Yes, letting u= 6x, du= 6dx so dx= (1/6)du. When x= 0 u= 6(0)= 0 and when x= 9 u= 6(9)= 54 so the first integral is \(\displaystyle (1/6)\int_0^{54} f(u)du\). Letting v= 3x, dv= 3dx so x= (1/3)dv. When x= 0, v= 3(0)= 0 and when x= 18, v= 3(18)= 54 so the second integral is \(\displaystyle (1/3)\int_0^{54} f(u)du\).
That's exactly what you have: \(\displaystyle \frac{1}{6}\int_0^{54}f(u)du+ \frac{1}{3}\int_0^{54} f(v)dv= 85\). Great!
Now, tkhunny's comment "why did you use u and v" is just referencing that your "u" and "v", as well as the "x" in the original integrals are "dummy" variables. That is, everyone of those integrals is a number, there is no "u" or "v" or "x" in the final value. You could as easily let u= 6x in one and u= 3x in the other. Since they are separate integrals there is no problem with that. And then you would have \(\displaystyle \frac{1}{6}\int_0^{54}f(u)du+ \frac{1}{3}\int_0^{54} f(u)du= 85\)
\(\displaystyle \frac{1}{6}\int_0^{54}f(u)du+ \frac{2}{6}\int_0^{54} f(u)du= 85\)
\(\displaystyle \frac{3}{6}\int_0^{54} f(u)du= \frac{1}{2}\int_0^{54} f(u)du= 85\)
So what is \(\displaystyle \int_0^{54} f(u)du\)?
