## Thomas' Calculus 13th Edition

$$5$$
We calculate the line integral by using substitution as follows: $$x=1+9t \\ \implies dx=9 dt \\ y= 1+2t \\ \implies dy=2 \space dt \\ z=1+2\space t \\ \implies dz=2 \space dt$$ We need to plug the above values in the given integral. $$\int_{0}^{1} 9 \space dt-\sqrt{\dfrac{1+2t}{1+2t}}(2 dt) -\sqrt{\dfrac{1+2t}{1+2t} } \times (2 \space dt)\\=\int_{0}^{1}9 dt-2 dt -2dt=5\int_0^1 dt\\=5$$