Answer
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Hint: We have to determine what is the antiderivative of every odd function. For finding whether the antiderivative of an odd function is odd or even, we will first let one function which is an odd function, and find the antiderivative of that odd function and see whether the obtained antiderivative is odd or even.
Complete step by step solution
The aim is to find whether the antiderivative of an odd function is odd or even.
We will first let function that is \[f\left( x \right) = x\] which represents the odd function as we know that the function is odd if and only if \[f\left( { - x} \right) = - f\left( x \right)\]
From this, we can verify that the function is an odd function,
\[
\Rightarrow f\left( { - 2} \right)\mathop = - f\left( 2 \right) \\
\Rightarrow - 2 = - 2 \\
\]
Next, we need to find the antiderivative of the above odd function that is the integral of the function \[f\left( x \right) = x\],
Thus, we get,
\[
\Rightarrow \int {f\left( x \right)dx = } \int {xdx} \\
\Rightarrow \int {f\left( x \right)dx = } \dfrac{{{x^2}}}{2} + c \\
\]
From this, we can let the antiderivative of the function \[f\left( x \right)\] as \[g\left( x \right)\].
Thus, we get,
\[ \Rightarrow g\left( x \right) = \dfrac{{{x^2}}}{2} + c\]
Here, the function \[g\left( x \right)\] is an even function as we know that the function is even if and only if \[g\left( { - x} \right) = g\left( x \right)\].
From this, we can verify that the function obtained is an even function.
Thus, we have,
\[
\Rightarrow g\left( { - 2} \right)\mathop = \left( 2 \right) \\
\Rightarrow \dfrac{{{{\left( { - 2} \right)}^2}}}{2}\mathop =\dfrac{{{{\left( 2 \right)}^2}}}{2} \\
\Rightarrow \dfrac{4}{2} = \dfrac{4}{2} \\
\]
Hence, we can conclude that the antiderivative of an odd function is an even function.
Thus, option B is correct.
Note: The antiderivative of a function is equal to the integration of that function. Odd and even functions are functions which satisfy particular symmetry relations with each other, with respect to taking additive inverses. After evaluating the antiderivative verify whether it is an odd function or even function.
Complete step by step solution
The aim is to find whether the antiderivative of an odd function is odd or even.
We will first let function that is \[f\left( x \right) = x\] which represents the odd function as we know that the function is odd if and only if \[f\left( { - x} \right) = - f\left( x \right)\]
From this, we can verify that the function is an odd function,
\[
\Rightarrow f\left( { - 2} \right)\mathop = - f\left( 2 \right) \\
\Rightarrow - 2 = - 2 \\
\]
Next, we need to find the antiderivative of the above odd function that is the integral of the function \[f\left( x \right) = x\],
Thus, we get,
\[
\Rightarrow \int {f\left( x \right)dx = } \int {xdx} \\
\Rightarrow \int {f\left( x \right)dx = } \dfrac{{{x^2}}}{2} + c \\
\]
From this, we can let the antiderivative of the function \[f\left( x \right)\] as \[g\left( x \right)\].
Thus, we get,
\[ \Rightarrow g\left( x \right) = \dfrac{{{x^2}}}{2} + c\]
Here, the function \[g\left( x \right)\] is an even function as we know that the function is even if and only if \[g\left( { - x} \right) = g\left( x \right)\].
From this, we can verify that the function obtained is an even function.
Thus, we have,
\[
\Rightarrow g\left( { - 2} \right)\mathop = \left( 2 \right) \\
\Rightarrow \dfrac{{{{\left( { - 2} \right)}^2}}}{2}\mathop =\dfrac{{{{\left( 2 \right)}^2}}}{2} \\
\Rightarrow \dfrac{4}{2} = \dfrac{4}{2} \\
\]
Hence, we can conclude that the antiderivative of an odd function is an even function.
Thus, option B is correct.
Note: The antiderivative of a function is equal to the integration of that function. Odd and even functions are functions which satisfy particular symmetry relations with each other, with respect to taking additive inverses. After evaluating the antiderivative verify whether it is an odd function or even function.
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