The Fundamental Theorem of Calculus and Definite Integrals AB & BC

The Fundamental Theorem of Calculus, Part 2, says that for continuous $f$ and any antiderivative $F$, $\int_{a}^{b} f(x)\,dx = F(b) - F(a)$. Any antiderivative works because two of them differ by a constant and the constant appears in both substitutions, so it cancels; that is also why $+ C$ is never written when evaluating a definite integral. Part 1 produces a function and Part 2 produces a number, so a definite integral answered with an expression in $x$ has used the wrong half.

Nearly all the losses are in the subtraction. Top minus bottom, with both substitutions in brackets: $\int_{-1}^{2} 3x^{2}dx = \big[x^{3}\big]_{-1}^{2} = 8 - (-1) = 9$, where writing $8 - 1$ gives $7$ for an integrand that is positive throughout. Applied to a derivative the theorem reads $\int_{a}^{b} F'(x)dx = F(b) - F(a)$, the Net Change Theorem, so an integral of a rate gives a change and the starting amount must be added separately. An absolute value or a piecewise integrand has no single antiderivative across the interval, so it is split at the break and reassembled by additivity.