integral of calculus
DESCRIPTION
INTEGRAL OF CALCULUS. Arc Length (Intro). Nikenasih B, M.Si Mathematics Educational Department Faculty of Mathematics and Natural Science State University of Yogyakarta. Contents. Problem and Aim Solution : A-R-L Method The Steps of Solution Examples Exercises. Problem and Aim. - PowerPoint PPT PresentationTRANSCRIPT
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INTEGRAL OF CALCULUSArc Length (Intro)
Nikenasih B, M.SiMathematics Educational Department
Faculty of Mathematics and Natural ScienceState University of Yogyakarta
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ContentsProblem and AimSolution : A-R-L MethodThe Steps of SolutionExamplesExercises
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Problem and Aim f function of xdomain a ≤ x ≤ b
x-axis
y-axis
f(x)
x = b
What is the length of this arc?
)(, afa )(, bfb
x = a
g(x)
x = b
)(, aga
)(, bgb
x = a
22
)()(, abagbgBAd
y-axis
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Solution : A-R-L MethodWhat we Know The length of a straight lineWhat we want to know The volume of an arbitrary curveHow we do it We approximate the curve with
polygonal (broken) lines.
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The steps of solution (1)
Approximate the region with n-straight lines, obtained by partitioning the interval [a,b]
Here we get n-subinterval [x0 = a,x1] , [x1,x2] , ... , [xn-1,xn = b]Determine points on the curve (x0,y0),
(x1,y1), (x2,y2), ... , (xn,yn), where
yi = f(xi), for all i = 1, 2, ..., n.
bxxxax n ,,,, 210
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The steps of solution (2)Construct the line segments joining
each consecutive pair of points, (x0,y0) with (x1,y1), (x1,y1) with (x2,y2), and so forth.
Compute the length of this polygonal path. Use the formula of distance between two points to do this. Thus
The first segment length is The second segment length is and so forth.
201
2
01xxyy
212
2
12xxyy
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Hence the total length of the polygon is
n
iiiiinxxyyL
1
2
1
2
1
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Example
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Exercise