Answer:
18
Step-by-step explanation:
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
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# 1
(1,1)(0,3)
slope = (3-1)/(0-1) = 2/-1 = -2
answer
-2 (first choice)
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#2
<span> Find the slope of a line that passes through (-2, -3) and (1, 1)
slope = (1 + 3)/(1 +2) = 4/3
answer
</span><span>d: 4/3</span>
.........................lol
⟨0,1⟩ could be taken to begin at the origin and end at (0,1). <span> ⟨0,1⟩ has the length/magnitude of 1 and direction j. So the desired unit vector is simply j.</span>