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How do you find the exact length of the curve y = ln(1 + x^2) from x = 0 to x = 18?

To find the exact length of the curve defined by the function y = ln(1 + x^2) from x = 0 to x = 18, we use the arc length formula. The formula for the length of a curve given by y = f(x) from a to b is:

L = ∫_a^b √(1 + (dy/dx)^2) dx

First, we need to find dy/dx. Differentiating y with respect to x, we get:

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

Now we can compute ((dy/dx)^2):

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

Next, we substitute this back into the arc length formula:

L = ∫_0^18 √(1 + (4x^2) / (1 + x^2)^2) dx

We simplify the expression:

L = ∫_0^18 √((1 + x^2)^2 + 4x^2) / (1 + x^2) dx

This simplifies further to:

L = ∫_0^18 (1 + x^2) / (1 + x^2) dx = ∫_0^18 dx

Thus, the actual length integral simplifies to:

L = [x]_0^18 = 18 - 0 = 18

Therefore, the exact length of the curve from x = 0 to x = 18 is 18 units.

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