## Calculus 8th Edition

$$\int_0^2f(2x)dx=5$$
To find the integral $$\int_0^2f(2x)dx$$ we will use substitution $2x=t$ which gives us $2dx=dt\Rightarrow dx=\frac{dt}{2}$ and integration bounds will be: for $x=0$ we have $t=0$ and for $x=2$ we have $t=4$. Putting this into the integral we get: $$\int_0^2f(2x)dx=\int_0^4f(t)\frac{dt}{2}=\frac{1}{2}\int_0^4f(t)dt$$ Based on the preposition that $$\int_0^4f(x)dx=10$$ (doesn't metter how we name the integration varialble), we now have: $$\int_0^2f(2x)dx=\frac{1}{2}\int_0^4f(t)dt=\frac{1}{2}\cdot10=5$$