Compute the length of the curve: y=((x^3)/6))+(1/2x) , 1 <_x<_2 Solution The len

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The length of a curve x = x(y) between y = a and y = b is given by integral (y = a ---> b) sqrt [1 + (dx/dy)^2] dy Here, dx/dy = (1/3)[sqrt y + (y - 3)/(2sqrt y)] (dx/dy)^2 = (1/9)[y + (y - 3)^2/(4y) + (y - 3)] = (1/9)[(y - 3)^2/(4y) + 2y - 3] The length of the curve = integral (y = 1 ---> 9) sqrt{1 + (1/9)[(y - 3)^2/(4y) + 2y - 3]} dy = integral (y = 1 ---> 9) sqrt{1 + (1/9)[y/4 - 3/2 + 9/(4y) + 2y - 3]} dy = integral (y = 1 ---> 9) sqrt{1 + (1/9)[9y/4 - 9/2 + 9/(4y)]} dy = integral (y = 1 ---> 9) sqrt{1 + [y/4 - 1/2 + 1/(4y)]} dy = integral (y = 1 ---> 9) sqrt[y/4 + 1/2 + 1/(4y)] dy = integral (y = 1 ---> 9) [sqrt y / 2 + 1/(2 sqrt y)] dy = [(1/3)y^(3/2) + y^(1/2)] (y = 1 ---> 9) = [(1/3)27 + 3] - [(1/3) + 1] = 12 - 4/3 = 32/3 or 10 2/3

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