IIT-JEE 2002 ScreeningQuestion 1
Let $T>0$ be a fixed real number. Suppose $f$ is a continuous function such that for all $x\in R$, $f(x+T)=f(x)$. If $I=\displaystyle\int_0^{T}f(x)\,dx$ then the value of $\displaystyle\int_3^{3+3T}f(2x)\,dx$ is
Definite Integration
Solution
Concept. Put $u=2x$; the new interval has length $6T$, i.e. $6$ full periods. Formulas used. (1) $u=2x$: $dx=\frac12du$(2) For period $T$: $\int_a^{a+nT}f=n\int_0^Tf=nI$Step 1 - Substitute. By (1), $x:3\to3+3T$ gives $u:6\to6+6T$: $$\int_3^{3+3T}f(2x)\,dx=\frac12\int_6^{6+6T}f(u)\,du.$$ Step 2 - Use (2). The interval is $6$ periods long, so the integral is $\frac12\cdot6I=3I$. Answer: (C)