Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
Then r' = (1, 2t, 3t^2)
General Line integral is:
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
After integrating using power rule, replace 'u' with function for 't' and evaluate limits: