Question

If $f(-x)+f\left(x\right)=0$ then ${\int }_0^{x}f\left(t\right)$ dt is

• A. an odd function
• B. an even function
• C. a periodic function
• D. none of these

Answer 1

Answer: (B)

an even function

Given that f is and odd function.
Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$
Now, $g(-x)={\int }_0^{-x}f\left(t\right)dt=-{\int }_0^{x}f(-t)dt={\int }_0^{x}f\left(t\right)dt=g\left(x\right)$ ...{ Since, $f$ is an odd function }
Therefore, $g$ is an odd function.