Let $f$ be a function such that $\int_{6}^{12}f(2x)dx=10$, demonstrate that $\int_{12}^{24}f(t)dt=20$

Question

Let $f$ be a function such that $\int_{6}^{12}f(2x)dx=10$, demonstrate that $\int_{12}^{24}f(t)dx=20$

I didn't know how to do this and this is what I thought:

Let $t = 2 x$, therefore, the limit start and end point values should be $12$ and $24$ because $2*6=12$ and $2*12=24$.

But what I don't understand is why it's equal to now $20$ instead of $10$.

Answer 1

$$10=\int_{6}^{12}f(2x)dx=\frac {1}{\color {red}2} \int_{6}^{12}f(2x)d ({\color {red} 2}x)=\frac {1}{2} \int_{12}^{24}f(t)dt\\ \Rightarrow \int_{12}^{24}f(t)dt=20. $$

Answer 2

The answer can be obtained by a single substitution $t=2x$ and so $dt=2dx$ and that's where the factor of $2$ comes from.

Intuitively, the two integrals are sums of the exact same values of the function. It's just that in the second integrals, the same values of the function are stretched out over a domain that twice as wide. Thus the factor of two.

If you would like to practice similar problems, check out this problem: https://lem.ma/Nz