Calculus 1 — 26.4: Fundamental Theorem of Calculus, Part 1 скачать в хорошем качестве

Calculus 1 — 26.4: Fundamental Theorem of Calculus, Part 1

The derivative of an integral just returns the original function — even when that integral has no elementary antiderivative. This video explains the Fundamental Theorem of Calculus, Part 1, showing why differentiation undoes integration and how to apply the theorem instantly to problems that standard techniques cannot solve. Key concepts covered: • Defining the area function A(x) = ∫ₐˣ f(t) dt as a function of the variable upper bound x • Statement of FTC Part 1: d/dx[∫ₐˣ f(t) dt] = f(x) • Why the lower bound vanishes under differentiation (it produces a constant, and d/dx of a constant is zero) • Dummy variables: why t, u, and s are interchangeable placeholders inside the integral • Verification with a concrete example: d/dx[∫₁ˣ t⁴ dt] = x⁴, confirmed both by the theorem and by direct antidifferentiation • Applying FTC Part 1 to integrands with no elementary antiderivative, such as sin(t)/t and e^(t²) • The continuity requirement: f(t) must be continuous on the interval for the theorem to apply • Common misconception: FTC Part 1 gives the derivative of the integral, not the value of the integral itself • Preview of the chain rule extension when the upper bound is a function g(x) instead of plain x • Practice problems: d/dx[∫₂ˣ e^(t²) dt] and d/dx[∫₀ˣ cos(t³) dt] ━━━━━━━━━━━━━━━━━━━━━━━━ SOURCE MATERIALS The source materials for this video are from    • Calculus 1 Lecture 4.5:  The Fundamental T...