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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Aicharya · Accepted Answer

$\mathrm{p}^{\prime}(\mathrm{x})=\mathrm{p}^{\prime}(1-\mathrm{x})$ $\Rightarrow \mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+\mathrm{c}$ at $x=0$ $\mathrm{p}(0)=-\mathrm{p}(1)+\mathrm{c} \quad \Rightarrow 42=\mathrm{c}$ now $\mathrm{p}(\mathrm{x})=-\mathrm{p}(1-\mathrm{x})+42$ $\Rightarrow \mathrm{p}(\mathrm{x})+\mathrm{p}(1-\mathrm{x})=42$ $\mathrm{I}=\int_0^1 \mathrm{p}(\mathrm{x}) \mathrm{dx}=\int_0^1 \mathrm{p}(1-\mathrm{x}) \mathrm{dx}$ $2 \mathrm{I}=\int_0^1(42) \mathrm{dx} \quad \Rightarrow \mathrm{I}=21$

Let $p(x)$ be a function defined on $R$ such that $p^{\prime}(x)=p^{\prime}(1-x)$, for all $x \in[0,1],…

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