HMMT 2002 Calculus #10: Integrating Without Finding f

Posted on Oct 21, 2026

Problem (Harvard–MIT Mathematics Tournament, March 2002, Calculus #10). A continuous real function $f$ satisfies the identity $f(2x) = 3f(x)$ for all $x$ . If $\int_0^1 f(x)\,dx = 1$ , what is $\int_1^2 f(x)\,dx$ ?

Solution. Substitute $x = 2u$ , so $dx = 2\,du$ :

\begin{aligned} \int_0^2 f(x)\,dx &= 2\int_0^1 f(2u)\,du \\ &= 2\int_0^1 3f(u)\,du \\ &= 6 \cdot 1 = 6. \end{aligned}

Therefore

\begin{aligned} \int_1^2 f(x)\,dx &= \int_0^2 f(x)\,dx - \int_0^1 f(x)\,dx \\ &= 6 - 1 = \boxed{5} \qquad \blacksquare \end{aligned}

Remark. There are many ways to do this, but this is the most direct. The step to be careful with is the change of variables: the integral over $[0,2]$ is six times the integral over $[0,1]$ , not three. The identity contributes a factor of $3$ , and stretching the interval from length $1$ to length $2$ contributes another factor of $2$ .

Originally posted on X in reply to @Math_files.