Which of the following is the arc length formula for y = f(x) from a to b?

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Multiple Choice

Which of the following is the arc length formula for y = f(x) from a to b?

Explanation:
The arc length of a curve comes from summing tiny straight segments. For a function y = f(x), a tiny horizontal change dx and vertical change dy form a small piece ds = sqrt(dx^2 + dy^2). Since dy = f′(x) dx, this becomes ds = sqrt(1 + (f′(x))^2) dx. Integrating these tiny lengths from a to b gives the arc length: ∫ from a to b of sqrt(1 + (f′(x))^2) dx. Why the others don’t fit: removing the square root or dropping the square changes the geometry; for instance, sqrt(1 + f′(x)) isn’t derived from ds = sqrt(dx^2 + dy^2) and isn’t the actual infinitesimal length. Likewise, integrating 1 + (f′(x))^2 ignores the necessary square root, misrepresenting the true length, and taking sqrt((f′(x))^2) gives |f′(x)|, missing the horizontal base dx and the 1 term under the radical.

The arc length of a curve comes from summing tiny straight segments. For a function y = f(x), a tiny horizontal change dx and vertical change dy form a small piece ds = sqrt(dx^2 + dy^2). Since dy = f′(x) dx, this becomes ds = sqrt(1 + (f′(x))^2) dx. Integrating these tiny lengths from a to b gives the arc length: ∫ from a to b of sqrt(1 + (f′(x))^2) dx.

Why the others don’t fit: removing the square root or dropping the square changes the geometry; for instance, sqrt(1 + f′(x)) isn’t derived from ds = sqrt(dx^2 + dy^2) and isn’t the actual infinitesimal length. Likewise, integrating 1 + (f′(x))^2 ignores the necessary square root, misrepresenting the true length, and taking sqrt((f′(x))^2) gives |f′(x)|, missing the horizontal base dx and the 1 term under the radical.

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