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If $k \int_{0}^{1} x \cdot f(3 x) d x=\int_{0}^{3} t \cdot f(t) d t$, then the value of $k$ is
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Verified Answer
The correct answer is:
9
Let $I=\int_{0}^{3} t f(t)$
Put $t=3 x \Rightarrow d t=3 d x$
When $t=0 \Rightarrow x=0$
and $t=3 \Rightarrow x=1$
$$
\begin{aligned}
\therefore \quad I &=\int_{0}^{1} 3 x f(3 x)(3 d x) \\
&=9 \int_{0}^{1} x f(3 x) d x \\
&=k \int_{0}^{1} x f(3 x) d x \\
\therefore \quad k &=9
\end{aligned}
$$
Put $t=3 x \Rightarrow d t=3 d x$
When $t=0 \Rightarrow x=0$
and $t=3 \Rightarrow x=1$
$$
\begin{aligned}
\therefore \quad I &=\int_{0}^{1} 3 x f(3 x)(3 d x) \\
&=9 \int_{0}^{1} x f(3 x) d x \\
&=k \int_{0}^{1} x f(3 x) d x \\
\therefore \quad k &=9
\end{aligned}
$$
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