The line integral is the path of the function along a line having a continuous value.
The integral solved gives the value 111.
The given question is
where C is the line from (1, 0, 0) to (4, 1, 3)
Here
x→ 1⇒ 4
y→ 0⇒1
z→ 0⇒ 3
Let t be defined be the range 0≤ t ≤ 1
Then x= 4t+1 : dx= 4dt
y= t ": dy= dt
z= 3t : dz= 3dt
Putting the values
= [36(1)²- 36(0)²]+ 3 [16(1)² +1+8(1)] - 3 [16(0)² +1+8(0)] + [3(1)²- 3(0)² ]
= 36 +75-3+3
= 111
