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Chapter 16.2, Problem 5E
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### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Evaluate the line integral, where C is the given curve.5. ∫C (x2y + sin x) dy, C is the arc of the parabola y = x2 from (0, 0) to (π, π2)

To determine

To Evaluate: The line integral C(x2y+sinx)dy for the arc of the parabola y=x2 from the point (0,0) to the point (π,π2) .

Explanation

Given data:

The given curve C is the arc of the parabola y=x2 from the point (0,0) to the point (π,π2) .

Formula used:

Write the expression to evaluate the line integral with respect to arc length.

Cf(x,y)dy=abf(x(t),y(t))y(t)dt (1)

Here,

a is the lower limit of x-coordinate of the curve C and

b is the upper limit of x-coordinate of the curve C .

As the equation of parabola is y=x2 , choose x as parameter and write the parametric equations of the curve as follows.

x=x,y=x2

From the given two points, choose x-coordinates as limits to evaluate the integral C(x2y+sinx)dy .

0xπ

Find the expression (x2y+sinx) as follows.

Substitute x2 for y in the expression (x2y+sinx) ,

x2y+sinx=x2(x2)+sinx=x4+sinx

Calculation y(t)dt :

Write the equation of parabola as follows.

y=x2

Differentiate on both sides of the expression with respect to x .

ddx(y)=ddx(x2)dydx=2xdy=2xdx

Evaluation of line integral C(x2y+sinx)dy :

Substitute (x2y+sinx) for f(x,y) , (x4+sinx) for f(x(t),y(t)) , 2xdx for y(t)dt , 0 for a , and π for b in equation (1),

C(x2y+sinx)dy=0π(x4+sinx)2xdx=(2)0π(x5+xsinx)dx=(2)[0π(x5)dx+0π(xsinx)dx]=(2)([x66]0π+0π(xsinx)dx)

Simplify the expression as follows.

C(x2y+sinx)dy=(2)([π66066]+0π(xsinx)dx)

C(x2y+sinx)dy=(2)[16π6+0π(xsinx)dx] (2)

Compute the integration part 0π(xsinx)dx as follows

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