Asked by maath
                Find the length of the curve
y = x^2 - (1/8)lnx, 1 ≤ x ≤ e
            
        y = x^2 - (1/8)lnx, 1 ≤ x ≤ e
Answers
                    Answered by
            oobleck
            
    ∫[1,e] √(1 + (2x - 1/(8x)))^2 dx
    
                    Answered by
            maath
            
    how would you simplify that to a fraction
    
                    Answered by
            Bosnian
            
    In google paste:
Find the length of the curve calculator
When you see list of results click on:
Arc Length Calculator for Curve - eMathHelp
When page be open in rectangle
Enter a function
paste:
x^2 - (1/8)lnx
Enter a lower limit: 1
Enter an upper limit: e
click on CALCULATE and follow steps
    
Find the length of the curve calculator
When you see list of results click on:
Arc Length Calculator for Curve - eMathHelp
When page be open in rectangle
Enter a function
paste:
x^2 - (1/8)lnx
Enter a lower limit: 1
Enter an upper limit: e
click on CALCULATE and follow steps
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