Evaluate $\int_{C} x d s,$ where $C$ is a. the straight-line segment $x=t, y=t / 2,$ from $(0,0)$ to $(4,2) .$ b. the parabolic curve $x=t, y=t^{2},$ from $(0,0)$ to $(2,4)$

So now we have to have sin function to path f equals to swear the explosives to oi So only path one we have our equals. Two two 40 auntie's from 0 to 1. Let's see, so the purity will be. I simply want four and the model of it will be kun swerves and tin so the patterns will will be transformed. They just want to do to one screw. The tea comes T plus 80 times swirl of 17 DT. The answer is simply to Times Square with a 17 let's see four dissident path, it's t zero. So the module of the derivative it's the module of 10 So which is what Here and over af I'll see will be transformed to 0 to 1 spirit of one plus two TD t plus Do too. Two. Sure, the one plus two Todt. It's simply title here. Sit here. Who has zero? Two of them, Cyril Mystery call that this is actually only for half of the path. So in our case, B, we have Ah si was to see one policy to and this is actually foresee war in no case be and we just forgot see, too. So let's read it out. We'll see. True, we have won t hints The directive for C two would be, um, still one. Okay, so plug in this, too. We have 0210 to 2. That's Chris wants to the range of our tea and the valued. This two intervals we have fear or two for us. Phantoms were five months, one over three. Silly answer will be five times sort of by plus one over three. That's for party, because it's a composite, composite path way have come up.

## Discussion

## Video Transcript

So now we have to have sin function to path f equals to swear the explosives to oi So only path one we have our equals. Two two 40 auntie's from 0 to 1. Let's see, so the purity will be. I simply want four and the model of it will be kun swerves and tin so the patterns will will be transformed. They just want to do to one screw. The tea comes T plus 80 times swirl of 17 DT. The answer is simply to Times Square with a 17 let's see four dissident path, it's t zero. So the module of the derivative it's the module of 10 So which is what Here and over af I'll see will be transformed to 0 to 1 spirit of one plus two TD t plus Do too. Two. Sure, the one plus two Todt. It's simply title here. Sit here. Who has zero? Two of them, Cyril Mystery call that this is actually only for half of the path. So in our case, B, we have Ah si was to see one policy to and this is actually foresee war in no case be and we just forgot see, too. So let's read it out. We'll see. True, we have won t hints The directive for C two would be, um, still one. Okay, so plug in this, too. We have 0210 to 2. That's Chris wants to the range of our tea and the valued. This two intervals we have fear or two for us. Phantoms were five months, one over three. Silly answer will be five times sort of by plus one over three. That's for party, because it's a composite, composite path way have come up.

## Recommended Questions

Evaluate $\int _ { C } \sqrt { x + 2 y } d s ,$ where $C$ is

a. the straight-line segment $x = t , y = 4 t ,$ from $( 0,0 )$ to $( 1,4 )$ .

b. $C _ { 1 } \cup C _ { 2 } ; C _ { 1 }$ is the line segment from $( 0,0 )$ to $( 1,0 )$ and $C _ { 2 }$ is

the line segment from $( 1,0 )$ to $( 1,2 )$ .

Evaluate $\int_{C} \sqrt{x+2 y} d s,$ where $C$ is

a. the straight-line segment $x=t, y=4 t,$ from (0,0) to (1,4).

b. $C_{1} \cup C_{2} ; C_{1}$ is the line segment from (0,0) to $(1,0),$ and $C_{2}$ is the line segment from (1,0) to (1,2).

Evaluate the line integral along the curve C.

$$

\begin{array}{l}{\int_{C}(x+2 y) d x+(x-y) d y} \\ {C: x=2 \cos t, y=4 \sin t \quad(0 \leq t \leq \pi / 4)}\end{array}

$$

Evaluate the line integral, where $C$ is the given curve.

$$\begin{array}{l}{\int_{C}(x+2 y) d x+x^{2} d y, \quad C \text { consists of line segments }} \\ {\text { from }(0,0) \text { to }(2,1) \text { and } \text { from }(2,1) \text { to }(3,0)}\end{array}$$

Evaluate the line integral.

$\int_{C} 2 y d s,$ where $C$ is the portion of $y=x^{2}$ from (0,0) to (2, 4), followed by the line segment to (3,0)

Evaluate the line integral, where $C$ is the given curve.

$$\begin{array}{l}{\int_{C} x^{2} d x+y^{2} d y, \quad C \text { consists of the arc of the circle }} \\ {x^{2}+y^{2}=4 \text { from }(2,0) \text { to }(0,2) \text { followed by the line }} \\ {\text { segment from }(0,2) \text { to }(4,3)}\end{array}$$

Evaluate the line integral.

$\int_{C} 2 x d x,$ where $C$ is the portion of $y=x^{2}$ from (2,4) to (0,0)

Evaluate the line integral.

$\int_{C} 3 y^{2} d y,$ where $C$ is the portion of $y=x^{2}$ from (2,4) to (0,0)

Evaluate the line integral with respect to $s$ along the curve $C .$

$$

\begin{array}{l}{\int_{C} \frac{x}{1+y^{2}} d s} \\ {C: x=1+2 t, \quad y=t \quad(0 \leq t \leq 1)}\end{array}

$$

Evaluate the line integral with respect to $s$ along the curve $C .$

$$

\begin{array}{l}{\int_{C} \frac{e^{-z}}{x^{2}+y^{2}} d s} \\ {C: \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}+t \mathbf{k} \quad(0 \leq t \leq 2 \pi)}\end{array}

$$