The Fundamental Theorem of Calculus
Both halves: differentiating an accumulation function returns the integrand, and a definite integral can be evaluated from any antiderivative. This is why the two calculus operations undo each other.
What a learner can do afterwards
- Differentiate an integral with a variable upper limit
- Evaluate a definite integral by antidifferentiation and justify the step
- Explain why an accumulation function is continuous even when the integrand is not
1 · Read
Part 1 builds an accumulation function g of x as the integral from a to x of f of t dt, then says its derivative is f of x. Differentiating the buildup returns the integrand with x in place of t, so the integral of sin from 0 to x differentiates back to sin of x.
Part 2 evaluates integrals from antiderivatives: find any F with derivative f, then the integral from a to b is F of b minus F of a. For 3x squared plus 1 from 0 to 2, use x cubed plus x: at 2 it is 10, at 0 it is 0, so the integral is 10. Added constants cancel in the subtraction, so any antiderivative works.
When the top limit is a function of x, pair the theorem with the chain rule. For the integral from 0 to x squared of cos, evaluate at the top to get cos of x squared, then multiply by the top derivative 2x, giving 2x cos of x squared. From a graph, f positive means g climbs, since g prime is f.
Accumulation smooths jumps: if f jumps at a point, g stays continuous there because the extra area over a tiny interval is tiny, though g may lose differentiability at the jump.
Differentiate accumulation to recover f, and evaluate integrals by subtracting one antiderivative at the limits.
2 · Watch
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.