Find the length of the arc

Question

Find the length of the arc y=2x^2 para 1<=x<=2.

Answer 1

Recall arc length formula and list given information:

L = ∫[a,b] √[1+f'(x)²]dx

f(x) = 2x²

f'(x) = 4x

f'(x)² = 16x²

[a,b] = [1,2]

Rewrite the integral so the integrand is in √(a²+x²)dx form:

L = ∫[1,2] √[1+16x²]dx

L = ∫[1,2] √[16(1/16 + x²)]dx

L = 4∫[1,2] √(1/16 + x²)dx

Use the integration formula for ∫√(a²+x²)dx and make substitutions:

∫√(a²+x²)dx = (x/2)√(a²+x²) + (a²/2)ln|x+√(a²+x²)| + C

∫√(1/16 + x²)dx = (x/2)√(1/16 + x²) + (1/32)ln|x+√(1/16 + x²)| + C

Now plug in the bounds and solve for L:

L = 4[(x/2)√(1/16 + x²) + (1/32)ln|x+√(1/16 + x²)|] [1,2]

L = 4[(2/2)√(1/16 + 2²) + (1/32)ln|2+√(1/16 + 2²)| - (1/2)√(1/16 + (1/2)²) + (1/32)ln|1+√(1/16 + (1/2)²)|]

L ≈ 6.086

Therefore, the length of the arc y = 2x² from x=1 to x=2 is about 6.086 units!

I hope this helped!