# Line integral w/ constants only only in integrand

1. Jul 27, 2006

### AO eye 5

Hello,

I want help in the line integration of:

Integral( 1 dy + 3 dx ), over the curve C. Where C is the union of two line segments:

Line 1 from point (0,0) to (1, -3)

Line 2 from point (1, -3) to (2,0)

The thing is I do not know what to do with the integrand being composed of constants, as I am only aware of line integral examples with functions in the integand and subsequent substitution with parameterizations.

Hints are welcome, I am stuck upon the integrand substitution with the parameterization of the lines into the constant integrand.

I have C parameterized as:

Line 1: x = t y = -3t t [0,1]
Line 2: x = t+1 y = 3t-3 t [1,2]

2. Jul 27, 2006

### benorin

line integrals...

$$\int_{C} 1 \, dy + 3 \, dx$$

where C is the union of line 1 and line 2. Now on line 1, we have x=t, y=-3t, $$t\in \left[ 0,1\right]$$ so that dx = dt and dy = -3dt on line 1, the integral over line 1 is then

$$\int_{\mbox{Line 1}} 1 \, dy + 3 \, dx = \int_{0}^{1} 1(-3 dt)+3 (dt) = 0$$

now you do the integral over Line #2.

3. Jul 27, 2006

### HallsofIvy

Staff Emeritus
So on line 1, dx= dt, dy= -3dt. The integral on that is:
$$\int_0^1 1(-3dt)+ 3(dt)= 0\int_0^1dt= 0$$

On line 2, dx= dt, dy= 3dt. The integral on the second line is:
$$\int_0^1 1(3dt)+ 3(dt)= 6\int_0^1dt= 6$$

Actually, there's an easy way to do that: $\int_a^b dx= b-a$ so the integral of dy as y goes from 0 to -3 is -3, the integral of 3dx as x goes from 0 to 1 is 3, the integral of dy as y goes from -3 to 0 is 3, the integral of 3dx as x goes from 1 to 2 is 3(1)= 3. The entire integral is
-3+ 3+ 3+ 3= 6.