## Calculus 8th Edition

$$\int_0^3xf(x^2)dx=2$$
To find the integral $$\int_0^3xf(x^2)dx$$ we will use substitution $x^2=t$ which gives us $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$ and the integration bounds would be: for $x=0$ we have $t=0$ and for $x=3$ we have $t=9$. Putting all this into the integral we get: $$\int_0^3f(x^2)xdx=\int_0^9f(t)\frac{dt}{2}=\frac{1}{2}\int_0^9f(t)dt$$ Based on the preposition that $$\int_0^9f(x)dx=4$$ (doesn't metter how we name the integration variable), we now have: $$\int_0^3xf(x^2)dx=\frac{1}{2}\int_0^9f(t)dt=\frac{1}{2}\cdot 4=2$$