   Chapter 16.2, Problem 4E

Chapter
Section
Textbook Problem

Evaluate the line integral, where C is the given curve.4. ∫C xey ds, C is the line segment from (2, 0) to (5, 4)

To determine

To Evaluate: The line integral C(xey)ds for the line segment from the point (2,0) to the point (5,4) .

Explanation

Given data:

The given curve C is a line segment from the point (2,0) to the point (5,4) .

Formula used:

Write the expression to evaluate the line integral for a function f(x,y) along the curve C .

Cf(x,y)ds=abf(x(t),y(t))(dxdt)2+(dydt)2dt (1)

Here,

a is the lower limit of the curve C and

b is the upper limit of the curve C .

Write the expression to find the parametric equations for a line through the point (x0,y0) and parallel to the direction vector v=a,b .

x=x0+at,y=y0+bt (2)

Write the expression to find direction vector v=a,b for a line segment from the point (x0,y0) to the point (x1,y1) .

a,b=x1x0,y1y0 (3)

Write the required differential formulae to evaluate the given integral.

ddttn=ntn1

Consider the points (x0,y0) as (2,0) and (x1,y1) as (5,4) .

Calculation of direction vector:

Substitute 5 for x1 , 4 for y1 , 2 for x0 , and 0 for y0 in equation (3),

a,b=52,40=3,4

Calculation of parametric equations of the curve:

Substitute 2 for x0 , 0 for y0 , 3 for a , and 4 for b in equation (2),

x=2+3t,y=0+4tx=2+3t,y=4t

Consider the limits of scalar parameter t are 0 to 1.

0t1

Find the expression (xey) as follows.

Substitute (2+3t) for x and 4t for y in the expression (xey) ,

xey=(2+3t)e4t

Evaluation of line integral C(xey)ds :

Substitute (xey) for f(x,y) , (2+3t)e4t for f(x(t),y(t)) , (2+3t) for x , 4t for y , 0 for a , and 1 for b in equation (1),

C(xey)ds=01(2+3t)e4t[ddt(2+3t)]2+(ddt4t)2dt

Rewrite and compute the expression as follows.

C(xey)ds=01(2+3t)e4t(0+3)2+(4)2dt=01(2+3t)e4t(5)dt

C(xey)ds=501(2+3t)e4tdt (4)

Compute the integration part 01(2+3t)e4tdt as follows

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