Definite Integration Question 346
Question: The smallest interval $ [a,b] $ such that $ \int_0^{1}{\frac{dx}{\sqrt{1+x^{4}}}}\in [a,b] $ is given by
Options:
A) $ [ \frac{1}{\sqrt{2}},1 ] $
B) $ [0,1] $
C) $ [ \frac{1}{2},2 ] $
D) $ [ \frac{3}{4},1 ] $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ I=\int_0^{1}{\frac{dx}{\sqrt{1+x^{4}}}} $
Here, $ 0\le x\le 1\Rightarrow 1\le (1+x^{4})\le 2 $
therefore $ 1\le \sqrt{1+x^{4}}\le \sqrt{2}\Rightarrow \frac{1}{\sqrt{2}}\le \frac{1}{\sqrt{1+x^{4}}}\le 1 $
therefore $ \frac{1}{\sqrt{2}}\le \int_0^{1}{\frac{dx}{\sqrt{1+x^{4}}}\le 1} $
Hence $ [ \frac{1}{\sqrt{2}},1 ] $ is the smallest interval, such that $ I\in [ \frac{1}{\sqrt{2}},1 ] $ . Note: If $ m= $ least value of $ f(x) $ and M= greatest value of $ f(x) $ in [a, b], then $ m(b-a)\le \int_a^{b}{f(x)dx\le M(b-a)} $ .
BETA
