Question

A curve passes through $(2,0)$ and slope at point $P(x,y)$ is $\frac{(x+1{)}^2+(y-3)}{(x+1)}$. The area between curve and $x$-axis in 4th quadrant is

• A. $2/3$
• B. $1/3$
• C. 2
• D. $4/3$

Answer 1

Answer: (D)

Here, slope of tangent
$\frac{dy}{dx}=\frac{(x+1{)}^2+y-3}{(x+1)}$
$\Rightarrow \frac{dy}{dx}=(x+1)+\frac{(y-3)}{(x+1)}$,
put $x+1=X$ and $y-3=Y\left(\text{ here }\frac{dy}{dx}=\frac{dY}{dX}\right)$
$\therefore \frac{dY}{dX}=X+\frac{Y}{X}\Rightarrow \frac{dY}{dX}-\frac{1}{X}Y=X$
where integrating factor
$=e^{\int -\frac{1}{x}dX}=e^{-\ell nX}=\frac{1}{X}$
$\therefore$ Solution is, $Y\cdot \frac{1}{X}=\int X\cdot \frac{1}{X}dX+c\Rightarrow \frac{Y}{X}=X+c$
$y-3=(x+1{)}^2+c(x+1)$, which passes through $(2,0)$

$-3=9+3c$
$\Rightarrow c=-4$
$\therefore$ Required curve
$\therefore y=(x+1{)}^2-4(x+1)+3$
$\Rightarrow y=x^2-2x$
Drawing curve
Thus, required area
$=\left|\underset{0}{\overset{2}{\int }}\left(x^2-2x\right)dx\right|=\left|{\left(\frac{x^3}{3}-x^2\right)}_0^2\right|=\frac{4}{3}$ sq. units