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Let $\mathrm{f}$ be derivable in $[0,1]$, then
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Verified Answer
The correct answer is:
there exists $c \in(0,1)$ such that $\int_0^c f(x) d x=(1-c) f(c)$
Let $g(x)=x \int_0^x f(t) d t-\int_0^x f(t) d t$
Now, $g(0)=0$
$g(1)=0$
By Rolle's Theorem
$g^{\prime}(x)=0$
for some $x \in(0,1)$
$g^{\prime}(x)=x f(x)+\int_0^x f(t) d t-f(x)=0$
Now, $g(0)=0$
$g(1)=0$
By Rolle's Theorem
$g^{\prime}(x)=0$
for some $x \in(0,1)$
$g^{\prime}(x)=x f(x)+\int_0^x f(t) d t-f(x)=0$
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