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INTEGRAL OF CALCULUS

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.

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INTEGRAL OF CALCULUS

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  1. INTEGRAL OF CALCULUS Arc Length (Intro) Nikenasih B, M.Si Mathematics Educational Department Faculty of Mathematics and Natural Science State University of Yogyakarta

  2. Contents • Problem and Aim • Solution : A-R-L Method • The Steps of Solution • Examples • Exercises

  3. Problem and Aim f function of x domain a ≤ x ≤ b y-axis y-axis f(x) g(x) x = b x = a x = a x = b x-axis What is the length of this arc?

  4. Solution : A-R-L Method • What we Know The length of a straight line • What we want to know The volume of an arbitrary curve • How we do it We approximate the curve with polygonal (broken) lines.

  5. 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.

  6. 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.

  7. Hence the total length of the polygon is

  8. Example

  9. Exercise

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