82. Find a curve through the point (1, 0) whose length from x=1 to x=2 is
L = ∫(from 1 to 2)√(1 + 1/x²)dx.
Verified step by step guidance
82. Find a curve through the point (1, 0) whose length from x=1 to x=2 is
L = ∫(from 1 to 2)√(1 + 1/x²)dx.
In Exercises 5–8, show that each function is a solution of the given initial value problem.
7. Differential Equation: xy' + y = -sin(x), x>0
Initial condition: y(π/2) = 0
Solution candidate: y = cos(x)/x
Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
75. lim (x → ∞) e^(x²) / (x e^x)
In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.
115. y = (x + 1)ˣ
Each of Exercises 1–4 gives a value of sinh x or cosh x. Use the definitions and the identity cosh²x - sinh²x = 1 to find the values of the remaining five hyperbolic functions.
2. sinh x = 4/3
Verify the integration formulas in Exercises 37–40.
39. ∫x coth⁻¹(x)dx = ((x²-1)/2)coth⁻¹(x) + x/2 + C