$ :
→The value of $\frac{d}{{d(\ln x)}}({e^x}{\ln ^2}x)$ at $x=e$, is
→The equation $x^2- x - 2 = 0$ in three dimensional space is represented by:
→$\int_0^{b - c} {\,\,f''(x + a)\,dx = } $
→Area bounded by parabola ${y^2} = x$ and straight line $2y = x$ is
→Which of the below given is a vector quantity:
The value of $\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ is:
→A pair of $12 -$ sided fair dice with faces numbered $1,2$ , $3, \ldots, 12$ is rolled. The probability that the sum of the numbers appearing has remainder $2$ when divided by $9$ is
→What is the value of $\int_{0}^{1}\frac{\text{d}}{\text{dx}}\{\sin^{-1}(\frac{2\text{x}}{1+\text{x}^2})\}\text{dx}\ ?$
→