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If \( f(x)=f(\Pi+e-x) \) and \( \int_{e}^{I I} f(x) d x=\frac{2}{e+\Pi^{\prime}} \), then \( \int_{e}^{I I} x f(x) d x \) is equal to
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Verified Answer
The correct answer is:
\( 1 \)
Given that
\[
\begin{array}{l}
f(x)=f(\Pi+e-x) \rightarrow(1) \\
\text { and } \int_{e}^{\Pi} f(x) d x=\frac{2}{e+\Pi} \rightarrow(2)
\end{array}
\]
Now,
\[
\begin{array}{l}
I=\int_{e}^{\Pi} x f(x) d x=\int_{e}^{\Pi}(e+\Pi-x) f(e+\Pi-x) d x \\
=\int_{e}^{\Pi}(e+\Pi-x) f(x) d x=\int_{e}^{\Pi}(e+\Pi) f(x)-I \\
\Rightarrow 2 I=(e+\Pi) \frac{2}{e+\Pi} \Rightarrow I=1
\end{array}
\]
\[
\begin{array}{l}
f(x)=f(\Pi+e-x) \rightarrow(1) \\
\text { and } \int_{e}^{\Pi} f(x) d x=\frac{2}{e+\Pi} \rightarrow(2)
\end{array}
\]
Now,
\[
\begin{array}{l}
I=\int_{e}^{\Pi} x f(x) d x=\int_{e}^{\Pi}(e+\Pi-x) f(e+\Pi-x) d x \\
=\int_{e}^{\Pi}(e+\Pi-x) f(x) d x=\int_{e}^{\Pi}(e+\Pi) f(x)-I \\
\Rightarrow 2 I=(e+\Pi) \frac{2}{e+\Pi} \Rightarrow I=1
\end{array}
\]
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