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# The anti-derivative of every odd function is-A.An odd functionB.An even functionC.Neither even nor oddD.Sometimes even, sometimes odd

Last updated date: 13th Jun 2024
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Hint: We have to determine whether the antiderivative of every odd function is odd or even function. We will use the definition of odd function which is given as-
A function is odd if and only if ${\text{f}}\left( {{\text{ - x}}} \right) = - {\text{f}}\left( {\text{x}} \right)$ . Let us assume the function to be x. Then integrate the given function to find the antiderivative and name it g(x). Then check whether this function is odd or even if it is odd then ${\text{f}}\left( {{\text{ - x}}} \right) = - {\text{f}}\left( {\text{x}} \right)$ and if it is even then ${\text{f}}\left( {{\text{ - x}}} \right) = {\text{f}}\left( {\text{x}} \right)$.

We have to find the anti-derivative of every odd function.
Let us assume that the function be ${\text{f}}\left( {\text{x}} \right){\text{ = x}}$ which represents an odd function. We know that a function is odd if and only if ${\text{f}}\left( {{\text{ - x}}} \right) = - {\text{f}}\left( {\text{x}} \right)$.
So suppose if x=$2$ then according to definition of odd function-
$\Rightarrow f\left( { - 2} \right) = - f\left( 2 \right)$
So we can write-
$\Rightarrow - 2 = - 2$
So f(x) =x is an odd function it is verified.
Now we have to find the anti-derivative of the given function. So to find the anti-derivative of the function, we will integrate the given function-
$\Rightarrow \int {{\text{f}}\left( {\text{x}} \right)} dx = \int x dx$
Now we know that formula of integration of x is given as-
$\Rightarrow \int {{x^n} = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
On using this formula we get-
$\Rightarrow \int {{\text{f}}\left( {\text{x}} \right)} dx = \dfrac{{{x^{1 + 1}}}}{{1 + 1}} + C$ ,Where C is integration constant.
On solving, we get-
$\Rightarrow \int {{\text{f}}\left( {\text{x}} \right)} dx = \dfrac{{{x^2}}}{2} + C$
Now we can write the anti-derivative of f(x) as g(x).
Then we can write-
$\Rightarrow {\text{g}}\left( {\text{x}} \right) = \dfrac{{{x^2}}}{2} + C$
$\Rightarrow {\text{g}}\left( {{\text{ - x}}} \right) = \dfrac{{{{\left( { - x} \right)}^2}}}{2} + C$
Now we know that square of any function will always give us an even function
$\Rightarrow {\text{g}}\left( {{\text{ - x}}} \right) = \dfrac{{{x^2}}}{2} + C$
So it is an even function as an even function is defined as-
A function is even if and only if ${\text{f}}\left( {{\text{ - x}}} \right){\text{ = f}}\left( {\text{x}} \right)$
We can also verify it by putting any number in place of x like-
$\Rightarrow {\text{g}}\left( {{\text{ - 2}}} \right) = \dfrac{{{{\left( { - 2} \right)}^2}}}{2} = \dfrac{4}{2}$
So it is proved that the anti-derivative of an odd function is an even function.