Finding the arc length of f(x)= 4√x^3 from x = 4 to x = 7

Question

Find the arc length of the graph of the function f(x) = 4√x³ from x=4 to x=7

I keep getting 42.3, but it's not the right answer and I don't know where I'm going wrong.

Answer 1

It's possible you might be rounding at an intermediate step or two. You're very close to the answer I get when you save the rounding for the end. This is the integral you're evaluating:

L = ∫√(1 + 36x) dx from x = 4 to x = 7

This is what this integral evaluates to (after doing u-substitution):

L = 1/54 * [253 * √253 - 145 * √145]

Using a calculator, this evaluates to 42.1885 rounded to four decimal places.

Hope this is helpful. Let me know if you need more steps, but I imagine you're already doing the u-substitution and set-up correctly.

Answer 2

f(x) = 4√(x³)

f(x) = 4x^3/2

f '(x) = (3/2)4x^1/2

f '(x) = 6x^1/2

[f '(x)]² = 36x

L_4→7 = ₄∫⁷√(1 + 36x) dx = ₄∫⁷(1 + 36x)^1/2 dx

let u = 1 + 36x

du/dx = 36

dx = du/36

So, since u = 1 + 36x. then when x = 4, u = 145 and when x = 7, u = 253

So the integral becomes:

₁₄₅∫²⁵³ u^1/2•du/36

(1/36) (2/3)(u^3/2) evaluated between 145 and 253

(1/54)(u^3/2) evaluated between 145 and 253

(1/54)[253^3/2 - 145^3/2) = 42.1885 rounded

The exact answer is messy having square roots but can be written as well:

(1/54)(√16194277 - √3048625)