Question

Let $f\left(x\right)$ and $\varphi \left(x\right)$ are two continuous functions on $R$ satisfying $\varphi \left(x\right)=\underset{a}{\overset{x}{\int }}f\left(t\right)dt,a\ne 0$. If $f\left(x\right)$ is an even function, then which of the following statements are correct?

• A. $\varphi \left(x\right)$ is always an even function
• B. $\varphi \left(x\right)$ is always an odd function
• C. $\varphi \left(x\right)$ is an even function if $f(a-x)=-f\left(x\right)$
• D. $\varphi \left(x\right)$ is an odd function if $f(a-x)=-f\left(x\right)$

Answer 1

The correct option is D $\varphi \left(x\right)$ is an odd function if $f(a-x)=-f\left(x\right)$

$\varphi \left(x\right)=\underset{a}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow \varphi (-x)=\underset{a}{\overset{-x}{\int }}f\left(t\right)dt$
Replacing $t$ with $-t$ we have:
$\Rightarrow \varphi (-x)=-\underset{-a}{\overset{x}{\int }}f(-t)dt\Rightarrow -\varphi (-x)=\underset{-a}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow -\varphi (-x)=\underset{-a}{\overset{a}{\int }}f\left(t\right)dt+\underset{a}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow -\varphi (-x)=2\underset{0}{\overset{a}{\int }}f\left(t\right)dt+\underset{a}{\overset{x}{\int }}f\left(t\right)dt$

If $f(a-x)=-f\left(x\right)$, then
$\underset{0}{\overset{a}{\int }}f\left(t\right)dt=0\therefore \varphi (-x)=-\underset{a}{\overset{x}{\int }}f\left(t\right)dt=-\varphi \left(x\right)$
Hence, $\varphi \left(x\right)$ is odd function.