Question

Let \[f\] and \[g\] be continuous functions on the interval \[\left[ 0,a \right]\], such that \[f\left( x \right)=f\left( a-x \right)\] and \[g\left( x \right)+g\left( a-x \right)=4\]. Then evaluate the integral \[\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\].
(a) \[4\int\limits_{0}^{a}{f\left( x \right)dx}\]
(b) \[2\int\limits_{0}^{a}{f\left( x \right)dx}\]
(c) \[-3\int\limits_{0}^{2}{f\left( x \right)dx}\]
(d) \[\int\limits_{0}^{a}{f\left( x \right)dx}\]

Answer 1

Hint: In this question, in order to evaluate the definite integral \[\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\] given that \[f\] and \[g\] be continuous functions on the interval \[\left[ 0,a \right]\], such that \[f\left( x \right)=f\left( a-x \right)\] and \[g\left( x \right)+g\left( a-x \right)=4\]. We will use the property of the definite integral that \[\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}\] in the integral \[\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\] to get a simplified form of the integral. We will then evaluate the same in order to get the desired answer.

Complete step by step answer:
We are given that \[f\] and \[g\] be continuous functions on the interval \[\left[ 0,a \right]\].
The function \[f\] satisfies the equation given by \[f\left( x \right)=f\left( a-x \right)...........(1)\].
And the function \[g\] satisfies the equation \[g\left( x \right)+g\left( a-x \right)=4..............(2)\].
Let us suppose that \[I\] denote the integral \[\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\].
That is, let \[I=\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\].
Since we know the property of the definite integral that \[\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}\] .
Using this in the integral \[I=\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\], we will have
\[\begin{align}
  & I=\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx} \\
 & =\int\limits_{0}^{a}{f\left( a-x \right)g\left( a-x \right)dx}
\end{align}\]
Now from equation (2), we have
\[g\left( a-x \right)=4-g\left( x \right)............(4)\]
On substituting the value of equation (4) in the above integral, we will get
\[\begin{align}
  & I=\int\limits_{0}^{a}{f\left( a-x \right)g\left( a-x \right)dx} \\
 & =\int\limits_{0}^{a}{f\left( a-x \right)\left( 4-g\left( x \right) \right)dx}
\end{align}\]
We know split the above integral to get

\[\begin{align}
  & I=\int\limits_{0}^{a}{f\left( a-x \right)\left( 4-g\left( x \right) \right)dx} \\
 & =4\int\limits_{0}^{a}{f\left( a-x \right)dx}-\int\limits_{0}^{a}{f\left( a-x \right)g\left( x \right)dx}
\end{align}\]
Now using equation (1) in the above integral, we will have
\[\begin{align}
  & I=4\int\limits_{0}^{a}{f\left( a-x \right)dx}-\int\limits_{0}^{a}{f\left( a-x \right)g\left( x \right)dx} \\
 & =4\int\limits_{0}^{a}{f\left( x \right)dx}-\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}
\end{align}\]
Using the fact that \[I=\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\] in the above equation we will get
\[\begin{align}
  & I=4\int\limits_{0}^{a}{f\left( x \right)dx}-\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx} \\
 & =4\int\limits_{0}^{a}{f\left( x \right)dx}-I
\end{align}\]
Now on taking \[I\] common at the left hand side of the above equation we get
\[2I=4\int\limits_{0}^{a}{f\left( x \right)dx}\]
We will now divide the above equation by 2 to get
\[\begin{align}
  & I=\dfrac{4}{2}\int\limits_{0}^{a}{f\left( x \right)dx} \\
 & =2\int\limits_{0}^{a}{f\left( x \right)dx}
\end{align}\]
Therefore we have \[\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}\].

So, the correct answer is “Option B”.

Note: In this problem, , in order to evaluate the definite integral \[\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)dx}\] given that \[f\] and \[g\] be continuous functions on the interval \[\left[ 0,a \right]\], such that \[f\left( x \right)=f\left( a-x \right)\] and \[g\left( x \right)+g\left( a-x \right)=4\], we are using the property that \[\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}\]. Please take care of the fact that we can use this property only in case of definite integrals where the integrand is continuous in the given range of lower limit and the upper limit of the integral.