Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:
[tex]\int_a^b f(r) |r'| dt[/tex]
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector [tex]\sqrt{(x')^2 + (y')^2 + (z')^2}[/tex]
[tex]\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt [/tex]
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
[tex]\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u} du \\ \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u} du \\ \\ \frac{1}{4} \int_0^1 \sqrt{u} du[/tex]
After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
[tex]=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\ \\ =\frac{1}{6} (14^{3/2} - 1)[/tex]
