HMMT 2002: Calculus Problem 10

Harvard–MIT Mathematics Tournament · March 17, 2002 · Individual Subject Test: Calculus · Problem 10. Transcribed from the supplied problem image.

A continuous real function $f$ satisfies

If $\int_0^1 f(x)\,dx=1$, find

We do not need an explicit formula for the function. Its scaling identity lets us compare integrals over two intervals directly.

1. Integrate over the larger interval and substitute $x=2u$. Then $dx=2\,du$, and the bounds become $u=0$ and $u=1$.

   $$\int_0^2 f(x)\,dx=2\int_0^1 f(2u)\,du$$

2. Apply $f(2u)=3f(u)$ and use the given integral.

   $$2\int_0^1 f(2u)\,du=6\int_0^1 f(u)\,du=6$$

3. Subtract the integral over the first interval.

   $$\int_1^2 f(x)\,dx=\int_0^2 f(x)\,dx-\int_0^1 f(x)\,dx=6-1=\boxed{5}$$

Doubling the input multiplies function values by $3$, while the substitution stretches interval widths by $2$. The integral therefore scales by $2\cdot3=6$. Using only the factor from the function identity would miss the change in the differential.

Continuity guarantees the definite integrals exist. The solution uses substitution and additivity, so no assumption about a particular formula or the sign of the function is needed.

Explore signed area and accumulation in the Integrals lab.