Find the length of the curve y = 2( x + 4)^3/2 on the interval [0,2] . (Round to 1 decimal place)

Question

Find the length of the curve y = 2( x + 4)^3/2 on the interval [0,2] . (Round to 1 decimal place)

Answer 1

L = ∫₀² √(1+(dy/dx)²)dx

dy/dx = 3(x+4)^1/2

(dy/dx)²=9(x+4) = 9x+36

L = ∫₀² √(1+9x+36) dx

Let u = 9x+37, so du = 9dx. When x = 0, u = 37, when x = 2, u = 55

L =1/9 ∫₃₇⁵⁵ u^1/2du

You should be able to take it from there. Confirm here:

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