Find the exact length of the curve, y = ln1 – x², 0≤ x ≤ 16.

The exact length of a curve is a geometrical concept that addressed by integral calculus. It is a method for calculating the exact lengths of line segments.

Answer: The exact length of the curve y = ln1 – x², 0≤ x ≤ 16 is 257.093 units

Let’s solve it step by step.

Explanation: 

For the given curve, We will use the formula for the length of the arc(L) it.

Given function ⇒ y = ln 1 – x²

Now, differentiate the given function (y = ln 1 – x²) with respect to “x”

dy/dx = d( ln 1 – x²)/dx

dy/dx = 0 - 2x ------------- (1)

For finding arc length (L), we will use the following formula,

L = \(\int_{a}^{b}\) √[1 + (dy/dx)²] dx.

Substituting the value of dy/dx in length of curve formula from (1)

L = √[1 + (-2x)²] dx.

\(\int_{a}^{b} \sqrt{1+4 x^{2}} d x\)

Evaluate the indefinite integral

\(\int \sqrt{1+4 x^{2}} d x\)

Using the formula for integration by parts, we have

\(x\sqrt{1+4 x^{2}}-\int \dfrac{4x^2}{\sqrt{1+4x^2}}d x\)

= \(x\sqrt{1+4 x^{2}}-\int \dfrac{1-1+4x^2}{\sqrt{1+4x^2}}d x\)

= \(x\sqrt{1+4 x^{2}}+\int \dfrac{1}{\sqrt{1+4x^2}}d x - \int\sqrt{1+4x^2} dx\)

= \(2\int\sqrt{1+4x^2} dx=x\sqrt{1+4 x^{2}}+\frac{1}{2}\ln|x+\sqrt{1+4x^2}|\)

= \(\int\sqrt{1+4x^2} dx=\frac{1}{2}x\sqrt{1+4 x^{2}}+\frac{1}{4}\ln|x+\sqrt{1+4x^2}| + C\)

Now, determine the definite integral 

\(\int_0^{16} \sqrt{1+4 x^{2}} d x \)

= \(\frac{1}{2}x\sqrt{1+4 x^{2}}+\frac{1}{4}\ln|x+\sqrt{1+4x^2}| |_0^{16}\)

= \(8\sqrt{1+4\times16^{2}}+\frac{1}{4}\ln|16+\sqrt{1+4\times 16^2}| | - 0\)

= 257.093 units

L ≈ 257.093 units

Therefore, the exact length of the curve is 257.093 units.