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Let $p(x)$ be a function defined on $R$ such that $p^{\prime}(x)=p^{\prime}(1-x)$, for all $x \in[0,1], p(0)=1$ and $p(1)=41$. Then $\int_0^1 p(x) d x$ equals
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The correct answer is:
21
21
$\mathrm{p}^{\prime}(\mathrm{x})=\mathrm{p}^{\prime}(1-\mathrm{x})$
$\Rightarrow \mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+\mathrm{c}$
at $x=0$
$\mathrm{p}(0)=-\mathrm{p}(1)+\mathrm{c} \quad \Rightarrow 42=\mathrm{c}$
now $\mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+42$
$\Rightarrow \mathrm{p}(\mathrm{x})+\mathrm{p}(1-\mathrm{x})=42$
$\mathrm{I}=\int_0^1 \mathrm{p}(\mathrm{x}) \mathrm{dx}=\int_0^1 \mathrm{p}(1-\mathrm{x}) \mathrm{dx}$
$2 \mathrm{I}=\int_0^1(42) \mathrm{dx} \quad \Rightarrow \mathrm{I}=21$
$\Rightarrow \mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+\mathrm{c}$
at $x=0$
$\mathrm{p}(0)=-\mathrm{p}(1)+\mathrm{c} \quad \Rightarrow 42=\mathrm{c}$
now $\mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+42$
$\Rightarrow \mathrm{p}(\mathrm{x})+\mathrm{p}(1-\mathrm{x})=42$
$\mathrm{I}=\int_0^1 \mathrm{p}(\mathrm{x}) \mathrm{dx}=\int_0^1 \mathrm{p}(1-\mathrm{x}) \mathrm{dx}$
$2 \mathrm{I}=\int_0^1(42) \mathrm{dx} \quad \Rightarrow \mathrm{I}=21$
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