Evaluating a rescaled integral via substitution

Let $u = 2x$, so $du = 2\,dx$, which means $dx = \frac{1}{2}\,du$. When $x=0$, $u=0$, and when $x=3$, $u=6$. Substituting gives $\int_0^3 f(2x)\,dx = \int_0^6 f(u)\cdot \frac{1}{2}\,du$.

This equals $\frac{1}{2}\int_0^6 f(u)\,du = \frac{1}{2}(15) = 7.5$, using linearity of the integral (pulling out the constant $\frac12$) and the given value of $\int_0^6 f(x)\,dx$.

Smaller. The rescaled integral equals one half times fifteen, which is seven point five. Since one half is less than one, multiplying fifteen by it produces a smaller number than the original value.