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Why is the derivative A'(x) equal to f(x)f(x) for the accumulation function A(x)A(x)?

The derivative A′(x)A'(x) equals f(x)f(x) because moving the endpoint from xx to x+hx+h adds a thin strip whose area increment is approximately f(x)hf(x)h. Dividing this increment by hh and taking the limit as hh approaches zero yields the local average height, which continuity ensures approaches the exact endpoint height f(x)f(x). Thus, the rate of change of the accumulated area is exactly the value of the function at that point.

Conditions

  • The function ff is continuous at xx.
  • A(x)A(x) is defined as the accumulation of ff from a fixed lower limit to xx.

Reasoning, step by step

  1. Consider the increment ΔA=A(x+h)−A(x)\Delta A = A(x+h) - A(x).
  2. Approximate ΔA\Delta A as f(x)h+o(h)f(x)h + o(h) for small hh.
  3. Form the difference quotient ΔAh≈f(x)+o(h)h\frac{\Delta A}{h} \approx f(x) + \frac{o(h)}{h}.
  4. Take the limit as h→0h \to 0.
  5. Conclude that A′(x)=f(x)A'(x) = f(x) due to the continuity of ff.

Example

The script explains: 'Divide the increment by h and let h approach zero. Continuity makes the local average height approach the endpoint height, so A′(x)=f(x)f(x). This does not claim a finite-width curved strip is already an exact rectangle.'

Common misconceptions

  • Believing that a finite-width curved strip is already an exact rectangle.
  • Confusing the finite increment ΔA\Delta A with the exact differential dAdA.

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