Question

If $\mathrm{f}\left(\mathrm{t}\right)$ is an odd function, then ${\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$ is

• A. An odd function
• B. An even function
• C. Neither even nor odd
• D. 0

Answer 1

The correct option is B

An even function

Explanation for the correct option:

Step 1: Determine the function

Given function $f\left(t\right)$ is an odd function

We can write

$\therefore -f\left(t\right)=f(-t)$

Let $g\left(-x\right)={\int }_0^{-\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$ and $\mathrm{t}=-\mathrm{y},\mathrm{dt}=-\mathrm{dy}$

Step 2: Applying limits,

$$\begin{gathered} \mathrm{t}=0\to \mathrm{y}=0 \\ \mathrm{t}=-\mathrm{x}\to \mathrm{y}=\mathrm{x} \end{gathered}$$

$\begin{array}{rcl}\therefore \mathrm{g}(-\mathrm{x})& =& {\int }_0^{\mathrm{x}}\mathrm{f}(-\mathrm{y})(-\mathrm{dy})\\ & =& {\int }_{-\mathrm{x}}^0\mathrm{f}\left(\mathrm{y}\right)\mathrm{dy}+{\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{y}\right)\mathrm{dy}\\ & =& 0+{\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{y}\right)\mathrm{dy}\left[\because \mathrm{g}(-\mathrm{x})={\int }_0^{-\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{g}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\right]\\ & =& \mathrm{g}\left(\mathrm{x}\right)\\ \therefore \mathrm{g}(-\mathrm{x})& =& \mathrm{g}\left(\mathrm{x}\right)\end{array}$

Therefore, ${\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$ is an even function.

Hence, the correct option is an option (B).