Find the length of the arc formed by y=(1/8)(-4x^(2)+2ln(x)) from x=2 to x=7?

Question

Can anyone help me with this problem and include step by step procedures? Thank you!

Answer 1

the answer is 45/ 2 + [1/4] ln [7/2]....just follow the formula, do your algebra, what is under √ will be a perfect square, easy to integrate. Do the work

Answer 2

the formula for the length of the curve (can also be called the arc) is

int (from a to b) sqrt(1+ (dy/dx)^2)dx  

(a and b are the limits and int means integrate)

y=(1/8)(-4x^(2)+2ln(x))

first simplify the equation

y=-0.5x^2+0.25ln(x)

dy/dx=-x+(1/4x)

(dy/dx)^2=((1/4x)-x)^2

(dy/dx)^2=((1/16x^2)-(2x/4x)+x^2)

(dy/dx)^2=((1/16x^2)-(1/2)+x^2)

now integrate

int ( from 2 to 7) sqrt(1+ (dy/dx)^2)dx  

int ( from 2 to 7) sqrt(1+ (1/16x^2)-(1/2)+x^2)dx  

int ( from 2 to 7) sqrt((1/16x^2)+(1/2)+x^2)dx  

int ( from 2 to 7) sqrt(((1/4x)+x)^2)dx  

int ( from 2 to 7) (1/4x)+x)dx  

( from 2 to 7)(0.25ln(x) +(x^2/2))

at 2 the answer is (2+(ln(2)/4))

at 7 the answer is ((49/2)+(ln(7)/4))

by subtracting the answers you will get  ((ln7-ln2)/4)+(45/2)