Definite Integration Question 342
Question: The value of the definite integral $ \int_0^{1}{\frac{xdx}{x^{3}+16}} $ lies in the interval $ [a,b]. $ The smallest such interval is
Options:
A) $ [ 0,\frac{1}{17} ] $
B) [0, 1]
C) $ [ 0,\frac{1}{27} ] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The function $ f(x)=\frac{x}{x^{3}+16} $ is an increasing function, so Min $ f(x)=f(0)=0 $ and Max $ f(x)=f(1)=\frac{1}{17} $
Therefore by the property $ m(b-a)\le \int_a^{b}{f(x)dx\le M(b-a)} $
(where m and M are the smallest and greatest values of function)
therefore $ 0\le \int_0^{1}{\frac{x}{x^{3}+16}dx\le \frac{1}{17}} $ .
BETA
