Topic 6.7 The Fundamental Theorem of Calculus and Definite Integrals: evaluate definite integrals using the second part of the Fundamental Theorem of Calculus.

What this topic is asking

The College Board (Topic 6.7) gives the second part of the Fundamental Theorem of Calculus (FTC), the evaluation tool: to compute $\int_a^b f(x)\,dx$, find any antiderivative $F$ of $f$ and evaluate $F(b) - F(a)$. This replaces the limit-of-Riemann-sums definition with a quick computation.

The evaluation theorem

The constant of integration is irrelevant here: if $F$ and $F + C$ are both antiderivatives, then $(F(b) + C) - (F(a) + C) = F(b) - F(a)$, so the $C$ cancels. For definite integrals you simply pick the cleanest antiderivative.

A worked evaluation

The net change theorem

A direct corollary of the FTC is the net change theorem: if $F' = f$ is a rate of change, then $\int_a^b f(x)\,dx = F(b) - F(a)$ is the net change in $F$ over $[a, b]$. So the integral of a velocity is the net displacement; the integral of a flow rate is the net amount accumulated. This connects the evaluation rule back to the Unit 6.1 accumulation idea: the area under a rate graph (now computed exactly via an antiderivative) is the net change in the quantity. The exam exploits this by giving a rate $f$ and asking for the net change, which you compute as $F(b) - F(a)$.

Why the two parts of the FTC fit together

Part 1 says differentiating an accumulation function recovers the integrand; Part 2 says a definite integral can be evaluated through any antiderivative. They are two faces of the same inverse relationship between differentiation and integration. Part 1 guarantees that antiderivatives exist (the accumulation function is one), and Part 2 lets you use whichever antiderivative is easiest. On the no-calculator section, evaluating definite integrals by Part 2 is among the most frequent tasks, so fluency with basic antiderivatives (the next topic) is what makes this fast. Always present the antiderivative in brackets with the limits, then the subtraction, so the grader can follow the work.

Combining the FTC with given function values

A frequent free-response pattern gives $f(a)$ and a rate $f'$, then asks for $f(b)$. By the net change theorem, $f(b) = f(a) + \int_a^b f'(x)\,dx$: the new value is the old value plus the accumulated change. You evaluate the integral (by an antiderivative, or numerically on the calculator part) and add it to the known starting value. This structure appears for position from velocity, amount from a flow rate, and temperature from a heating rate. The recurring error is reporting the integral alone, which is only the change, and forgetting to add the initial value $f(a)$. Writing the starting-value-plus-integral form explicitly keeps the two pieces straight.

Substitution with definite integrals

When the antiderivative requires a $u$-substitution, you have two clean options on a definite integral: change the limits to $u$-values and evaluate entirely in $u$, or find the antiderivative in $x$ and use the original $x$-limits. Both are valid; the limit-changing route avoids back-substitution and is usually faster. The pairing with the FTC is seamless, since once you have any antiderivative, Part 2 evaluates the definite integral as upper-minus-lower. Keeping the limits matched to the variable you evaluate in, $u$-limits with a $u$-antiderivative, $x$-limits with an $x$-antiderivative, prevents the classic mismatch error.