Okay. What we want to go ahead and do is step through how to integrate the function over the given curve. And for this particular case, we have the function x and y equal to, um X plus Why squared divided by the square root of one plus X squared and the curb is gonna be defined by why equal to X squared over two. And we're going from 1 1/2 two 00 Okay. And so what we need to do, first of all, is to define r of t. Um and so that is going to be equal to t I plus t squared over two. Okay. And so, um, then we need to take the derivative of our which is gonna give us V of tea. And so this is gonna be I plus TJ. And so the magnitude of e of t is gonna be the square root of one squared plus t squared. And so D s is equal to the square root of one place. T squared TT Okay. And so now I think we're okay with setting up our integral. So our inter goal is actually going to go from 1 to 0 or typically we like or smaller number two B or lower limit. So what we can go ahead and do is to, um this is gonna be negative, um, zero toe one. And it's because we're going from 1 to 0 in the X value, and so t goes from 1 to 0. And so this is gonna be, um, the integral. So this is gonna be t plus, um, t squared over, too. And that will be squared, divided by the square root of one plus t squared. And this is gonna be times the square root of one plus t squared T t. Okay. And so if you notice now, the one plus t squared, I'm reduced down to a one. And so my integral really becomes, um, negative, um, zero toe one of, um t plus, um, 1/4 t to the fourth. And we're integrating with respect to t. So this is gonna be equal to 1/2 and I have course, Have my negative sign out there. Um, t squared. Um, plus 1 20 year t to the fifth. Um, and we're gonna evaluate it at one and then again at zero. And so when we evaluated at one. Um, we get negative of 1/2 plus a 1/20 which is going to give me negative 11/20.

## Discussion

## Video Transcript

Okay. What we want to go ahead and do is step through how to integrate the function over the given curve. And for this particular case, we have the function x and y equal to, um X plus Why squared divided by the square root of one plus X squared and the curb is gonna be defined by why equal to X squared over two. And we're going from 1 1/2 two 00 Okay. And so what we need to do, first of all, is to define r of t. Um and so that is going to be equal to t I plus t squared over two. Okay. And so, um, then we need to take the derivative of our which is gonna give us V of tea. And so this is gonna be I plus TJ. And so the magnitude of e of t is gonna be the square root of one squared plus t squared. And so D s is equal to the square root of one place. T squared TT Okay. And so now I think we're okay with setting up our integral. So our inter goal is actually going to go from 1 to 0 or typically we like or smaller number two B or lower limit. So what we can go ahead and do is to, um this is gonna be negative, um, zero toe one. And it's because we're going from 1 to 0 in the X value, and so t goes from 1 to 0. And so this is gonna be, um, the integral. So this is gonna be t plus, um, t squared over, too. And that will be squared, divided by the square root of one plus t squared. And this is gonna be times the square root of one plus t squared T t. Okay. And so if you notice now, the one plus t squared, I'm reduced down to a one. And so my integral really becomes, um, negative, um, zero toe one of, um t plus, um, 1/4 t to the fourth. And we're integrating with respect to t. So this is gonna be equal to 1/2 and I have course, Have my negative sign out there. Um, t squared. Um, plus 1 20 year t to the fifth. Um, and we're gonna evaluate it at one and then again at zero. And so when we evaluated at one. Um, we get negative of 1/2 plus a 1/20 which is going to give me negative 11/20.

## Recommended Questions

In Exercises $27 - 30 ,$ integrate $f$ over the given curve.

$$

\begin{array} { l } { f ( x , y ) = \left( x + y ^ { 2 } \right) / \sqrt { 1 + x ^ { 2 } } , \quad C : \quad y = x ^ { 2 } / 2 \text { from } ( 1,1 / 2 ) \text { to } } \\ { ( 0,0 ) } \end{array}

$$

In Exercises $27 - 30 ,$ integrate $f$ over the given curve.

$$

\begin{array} { l } { f ( x , y ) = x ^ { 2 } - y , \quad C : \quad x ^ { 2 } + y ^ { 2 } = 4 \text { in the first quadrant from } } \\ { ( 0,2 ) \text { to } ( \sqrt { 2 } , \sqrt { 2 } ) } \end{array}

$$

In Exercises $27 - 30$ , integrate $f$ over the given curve.

$$

\begin{array} { l } { f ( x , y ) = x + y , \quad C : \quad x ^ { 2 } + y ^ { 2 } = 4 \text { in the first quadrant from } } \\ { ( 2,0 ) \text { to } ( 0,2 ) } \end{array}

$$

Integrate $f$ over the given curve.

$f(x, y)=x^{3} / y, \quad C: \quad y=x^{2} / 2, \quad 0 \leq x \leq 2$

Apply Green's Theorem to evaluate the integrals in Exercises $27-30 .$

$$\oint(3 y d x+2 x d y)$$

$C:$ The boundary of $0 \leq x \leq \pi, 0 \leq y \leq \sin x$

(a) Find a function $ f $ such that $ \textbf{F} = \nabla f $ and (b) use part (a) to evaluate $ \int_C \textbf{F} \cdot d \textbf{r} $ along the given curve $ C $.

$ \textbf{F}(x, y) = x^2y^3 \, \textbf{i} + x^3y^2 \, \textbf{j} $,

$ C $: $ \textbf{r}(t) = \langle t^3 - 2t, t^3 + 2t \rangle $, $ 0 \leqslant t \leqslant 1 $

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

$$\int_{C} y^{3} d s, \quad C : x=t^{3}, y=t, 0 \leqslant t \leqslant 2$$

Evaluate $\int _ { C } \frac { x ^ { 2 } } { y ^ { 4 / 3 } } d s ,$ where $C$ is the curve $x = t ^ { 2 } , y = t ^ { 3 } ,$ for $1 \leq t \leq 2$

Evaluate each line integral using the given curve $C$.

$\int x^{2} d x+d y+y d z ; \mathbf{C}$ is the curve $\mathbf{r}(t)=\left\langle t, 2 t, t^{2}\right\rangle,$ for

$0 \leq t \leq 3$

Find the line integral of $f ( x , y ) = y e ^ { x ^ { 2 } }$ along the curve $\mathbf { r } ( t ) = 4 t \mathbf { i } - 3 t \mathbf { j } , - 1 \leq t \leq 2$