Find the length of the arc formed by `y= 1/(8(-2x^2+4ln(x)))` from x = 3 to x = 8.
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You need to use the following formula to evaluate the given arclength such that:
`l = int_a^b sqrt(1 + (f'(x))^2)dx`
Reasoning by analogy yields:
`int_3^8 sqrt(1 + ((1/8(-2x^2+4ln(x)))')^2)dx`
You need to evaluate the derivative of the function `f(x) = 1/8(-2x^2+4ln(x))` , using quotient rule, such that:
`f'(x) = (-(-16x^2+32lnx)')/(64(-2x^2+4lnx)^2)`
`f'(x) = (32x- 32/x)/(64(-2x^2+4lnx)^2)`
Factoring out 32 yields:
`f'(x) = (x - 1/x)/(2(-2x^2+4lnx)^2)`
Raising to square yields:
`f'(x)^2 = (x - 1/x)^2/(4(-2x^2+4lnx)^4)`
You need to evaluate the integral`int_3^8 sqrt(1 + ((1/8(-2x^2+4ln(x)))')^2)dx` such that:
`int sqrt(1 + (x - 1/x)^2/(4(-2x^2+4lnx)^4))dx`
`int (sqrt (4(-2x^2+4lnx)^4 + (x - 1/x)^2))/(2(-2x^2+4lnx)^2)dx`
Hence, evaluating the arclength of the given curve yields `l =int_3^8 (sqrt (4(-2x^2+4lnx)^4 + (x - 1/x)^2))/(2(-2x^2+4lnx)^2)dx.`
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