Definite Integration Question 145
Question: The value of a (a > 0) for which the area bounded by the curves $ y=\frac{x}{6}+\frac{1}{x^{2}},y=0,x=a $ and $ x=2a $ has the least value is
Options:
A) 2
B) $ \sqrt{2} $
C) $ {2^{1/3}} $
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ f(a)=\int\limits_a^{2a}{( \frac{x}{6}+\frac{1}{x^{2}} )dx=( \frac{x^{2}}{12}-\frac{1}{x} )_a^{2a}} $
$ =( \frac{4a^{2}}{12}-\frac{1}{2a}-\frac{a^{2}}{12}+\frac{1}{a} )=\frac{a^{2}}{4}+\frac{1}{2a} $ Let $ f’(a)=\frac{2a}{4}-\frac{1}{2a^{2}}=0 $
$ \Rightarrow a=1 $ which is a point of minima.
BETA
