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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