Fundamental Theorem of Calculus

It connects the two great operations of calculus: differentiation and integration are inverses of each other. Part 1 says: the integral function I(x) = ∫ₐˣ f(t) dt of a continuous function f is differentiable, and its derivative is f again. Part 2 provides the computation rule: ∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F of f. The practical significance is enormous: instead of laboriously determining areas as limits of rectangle sums, you find an antiderivative and evaluate it at two points. Example: ∫₁³ 2x dx = [x²]₁³ = 9 − 1 = 8. Without the fundamental theorem there would be no practicable integral calculus at school.

In three steps. First: find an antiderivative F of the integrand using the basic integrals; you may drop the +C. Second: write the bracket notation, [F(x)]ₐᵇ. Third: insert the upper limit, insert the lower limit, subtract: F(b) − F(a). Example: ∫₀² (3x² + 1) dx = [x³ + x]₀² = (8 + 2) − (0 + 0) = 10. Watch the order, upper minus lower limit, and put negative limits in brackets when substituting so signs do not slip: [x²]₋₁¹ = 1 − 1 = 0, not 1 + 1. The result can be negative; that is not an error but means area below the x-axis.

Because it cancels itself during the subtraction. If you take the antiderivative F + C instead of F, you compute (F(b) + C) − (F(a) + C) = F(b) − F(a); the C cancels exactly, whatever its value. That is why the value of a definite integral does not depend on which antiderivative from the family you choose, and for simplicity one takes the one with C = 0. This also explains the formal difference: the indefinite integral ∫f(x) dx is a set of functions and needs the +C, the definite integral ∫ₐᵇ f(x) dx is a single number and does not. Carrying a C through a definite integral in an exam rather signals uncertainty.

An integral function fixes the lower limit and lets the upper one run: I(x) = ∫ₐˣ f(t) dt. It measures the signed area under f accumulated up to x. The fundamental theorem (part 1) says: I is differentiable with I′(x) = f(x), so every integral function is an antiderivative of f. The converse does not fully hold: not every antiderivative is an integral function, because integral functions always have a zero at x = a (the area there is 0). Typical exam questions: "Justify that I(a) = 0", "Where does I have extrema?" (at the zeros of f with sign change) and "Where is I increasing?" (where f is positive).

The theorem requires a continuous integrand on the entire interval of integration. If f has a pole there, you must not simply evaluate an antiderivative at the limits. Cautionary example: ∫₋₁¹ 1/x² dx "computed" with F(x) = −1/x would give −1 − 1 = −2, a negative value for an everywhere positive function, obviously absurd. The error: 1/x² is not defined at x = 0, the integral does not even exist there (it diverges). Therefore always check before computing whether the integrand is defined and continuous throughout [a; b]. Jump discontinuities are less dramatic: there you split the integral into subintervals and apply the theorem piecewise.