🤔 Find out what you don't know with free Quizzes 🤔Start Quiz Now! # Evaluate $\int _ { C } ( x y + y + z ) d s$ along the curve $\mathbf { r } ( t ) = 2 \mathrm { ti } +$ $t \mathbf { j } + ( 2 - 2 t ) \mathbf { k } , 0 \leq t \leq 1$

## $\frac{13}{2}$

Integrals

Vectors

Vector Functions

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### Video Transcript

okay. What we want to dio is we want to evaluate, um, over the curve on the line integral of X y plus, why play Z d s? Um, along the curve R t is equal to to t i plus t j plus to minus two tea. Okay. And t t goes from 0 to 1 inclusive. Okay, so the first thing we need to do is we know that d s is equal to the magnitude of e of T. Um d t. And so first thing we need to do is we know that the of t is equal to the derivative, um, of that curve r t. So this is gonna be equal to two i plus J minus two K. And so, um, the magnitude of e of t is equal to the square root of t squared plus one squared plus a native to squared. And so this is gonna give me, um three. Um, and so Gs is equal to three d t. Okay, so this is gonna be the integral from 0 to 1. Um, X is to t. Why is t and Z is T minus two t and three D t so that seeing a girl we want to do, um, so this is gonna be equal to, um, three times, integral from 0 to 1. Ah, to t squared minus t plus two tt. So this is gonna be three times 2/3 t cute, minus 1/2 t squared, plus to t evaluated, um, 0 to 1. And so when we do that, um, we should get 13 house. University of Central Arkansas

#### Topics

Integrals

Vectors

Vector Functions

##### Top Calculus 3 Educators ##### Lily A.

Johns Hopkins University   ##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp