Definite Integration Question 317
Question: The area bounded by the curves $ y=\sqrt{x}, $
$ 2y+3=x $ and $ x- $ axis in the 1st quadrant is
[IIT Screening 2003]
Options:
A) 9
B) $ \frac{27}{4} $
C) 36
D) 18
Show Answer
Answer:
Correct Answer: A
Solution:
Solving $ y^{2}=x $ and $ x=2y+3 $
$ 4y^{2}={{(x-3)}^{2}} $ , $ 4x=x^{2}-6x+9 $
therefore $ x^{2}-10x+9=0 $
therefore $ (x-1)(x-9)=0 $
therefore $ x=1,9 $ = $ -4[x\log x-x]_0^{1}=-4(-1)=4 $ sq. unit, $ (\because \underset{x\to 0}{\mathop{lim}}x\log x=0) $ . Required area = A+B $ =\int_0^{3}{\sqrt{x}dx+\int_3^{9}{[ \sqrt{x}-( \frac{x-3}{2} ) ]}}dx $
$ =\frac{2}{3}[{x^{3/2}}]_0^{3}+\frac{2}{3}[{x^{3/2}}]_3^{9}-\frac{1}{2}[ \frac{x^{2}}{2}-3x ]_3^{9} $
$ =\frac{2}{3}3\sqrt{3}+\frac{2}{3}[9\times 3-3\sqrt{3}]-\frac{1}{2}[ ( \frac{81}{2}-27 )-( \frac{9}{2}-9 ) ] $
$ =18-\frac{1}{2}[36-18]=18-9=9 $ sq. unit.
BETA
