# Find the exact length of the curve. y = (1 - x^{2} ), 0 ≤ x ≤ 1/2.

**Solution:**

Given, y = (1-x^{2} ), 0 ≤ x ≤ 1/2

Length of the curve, y = f(x) from x =a to x = b is given by:

\( \int_{a}^{b}\sqrt{1+f'(x)^{2}} .dx \)

y = (1-x^{2})

Differentiating w.r.t. x,

dy/dx = (-2x)

Length of the curve,

= \( \int_{0}^{1/2}\sqrt{1+ [-2x]^{2}}.dx \)

= \( \int_{0}^{1/2}\sqrt{1+ 4x^{2}}.dx\)

= \( \frac{\ \frac{2x}{2} \times \sqrt{1 + 4x^{2}}\Biggr|_{0}^{1/2}\ + \frac{1}{2}\times log \left | 2x + \sqrt{1 + 4x^{2}} \right |\Biggr|_{0}^{1/2}}{2} \)

= \( \left \{ \left ( \frac{1}{4}\times \sqrt{2} \right ) -0 \right \}+ \frac{1}{4}\left \{ log(1+\sqrt{2})-log 0 \right \} \)

= \( \frac{\sqrt{2}}{4}+ \left ( \frac{1}{4}log\left ( \sqrt{2}+1 \right ) \right ) \)

## Find the exact length of the curve. y = (1 - x^{2} ), 0 ≤ x ≤ 1/2.

**Summary:**

The exact length of the curve, y = (1-x^{2}), 0 ≤ x ≤ 1/2 is √2/4 + (1/4)log[(√2+1)].

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