Area Ques 43
Question
- 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)
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Answer:
Correct Answer: 43.$( y= x^2 -2x, \frac{4}{3} \text { sq units })$
Solution:
Formula:
- 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 $y-3=Y$
$\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 $
$\text { IF }=e^{\int-\frac{1}{X} d X} =e^{-\log X}=\frac{1}{X}$
$\therefore$ 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 } $
BETA
