Question

The function $f(x)=\underset{0}{\overset{x}{\int }}\sqrt{1-t^4}dt$ is such that :

• A. it is defined on the interval $[-1,1]$
• B. it is an increasing function
• C. it is an odd function
• D. the point $(0,0)$ is the point of inflection

Answer 1

Answer: (A), (B), (C), (D)

$f(x)=\underset{0}{\overset{x}{\int }}\sqrt{1-t^4}dt$
$f(-x)=\underset{0}{\overset{-x}{\int }}\sqrt{1-t^4}dt$
$=-\underset{0}{\overset{x}{\int }}\sqrt{1-u^4}du\text{ (Put }t=-u)$
$f(-x)=-f(x)\Rightarrow$ ' $f$ ' is odd function.
Check other options.