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Arc length Find the length of the curve y = x5/4 on the interval [0, 1]. (Hint: Write the arc length integral and let u² = 1+ (²)√x.)

Arc length Find the length of the curve y = x5/4 on the interval [0, 1]. (Hint: Write the arc length integral and let u² = 1+ (²)√x.)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I dont even know how to start this problem. can you show me the steps plz don't oversimplifed so I can follow and learn.

# 19) 79. Arc length Find the length of the curve y = x5/4 on the interval [0, 1]. (Hint: Write the arc length integral and let (2) √5².) u² = 1 +
Transcribed Image Text:# 19) 79. Arc length Find the length of the curve y = x5/4 on the interval [0, 1]. (Hint: Write the arc length integral and let (2) √5².) u² = 1 +
8.1.79 L = So and that da = L 64√x 25 = = 1+ = 25x¹/2 16 -u du 1024 625 - dx. Let u² = 1+ 1024 625 41/16 1024 625 ·(u² − u²) du (u³ – u) du. Substituting gives 2048 + 1763√//41 9375 25x¹/2 16 = ≈ 1.423. Then 2u du 1024 625 (u³/5 - u³/3) = ((√41/16) 5/5 - (√/41/16)³/3 - (1/5 - 1/3)) = = 25 32√x /41/16 dr. Note that √√x 1024 (12/13 625 15 + 1763√41 15360 - 16 25 (u² − 1),
Transcribed Image Text:8.1.79 L = So and that da = L 64√x 25 = = 1+ = 25x¹/2 16 -u du 1024 625 - dx. Let u² = 1+ 1024 625 41/16 1024 625 ·(u² − u²) du (u³ – u) du. Substituting gives 2048 + 1763√//41 9375 25x¹/2 16 = ≈ 1.423. Then 2u du 1024 625 (u³/5 - u³/3) = ((√41/16) 5/5 - (√/41/16)³/3 - (1/5 - 1/3)) = = 25 32√x /41/16 dr. Note that √√x 1024 (12/13 625 15 + 1763√41 15360 - 16 25 (u² − 1),
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