Let $$p(x)$$ be a function defined on R such that p'(x) $$=$$ p'(1$$ – $$x), for all x∈[$$0$$,$$1$$],$$p(0)=1$$ and $$p(1)$$ $$=$$ $$41$$. Then ∫$$01$$ $$p(x)dx$$ equals

Question

Let $$p(x)$$ be a function defined on R such that  p'(x) $$=$$ p'(1$$ – $$x), for all x∈[$$0$$,$$1$$],$$p(0)=1$$ and $$p(1)$$ $$=$$ $$41$$. Then ∫$$01$$ $$p(x)dx$$ equals

Infinity Learn · Accepted Answer

∫$$01$$ $$p(x)dx=$$∫$$01$$ $$1$$⋅$$p(x)dx=$$[$$xp(x)$$]$$01$$−$$I1=p(1)$$−$$I1where$$ $$I1=$$∫$$01$$ xp′(x)dx=$$∫$$01$$ $$(1$$−$$x)p$$′$$(1$$−$$x)dx=$$∫$$01$$ $$(1$$−$$x)p$$′$$(x)dx=$$∫$$01$$ p′$$(x)dx$$−$$I1$$⇒ $$2I1=$$[$$p(x)]$$01=p(1)$$−$$p(0)Thus$$, ∫$$01$$ $$p(x)dx=p(1)$$−$$12(p(1)$$−$$p(0))=12(p(1)+p(0))=12(41+1)=21$$

Let $p (x)$ be a function defined on R such that $p' (x) = p' (1 - x)$ , 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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Let p(x) be a function defined on R such that p'(x) = p'(1 – x), for all x∈[0,1],p(0)=1 and p(1) = 41. Then ∫01 p(x)dx equals

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Let $p (x)$ be a function defined on R such that $p' (x) = p' (1 - x)$ , for all $x \in [0, 1], p (0) = 1$ and $p (1) = 41 .$ Then $\int_{0}^{1} p (x) d x$ equals