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
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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.
More to read
Proof of fundamental theorem of calculus
The fundamental theorem of calculus connects derivatives and integrals. It has two parts that say the same connection in two directions.
Part 1: if you build a function by accumulating area, g(x) = the integral from a to x of f(t) dt, then differentiating g gives you f back. So g'(x) = f(x).
Part 2: if you want the definite integral from a to b of f(x) dx, find any antiderivative F of f and compute F(b) minus F(a).
The first part says an accumulation function is always an antiderivative of the function you accumulated. The second part says you can evaluate any definite integral once you know one antiderivative, with no need for limits of sums. Together they explain why differentiation and integration undo each other.
5.3 The Fundamental Theorem of Calculus
This section is all about one big idea: differentiation and integration are inverse operations.
The Mean Value Theorem for Integrals says a continuous function on a closed interval hits its average value at least once.
Part 1 of the Fundamental Theorem says the accumulation function g(x) = the integral from a to x of f(t) dt has derivative f(x). Use it to differentiate integrals.
Part 2 says the definite integral from a to b of f equals F(b) minus F(a) for any antiderivative F. Use it to evaluate definite integrals fast.
Interpreting the behavior of accumulation functions
In differential calculus you studied a function f by looking at its derivative f'. In integral calculus you flip the direction: you study an accumulation function g(x) = the integral from 0 to x of f(t) dt by looking at the graph of f, because f is the derivative of g.
If f is positive on an interval, g is increasing there, since area keeps being added. If f is negative, g decreases.
Even where f has a jump or a corner, g stays continuous, because the extra area over a tiny interval is tiny. Where f crosses the x axis, g can switch from increasing to decreasing, which is where g has a local maximum or minimum.
So the whole toolkit from differential calculus, slopes, intervals of increase, concavity, comes back, just read from the graph of f.
Key Terms
Definite integral: a primary operation of calculus. The signed area between a curve and the x axis over an interval is a definite integral.
Antiderivative of a function f: a function F whose derivative is f, so F'(x) = f(x).
Fundamental theorem of calculus: the theorem that ties it all together. Its first part says differentiating an accumulation function recovers the original function. Its second part says a definite integral can be computed from any antiderivative as F(b) minus F(a).
Average value of a function: the definite integral over an interval divided by the length of the interval.
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Where it sits
20 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.