Top minus bottom AB & BC

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then

$$\int_{a}^{b} f(x)\,dx = F(b) - F(a).$$

The standard notation for the middle step is a bracket with the limits attached, $\Big[F(x)\Big]_{a}^{b}$, and writing it is worth the two seconds it costs, because it puts the two substitutions side by side where a sign error is visible.

The word any is doing real work. Two antiderivatives of the same $f$ differ by a constant, so if $G(x) = F(x) + C$ then

$$G(b) - G(a) = \left[F(b) + C\right] - \left[F(a) + C\right] = F(b) - F(a).$$

The constant cancels. That is why nobody writes $+ C$ when evaluating a definite integral: it would appear twice and subtract itself away. Including it is not wrong so much as pointless, and it becomes wrong the moment it is added once instead of twice.

Two structural facts to keep straight from Topic 6.4. Part 1 differentiates and produces a function; Part 2 evaluates and produces a number. A definite integral answered with an expression in $x$ has used the wrong half of the theorem.

Top minus bottom, always in that order. $\int_{1}^{3} 2x\,dx = \Big[x^{2}\Big]_{1}^{3} = 9 - 1 = 8$, and $F(a) - F(b)$ would give $-8$, which is the answer to a question nobody asked.

The failure mode that actually costs marks is a negative value at the lower limit. Take $\int_{-1}^{2} 3x^{2}\,dx$:

$$\Big[x^{3}\Big]_{-1}^{2} = (2)^{3} - (-1)^{3} = 8 - (-1) = 9.$$

Written without the brackets, "$8 - 1$" appears and $7$ comes out. The integrand $3x^{2}$ is positive on the whole interval, so the answer has to exceed $8$, and $7$ fails that check instantly.

Three habits that close this off.

1. Parenthesize both substitutions before subtracting, especially when the antiderivative has more than one term. $\left(\frac{64}{3} - 48\right) - \left(\frac{1}{3} - 3\right)$ is safe; the same thing without brackets loses a sign on the last term.
2. Sanity-check the sign. A positive integrand on the interval must give a positive answer, and a negative one a negative answer.
3. Substitute into $F$, not into $f$. $f(b) - f(a)$ is a different number entirely, and it is what a hurried answer usually is.

Apply Part 2 to a derivative. If $F' = f$, then

$$\int_{a}^{b} F'(x)\,dx = F(b) - F(a),$$

which is the Net Change Theorem: integrating a rate of change over an interval gives the net change in the quantity. It is Topic 6.1's sentence with a symbol attached, and it is how almost every applied integral on the exam is set up.

1. $\int_{a}^{b} v(t)\,dt$ is the displacement, and $s(b) = s(a) + \int_{a}^{b} v(t)\,dt$.
2. $\int_{a}^{b} r(t)\,dt$ where $r$ is a flow rate is the net volume gained.
3. $\int_{a}^{b} C'(q)\,dq$ is the change in cost across a change in quantity.

Notice that the integral supplies a change. If a question asks for the amount at the end, the starting amount has to be added on separately, and it will be given.

The total-versus-net distinction survives intact. $\int_{a}^{b} v(t)\,dt$ is displacement; total distance is $\int_{a}^{b} |v(t)|\,dt$, and that absolute value is not decoration, it is an instruction to split.

Part 2 needs one antiderivative that works across the whole interval. An absolute value, or a function defined in pieces, does not have one, so the integral is split at the break and additivity from Topic 6.6 puts the pieces back together.

For $\int_{0}^{3} |2x - 4|\,dx$, the inside changes sign at $x = 2$:

$$\int_{0}^{2}(4 - 2x)\,dx + \int_{2}^{3}(2x - 4)\,dx = \Big[4x - x^{2}\Big]_{0}^{2} + \Big[x^{2} - 4x\Big]_{2}^{3} = 4 + 1 = 5.$$

Dropping the bars gives $\int_{0}^{3}(2x-4)\,dx = -3$, and taking the size of that gives $3$. Neither is the answer, because the two pieces have to be made positive before they are added, not after.

Two smaller property questions come up in the same breath.

1. Reversed limits. $\int_{5}^{2} f = F(2) - F(5)$, which is the same as negating $\int_{2}^{5} f$. Either route is fine; doing both is not.
2. A break outside the interval does not need a split. If a piecewise function changes formula at $x = 7$ and the integral runs from $0$ to $4$, only one formula is ever in play.