find arc length of the curve y=(2/3)x*(3/2)-4 from x=1 to x=4 Solution below pro

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below procedure is correct . plese change the equation to get correct result. The problem is that the resulting integrand is undefined at x = 0 (since dy/dx = (2/3)x^(-1/3)). However, let's instead write this in terms of y: x = ± y^(3/2) Let's express the bounds in terms of y: Since y = x^(2/3), x = 0 ==> y = 0 x = -1 ==> y = 1 x = 8 ==> y = 4 Since this graph is symmetric about the y-axis, it suffices to compute the arc length from x = 0 to x = 1 plus that from x = 0 to x = 8. In terms of y, this is computing the arc length from y = 0 to y = 1 plus that from x = 0 to y = 4. (Now, we can ignore the minus signs, and also makes sure we get the arc to the left of the y-axis too.) Since the arc length formula is given by ? sqrt[1 + (dx/dy)^2] dy, for this problem, we have L = ?(y = 0 to 1) sqrt[1 + {(3/2)y^(1/2)}^2] dy + ?(y = 0 to 4) sqrt[1 + {(3/2)y^(1/2)}^2] dy = ?(y = 0 to 1) sqrt(1 + 9y/4) dy + ?(y = 0 to 4) sqrt(1 + 9y/4) dy = (1/2) [?(y = 0 to 1) sqrt(4 + 9y) dy + ?(y = 0 to 4) sqrt(4 + 9y) dy] = (1/2) [(1/9)(2/3)(4 + 9y)^(3/2) {for y = 0 to 1} +(1/9)(2/3)(4 + 9y)^(3/2) {for y = 0 to 4} = (1/27)[13^(3/2) + 40^(3/2) - 16]

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