Question

A curve ${}^{\prime }{C}^{\prime }$ passes through $(2,0)$ and the slope at $(x,y)$ is $\frac{(x+1{)}^2+(y-3)}{x+1}.$ The area(in $\text{sq.units}$) bounded by curve and $x-$axis in fourth quadrant is:

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{8}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{4}{3}$

According to given data, slope of curve $C$ at $(x,y)$ is $\frac{dy}{dx}=\frac{(x+1{)}^2+(y-3)}{x+1}$
$x\ge 0$ in $I{V}_{th}$ quadrent.Now
$\frac{dy}{dx}=(x+1)+\frac{y-3}{x+1}$
$\Rightarrow \frac{dy}{dx}-\left(\frac{1}{x+1}\right)y=x+1-\frac{3}{x+1}$
$I.F.=e^{\int \frac{-1}{x+1}dx}=e^{-\ln (x+1)}=\frac{1}{x+1}$
$\therefore$ Solution is $y\frac{1}{x+1}=\int [1-\frac{3}{(x+1{)}^2}]dx$
$\frac{y}{x+1}=x+\frac{3}{x+1}+C$
$y=x(x+1)+3+C(x+1)\cdots \left(1\right)$
As the curve passes throgh $(2,0)$
$\therefore 0=2\cdot 3+3+C\cdot 3$
$\Rightarrow C=-3$
$\therefore$ Equation $\left(1\right)$ becomes
$y=x(x+1)+3-3x-3$
$y=x^2-2x\cdots \left(2\right)$
which is the required equation of curve.
This can be written as $(x-1{)}^2=(y+1)$
It is an upward parabola with vertex at $(1,-1)$ meeting $x-$axis at $(0,0)$ and $(2,0)$
Area bounded by curve and $x-$axis in fourth quadrant is as shown in the shaded region in figure and is given by:
$A=\left|\underset{0}{\overset{2}{\int }}ydx\right|=\left|\underset{0}{\overset{2}{\int }}(x^2-2x)dx\right|={[\frac{x^3}{3}-x^2]}_0^2$
$\Rightarrow A=|\frac{8}{3}-4|=\frac{4}{3}$ sq. units