Definite Integration Question 293
Question: If the ordinate $ x=a $ divides the area bounded by the curve $ y=( 1+\frac{8}{x^{2}} ), $
$ x- $ axis and the ordinates $ x=2, $
$ x=4 $ into two equal parts, then $ a= $
[IIT 1983]
Options:
A) 8
B) $ 2\sqrt{2} $
C) 2
D) $ \sqrt{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let the ordinate at $ x=a $ divide the area into two equal parts Area of $ AMNB $
$ =\int_2^{4}{( 1+\frac{8}{x^{2}} )dx=[ x-\frac{8}{x} ]}_2^{4}=4 $
Area of $ ACDM=\int_2^{a}{( 1+\frac{8}{x^{2}} )}dx=2 $
On solving, we get $ a=\pm 2\sqrt{2} $ ;Since $ a>0 $
therefore $ a=2\sqrt{2} $ .
BETA
