Definite Integration Question 185
Question: If the ordinate $ x=a $ divides the area bounded by x-axis, part of the curve $ y=1+\frac{8}{x^{2}} $ and the ordinates $ x=2,x=4 $ into two equal parts, then a is equal to
Options:
A) $ \sqrt{2} $
B) $ 2\sqrt{2} $
C) $ 3\sqrt{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The area bounded by the curve $ y=1+\frac{8}{x^{2}}, $ x-axis and the ordinates $ x=2,x=4 $ is $ =\int_2^{4}{ydx} $ $ =\int_2^{4}{( 1+\frac{8}{x^{2}} )dx} $ $ =[ x-\frac{8}{x} ]_2^{4}=4 $ . Since, $ x=a $ divides this area into two equal parts,
$ \therefore $ Required area $ =2\int_2^{a}{ydx} $
$ \therefore 4=2\int_2^{a}{( 1+\frac{8}{x^{2}} )dx} $
$ \Rightarrow 2=[ x-\frac{8}{x} ]_2^{a}=( a-\frac{8}{a} )-(2-4) $
$ \Rightarrow a^{2}=8\therefore a=2\sqrt{2} $
BETA
