AIEEE 2010 (Offline)Question 1
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 $\int_{0}^{1}p(x)\,dx$ equals
Definite Integration
Solution
Concept. Integrating $p'(x)=p'(1-x)$ gives $p(x)+p(1-x)=$ constant; then King's rule on $[0,1]$. Formulas used. (1) $\frac{d}{dx}\big(p(x)+p(1-x)\big)=p'(x)-p'(1-x)=0$(2) $\int_0^1p(x)\,dx=\int_0^1p(1-x)\,dx$Step 1 - Constant. By (1), $p(x)+p(1-x)=c$. At $x=0$: $c=p(0)+p(1)=42$. Step 2 - Add. By (2), $$2\int_0^1p(x)\,dx=\int_0^1\big(p(x)+p(1-x)\big)dx=42.$$ Step 3 - Value. $\int_0^1p(x)\,dx=21$. Answer: (A)