Area Question 33
Question 33
- A curve passes through $(2,0)$ and the slope of tangent at point $P(x, y)$ equals $\frac{(x+1)^{2}+y-3}{(x+1)}$.
Find the equation of the curve and area enclosed by the curve and the $X$-axis in the fourth quadrant. (2004, 5M)
Show Answer
Solution:
- Here, slope of tangent,
$$ \begin{aligned} \quad \frac{d y}{d x} & =\frac{(x+1)^{2}+y-3}{(x+1)} \ \Rightarrow \quad \frac{d y}{d x} & =(x+1)+\frac{(y-3)}{(x+1)}, \end{aligned} $$
Put $x+1=X$ and $y-3=Y$
$$ \begin{array}{llrl} & & \frac{d y}{d x} & =\frac{d Y}{d X} \ & & \frac{d Y}{d X} & =X+\frac{Y}{X} \ & & \frac{d Y}{d X} & =\frac{1}{X} Y=X \ & \text { IF }=e^{\int-\frac{1}{X} d X} & =e^{-\log X}=\frac{1}{X} \end{array} $$
$\therefore$ Solution is, $Y \cdot \frac{1}{X}=\int X \cdot \frac{1}{X} d X+c$
$\Rightarrow \quad \frac{Y}{X}=X+c$
$y-3=(x+1)^{2}+c(x+1)$, which passes through $(2,0)$.
$\Rightarrow \quad-3=(3)^{2}+3 c$
$\Rightarrow \quad c=-4$
$\therefore$ Required curve
$$ \begin{aligned} & & y & =(x+1)^{2}-4(x+1)+3 \ \Rightarrow & & y & =x^{2}-2 x \end{aligned} $$
$\therefore$ Required area $=\int_{0}^{2}\left(x^{2}-2 x\right) d x=\frac{x^{3}}{3}-x^{2}{ }_{0}^{2}$
$$ =\frac{8}{3}-4=\frac{4}{3} \text { sq units } $$
