Recent content by maogden

Ok I think I understand.. So what you're saying is... (from 0 to -1)∫ (2 + (3-2t)(1+2t))(0)dt + (from 0 to -1)∫ (2(2t))(3)dt + (from 0 to -1)∫ (2)(3-2t)(2)dt simplified: 4 * (from 0 to -1)∫ -4t^2 + 4t + 1 dt 4[(-4t^3)/3 + 2t^2 + t] dt <== from 0 to -1 4(4/3 + 2 - 1 - (0 + 0 +...

Ok I think I understand.. So what you're saying is... 0,-1\int

Sorry I didn't explain my work. I integrated both C1 and C2 from 0 to 1.. why are you integrating it from -1 to 0? And here's the rest of my work, C1: r=(1-t)<1,0,1> + t<2,3,1> r=<1+t,3t,1> (from 0 to 1)∫ (1+4t) (1)dt + (from 0 to 1)∫ (2) (1+t) (3) dt + (from 0 to 1)∫...

Homework Statement Find the line integral of ∫ x+yz dx + 2x dy + xyz dz C consists of line segments from (1,0,1) to (2,3,1) and from (2,3,1) to (2,5,2). Homework Equations r=(1-t)<r0> + t<r1> 0<t<1 The Attempt at a Solution I split up the two line segments into C1 and C2...