How do you find the arc length of the curve #f(x)=x^2-1/8lnx# over the interval [1,2]?

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Answer 1 · 2018-05-21T13:00:14

Length of arc is # 3 +1/8 ln (2)#

#f(x) = x^2- 1/8 ln |x| :.f^'(x) = 2 x - 1/( 8 x)#

# :.[f^'(x)]^2+1 = (2 x - 1/( 8 x))^2+1# or

# :.[f^'(x)]^2+1 = (2 x)^2 - 2*2 cancelx*1/(8cancelx)+(1/( 8 x))^2+1# or

# :.[f^'(x)]^2+1 = (2 x)^2 - 1/2+(1/( 8 x))^2+1# or

# :.[f^'(x)]^2+1 = (2 x)^2 + 1/2+(1/( 8 x))^2# or

# :.[f^'(x)]^2+1 = (2 x + 1/( 8 x))^2#

Length of arc is #L=int_1^2 (sqrt [f^'(x)^2+1] dx#

#L=int_1^2 sqrt ((2 x + 1/( 8 x))^2 ) dx# or

#L=int_1^2 (2 x + 1/( 8 x)) dx# or

#L= [x^2 +1/8 ln(x)]_1^2# or

#L= 4-1 +1/8 [ln(2)- ln(1)]# or

#L= 3 +1/8 ln (2) [ln 1=0]#

Length of arc is # 3 +1/8 ln (2)# [Ans]