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If $f(a+b-x)=f(x)$ then $\int_a^b x f(x) d x$ is equal to
(a) $\frac{a+b}{2} \int_a^b f(b-x) d x$
(b) $\frac{a+b}{2} \int_a^b f(b+x) d x$
(c) $\frac{b-a}{2} \int_a^b f(x) d x$
(d) $\frac{a+b}{2} \int_a^b f(x) d x$
If $f(a+b-x)=f(x)$ then $\int_a^b x f(x) d x$ is equal to
(a) $\frac{a+b}{2} \int_a^b f(b-x) d x$
(b) $\frac{a+b}{2} \int_a^b f(b+x) d x$
(c) $\frac{b-a}{2} \int_a^b f(x) d x$
(d) $\frac{a+b}{2} \int_a^b f(x) d x$
Solution:
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Verified Answer
(d) Let $\mathrm{I}=\int_a^b x f(x) d x$
Let $a+b-x=z \Rightarrow-d x=d z$
When $x=a, z=b$ and when $x=b, z=a$
$\begin{aligned}
&\therefore \quad \mathrm{I}=-\int_b^a(a+b-z) f(z) d z \\
&=(a+b) \int_b^a f(x) d x-\int_a^b x f(x) d x \\
&\mathrm{I}=(a+b) \int_b^a f(x) d x-\mathrm{I} \Rightarrow \mathrm{I}=\left(\frac{a+b}{2}\right) \int_a^b f(x) d x
\end{aligned}$
Let $a+b-x=z \Rightarrow-d x=d z$
When $x=a, z=b$ and when $x=b, z=a$
$\begin{aligned}
&\therefore \quad \mathrm{I}=-\int_b^a(a+b-z) f(z) d z \\
&=(a+b) \int_b^a f(x) d x-\int_a^b x f(x) d x \\
&\mathrm{I}=(a+b) \int_b^a f(x) d x-\mathrm{I} \Rightarrow \mathrm{I}=\left(\frac{a+b}{2}\right) \int_a^b f(x) d x
\end{aligned}$
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