Chapter 14 — Quiz

10 questions on the Fundamental Theorem of Calculus. Try each before opening the answer. Each answer cites the section in index.md to review.

Q1. Evaluate $\displaystyle\int_0^2 (3x^2 + 1)\,dx$ using FTC Part 2.

Answer An antiderivative of $3x^2 + 1$ is $x^3 + x$. By the evaluation bar, $$\int_0^2 (3x^2 + 1)\,dx = \big[x^3 + x\big]_0^2 = (8 + 2) - 0 = 10.$$ **Review §14.4 (FTC Part 2) and §14.6 (worked examples).**

Q2. State FTC Part 1 in your own words, then use it to find $F'(x)$ for $F(x) = \displaystyle\int_3^x \ln(t^2 + 1)\,dt$.

Answer FTC Part 1: the derivative of an accumulation function $\int_a^x f(t)\,dt$ is just the integrand evaluated at the upper limit, $f(x)$ — differentiation undoes integration. Here $f(t) = \ln(t^2+1)$, so $$F'(x) = \ln(x^2 + 1).$$ You never compute the integral itself. **Review §14.3 (FTC Part 1).**

Q3. Find $\dfrac{d}{dx}\displaystyle\int_0^{x^3} \cos t\,dt$.

Answer The upper limit is a function $u(x) = x^3$, so FTC Part 1 combines with the chain rule: $$\frac{d}{dx}\int_0^{x^3}\cos t\,dt = \cos(x^3)\cdot \frac{d}{dx}(x^3) = 3x^2\cos(x^3).$$ Forgetting the factor $3x^2$ is the most common error. **Review §14.8 (variable limits).**

Q4. A car's velocity is $v(t) = 4t$ m/s on $[0,3]$. What is its net displacement?

Answer By the Net Change Theorem, displacement is the integral of velocity: $$\int_0^3 4t\,dt = \big[2t^2\big]_0^3 = 18\ \text{m}.$$ **Review §14.7 (Net Change Theorem).**

Q5. Find the average value of $f(x) = x^2$ on $[0, 6]$.

Answer $$\bar f = \frac{1}{6 - 0}\int_0^6 x^2\,dx = \frac{1}{6}\cdot\frac{x^3}{3}\bigg|_0^6 = \frac{1}{6}\cdot\frac{216}{3} = \frac{1}{6}\cdot 72 = 12.$$ **Review §14.9 (average value).**

Q6. True or false: in $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$, the answer depends on which antiderivative $F$ you choose. Explain.

Answer **False.** Any two antiderivatives differ by a constant $C$, and $\big(G(b)+C\big) - \big(G(a)+C\big) = G(b) - G(a)$ — the constant cancels. That is exactly why the "$+C$" never appears in a definite integral. **Review §14.4 ("Why it follows from Part 1") and §14.14 Error 5.**

Q7. Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$.

Answer An antiderivative of $\sin x$ is $-\cos x$: $$\int_0^\pi \sin x\,dx = \big[-\cos x\big]_0^\pi = -\cos\pi - (-\cos 0) = -(-1) + 1 = 2.$$ The area under one hump of the sine curve is exactly $2$. **Review §14.6 (Example 2).**

Q8. Why is $\displaystyle\int_{-1}^{1}\frac{1}{x^2}\,dx = \big[-\tfrac{1}{x}\big]_{-1}^{1} = -2$ an invalid application of FTC?

Answer FTC Part 2 requires $f$ to be **continuous on the entire interval** $[a,b]$. The integrand $1/x^2$ blows up at $x=0$, which lies inside $[-1,1]$, so the hypothesis fails and the evaluation bar does not apply. (The tell-tale sign: a positive integrand cannot have a negative integral.) The correct framework is improper integrals — **Chapter 17**. **Review §14.4 (Common Pitfall) and §14.14 Error 1.**

Q9. A particle has velocity $v(t) = t - 2$ m/s on $[0,3]$. Find (a) its net displacement and (b) its total distance traveled. Why do they differ?

Answer (a) Net displacement: $$\int_0^3 (t-2)\,dt = \Big[\tfrac{t^2}{2} - 2t\Big]_0^3 = \left(\tfrac{9}{2} - 6\right) - 0 = -\tfrac{3}{2}\ \text{m}.$$ (b) $v < 0$ on $[0,2)$ and $v > 0$ on $(2,3]$, so total distance is $$\int_0^2 (2-t)\,dt + \int_2^3 (t-2)\,dt = 2 + \tfrac{1}{2} = \tfrac{5}{2}\ \text{m}.$$ They differ because the particle moves backward then forward; net change lets the legs cancel, while distance integrates $|v|$. **Review §14.7 and §14.14 Error 3.**

Q10. Explain why FTC guarantees that $\Phi(x) = \displaystyle\int_0^x e^{-t^2}\,dt$ exists and is differentiable, even though $\Phi$ has no elementary formula.

Answer Because $e^{-t^2}$ is continuous everywhere, FTC Part 1 guarantees its accumulation function $\Phi$ exists, is differentiable, and satisfies $\Phi'(x) = e^{-x^2}$ — that is the *existence* content of FTC Part 1. FTC Part 2 promises only that *if* you have an antiderivative you can subtract; it never promises that antiderivative is elementary. Liouville's theorem proves $e^{-t^2}$ has no elementary antiderivative, so we evaluate $\Phi$ numerically or by Taylor series (Chapter 23) — but $\Phi$ is a perfectly good, well-defined function. **Review §14.3 and §14.12.**

Scoring Guide

Score Interpretation
9–10 Excellent. You command both forms of FTC, variable-limit differentiation, and the continuity hypothesis. Move on to Chapter 15.
7–8 Solid. Re-read the sections cited by any questions you missed — most likely §14.8 (chain rule on limits) or §14.7 (distance vs. displacement).
5–6 Developing. Re-work §14.6 examples by hand, then redo Q1, Q4, Q5, Q7 before proceeding.
0–4 Revisit §14.3, §14.4, and §14.6 from the top. FTC is the floor every later chapter stands on; invest the time now.