Area Ques 29
Question
- The area (in sq units) bounded by the curves $y=\sqrt{x}, 2 y-x+3=0, X$-axis and lying in the first quadrant, is
(2013 Main, 03)
(a) $ 9$
(b) $ 6$
(c) $18$
(d) $\frac{27}{4}$
Show Answer
Answer:
Correct Answer: 29.(a)
Solution:
Formula:
- Given curves are $y=\sqrt{x}$ $\quad$ …….(i)
and $2 y-x+3=0$ $\quad$ …….(ii)
On solving Eqs. (i) and (ii), we get
$2 \sqrt{x}-(\sqrt{x})^{2}+3 =0 $
$\Rightarrow \quad (\sqrt{x})^{2}-2 \sqrt{x}-3 =0 $
$\Rightarrow \quad (\sqrt{x}-3)(\sqrt{x}+1) =0 \quad \Rightarrow \quad \sqrt{x}=3 $
$\therefore \quad y =3 $
Hence, required area
$ \begin{aligned} & =\int _{0}^{3}\left(x _{2}-x _{1}\right) d y=\int _{0}^{3}\left\{(2 y+3)-y^{2}\right\} d y \\ & =[y^{2}+3 y-{\frac{y^{3}}{3}}] _{0}^{3}=9+9-9=9 \text { sq units } \end{aligned} $
BETA
