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The length of the arc of \( y = f(x) \) from \( x = a \) to \( x = b \) is

  • A.\( \displaystyle\int_{a}^{b}\sqrt{1 + y'^{2}}\,dx \)
  • B.\( \displaystyle\int_{a}^{b}\sqrt{1 + y^{2}}\,dx \)
  • C.\( \displaystyle\int_{a}^{b} y'\,dx \)
  • D.\( \displaystyle\int_{a}^{b}\sqrt{1 + y''^{2}}\,dx \)

Answer: A

An element of arc satisfies \( ds^{2} = dx^{2} + dy^{2} \), so

\[ ds \;=\; \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}\,dx, \]

and integrating gives the length. The slope, not the function value or the second derivative, is what enters, since it is the slope that fixes how much longer the curve is than its projection

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