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

Question

Isaiah Allen · Accepted Answer

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]

Isaiah Allen · Answer

undefined To find the arc length of the curve $ f(x) = x^2 - \frac{1}{8} \ln x $ over the interval $[1, 2]$, you can use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{{dy}}{{dx}}\right)^2} dx $$

where $a$ and $b$ are the endpoints of the interval, and $\frac{{dy}}{{dx}}$ represents the derivative of $f(x)$ with respect to $x$.

1. First, find the derivative of $f(x)$ which is $f'(x)$.
2. Then, plug $f'(x)$ into the formula for the integrand.
3. Integrate the expression from $x = 1$ to $x = 2$ to find the arc length.

This gives you the arc length of the curve over the interval $[1, 2]$.