Definite Integration Question 316
Question: If the area bounded by $ y=ax^{2} $ and $ x=ay^{2} $ , $ a>0 $ , is 1, then $ a= $
[IIT Screening 2004]
Options:
A) 1
B) $ \frac{1}{\sqrt{3}} $
C) $ \frac{1}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The x-coordinate of A is $ \frac{1}{a} $
According to the given condition, $ 1=\int_0^{1/a}{( \sqrt{\frac{x}{a}}-ax^{2} )}dx $
therefore $ 1=\frac{1}{\sqrt{a}}.\frac{2}{3}[{x^{3/2}}]_0^{1/a}-\frac{a}{3}[x^{3}]_0^{1/a} $
therefore $ a^{2}=\frac{1}{3}\Rightarrow a=\frac{1}{\sqrt{3}} $ .
BETA
