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: $\int{}_{1}^{3} 2x dx = [x^{2}]_{1}^{3} = 9 - 1 = 8$. 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: $\int{}_{0}^{2} (3x^{2} + 1) dx = [x^{3} + x]_{0}^{2} = (8 + 2) - (0 + 0) = 10$. Watch the order, upper minus lower limit, and put negative limits in brackets when substituting so signs do not slip: $[x^{2}]_{-1}^{1} = 1 - 1 = 0$, 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: $\frac{\int{}_{-1}^{1} 1}{x^{2}} dx$ "computed" with $F(x) = -\frac{1}{x}$ would give $-1 - 1 = -2$, a negative value for an everywhere positive function, obviously absurd. The error: $\frac{1}{x^{2}}$ 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.