Answer: $\frac{d}{dx}\int_a^{g(x)} f(t)\, dt = f(g(x)) \cdot g'(x)$

Apply FTC Part 1 with the Chain Rule: Let $F(u) = \int_a^u f(t)\,dt$. Then $F'(u) = f(u)$ by FTC Part 1. With $u = g(x)$, the chain rule gives $\frac{d}{dx}F(g(x)) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)$. Why Option B is correct: It correctly combines FTC Part 1 with the chain rule by evaluating the integrand at $g(x)$ and multiplying by $g'(x)$. Why distractors fail: Option A omits the chain rule factor $g'(x)$. Option C incorrectly differentiates $f$ to get $f'(g(x))$ — FTC Part 1 returns the integrand, not its derivative. Option D evaluates $f$ at $x$ instead of at $g(x)$.

Answer: $60$ liters

Set up the integral using the Net Change Theorem: Total water = $\int_0^5 (4t + 2)\,dt$. Evaluate using FTC Part 2: Antiderivative: $F(t) = 2t^2 + 2t$. Then $F(5) - F(0) = (2 \cdot 25 + 10) - 0 = 50 + 10 = 60$ liters. Why distractors fail: Option A ($50$) omits the $+2t$ term. Option C ($22$) may come from evaluating the rate at $t = 5$ as $4(5)+2 = 22$, which is instantaneous rate, not total. Option D ($110$) could be an arithmetic error.

Answer: $1000 + \int_0^{10} 50e^{0.02t}\, dt$

Apply the Net Change Theorem: By the Net Change Theorem, $\int_0^{10} P'(t)\,dt = P(10) - P(0)$. Therefore $P(10) = P(0) + \int_0^{10} P'(t)\,dt$. Why Option B is correct: It correctly expresses $P(10)$ as the initial value plus the accumulated change: $1000 + \int_0^{10} 50e^{0.02t}\,dt$. Why distractors fail: Option A is just the rate evaluated at $t = 10$. Option C gives only the net change without adding the initial population. Option D assumes exponential growth $P = P_0 e^{kt}$, which is not the same as adding the integral of $P'(t)$ to $P(0)$.

Answer: $-(x^2 - 4)$

Reverse the limits: $H(x) = \int_x^5 (t^2 - 4)\,dt = -\int_5^x (t^2 - 4)\,dt$. Apply FTC Part 1: $\frac{d}{dx}\left[-\int_5^x (t^2 - 4)\,dt\right] = -(x^2 - 4)$. Why distractors fail: Option A forgets the negative sign from reversing limits. Option C evaluates the integral at $x = 5$ and $x$, computing the value rather than the derivative. Option D differentiates $x^2 - 4$ again instead of applying FTC Part 1.

Answer: Integration and differentiation are inverse operations.

Core idea of FTC Part 1: FTC Part 1 shows that differentiating the integral $\int_a^x f(t)\,dt$ returns $f(x)$, demonstrating that differentiation undoes integration. Why Option A is correct: The theorem establishes that differentiation and integration are inverse processes — this is the fundamental connection. Why distractors fail: Option B is false; antiderivatives differ by a constant, so there are infinitely many. Option C is a garbled version of FTC Part 2. Option D is false; a function can be integrable without being differentiable.

Answer: $\frac{1}{2}$

Apply FTC Part 1: $F'(x) = \frac{1}{1+x^4}$ by the Fundamental Theorem of Calculus Part 1. Evaluate at $x = 1$: $F'(1) = \frac{1}{1+1^4} = \frac{1}{2}$. Why distractors fail: Option A confuses $F(1) = 0$ (since the integral from 1 to 1 is 0) with $F'(1)$. Option C evaluates $\frac{1}{1+0^4}$ instead of $\frac{1}{1+1^4}$. Option D is wrong because FTC Part 1 gives $F'(x)$ directly without computing the integral.

Answer: FTC Part 1 requires the integrand to be continuous, and the variable of differentiation must appear as the upper limit of the integral.

Identify the missing conditions: FTC Part 1 states: if $f$ is continuous on an interval containing $a$ and $x$, then $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. The lower limit must be constant and the upper limit must be the variable of differentiation. Why Option B is correct: It correctly identifies both the continuity requirement and the structural requirement that $x$ serves as the upper limit. Why distractors fail: Option A is too restrictive — FTC Part 1 works for any continuous integrand. Option C is partially right about the lower limit but wrong that the upper limit must also be constant (the upper limit must be the differentiation variable). Option D has no basis in the theorem.

Answer: $3x^2 + 1$

Apply FTC Part 1 directly: By FTC Part 1, $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. Here $f(t) = 3t^2 + 1$, so $f(x) = 3x^2 + 1$. Why Option A is correct: Replacing $t$ with $x$ in the integrand gives $3x^2 + 1$. Why distractors fail: Option B computes the antiderivative and evaluates it — that's the integral itself, not its derivative. Option C differentiates $3x^2 + 1$ again, applying an extra differentiation step. Option D leaves the dummy variable $t$ instead of substituting $x$.

Answer: $10{,}000$ dollars

Set up: By the Net Change Theorem, $\int_{100}^{200} C'(q)\,dq = C(200) - C(100)$. Evaluate: $\int_{100}^{200}(0.6q+10)\,dq = [0.3q^2 + 10q]_{100}^{200} = (0.3(40000)+2000)-(0.3(10000)+1000) = 14000 - 4000 = 10{,}000$. Why distractors fail: Option B ($9{,}000$) results from an error in the constant term evaluation. Option C ($14{,}000$) uses only the upper-limit value $F(200) = 14{,}000$ without subtracting $F(100)$. Option D ($4{,}000$) uses only the lower-limit value $F(100) = 4{,}000$ without computing the difference correctly.

Answer: $F$ is increasing and concave up on $[3, 10]$.

Determine $F'(x)$: By FTC Part 1, $F'(x) = \sqrt{x^2 + 5}$. Since $\sqrt{x^2 + 5} > 0$ for all $x$, $F$ is increasing. Determine $F''(x)$: $F''(x) = \frac{d}{dx}\sqrt{x^2 + 5} = \frac{x}{\sqrt{x^2 + 5}}$. For $x > 0$ (i.e., on $[3, 10]$), $F''(x) > 0$, so $F$ is concave up. Why distractors fail: Option A is wrong because $F' > 0$ means $F$ is increasing, not decreasing. Option C is wrong because $F'' > 0$ on $[3,10]$ means concave up, not down. Option D is wrong because $F' > 0$ everywhere on the interval.

Answer: $A(x)$ has a local maximum at $x = 4$.

Apply FTC Part 1: $A'(x) = f(x)$. Since $f(4) = 0$, $A'(4) = 0$, making $x = 4$ a critical point. Use the first derivative test: For $t \in (0, 4)$, $f(t) > 0$ so $A'(x) > 0$ (increasing). For $t \in (4, 6)$, $f(t) < 0$ so $A'(x) < 0$ (decreasing). Since $A'$ changes from positive to negative, $x = 4$ is a local maximum. Why distractors fail: Option A says local minimum, but $A'$ goes from $+$ to $-$, indicating a max. Option C is false since $A$ increases then decreases. Option D incorrectly denies the extremum despite the sign change in $A'$.

Answer: Both methods are valid and yield the same answer.

Check Student X: FTC Part 1 + chain rule: $\frac{d}{dx}\int_0^{x^2+1} \cos(t)\,dt = \cos(x^2+1) \cdot 2x$. ✓ Check Student Y: Evaluate: $\int_0^{x^2+1} \cos(t)\,dt = \sin(x^2+1) - 0 = \sin(x^2+1)$. Differentiate: $\frac{d}{dx}\sin(x^2+1) = \cos(x^2+1) \cdot 2x$. ✓ Both agree: Both methods yield $2x\cos(x^2+1)$. Student Y's works here because $\cos(t)$ has a known antiderivative. Student X's method is more general (works even when no closed-form antiderivative exists). Why distractors fail: Options A and B each exclude a valid method. Option D is wrong since both approaches work.

Answer: $6$

Find the antiderivative: $F(x) = x^3 - 2x^2 + x$. Evaluate at the limits: $F(2) = 8 - 8 + 2 = 2$. $F(-1) = -1 - 2 - 1 = -4$. Compute the difference: $F(2) - F(-1) = 2 - (-4) = 6$. Why distractors fail: Option B ($3$) may result from an error in evaluating $F(-1)$. Option C ($0$) might come from assuming symmetry that doesn't exist. Option D ($9$) could arise from an arithmetic mistake.

Answer: $\int_0^4 |t^2 - 4t + 3|\, dt$

Distinguish displacement from distance: Displacement is $\int_0^4 v(t)\,dt$, which can be negative. Total distance requires integrating the absolute value: $\int_0^4 |v(t)|\,dt$. Why Option B is correct: Total distance = $\int_0^4 |t^2 - 4t + 3|\,dt$. The absolute value ensures all contributions are positive. Why distractors fail: Option A gives net displacement, which could be less than total distance. Option C takes the absolute value of the net displacement — this gives a single non-negative number but does not equal total distance when velocity changes sign. Option D squares the velocity, which changes the units and the meaning.

Answer: The integrand must be $f'(x)$ — the derivative of some function $f$ — not an arbitrary function.

Identify the misunderstanding: The Net Change Theorem says $\int_a^b f'(x)\,dx = f(b) - f(a)$. The key is that the integrand is specifically $f'$, the derivative of some function $f$. Why Option B is correct: The student's error is generalizing the result to any function. The theorem applies when the integrand is a derivative. For a general function $g$, we need an antiderivative $G$ of $g$ to write $\int_a^b g(x)\,dx = G(b) - G(a)$, which is FTC Part 2. Why distractors fail: Option A is wrong; the theorem holds regardless of the sign of $f'$. Option C is wrong; the theorem applies to any continuously differentiable function. Option D is wrong; there is no restriction on $a$.

Answer: No, the antiderivative of $|2x - 4|$ is not $x^2 - 4x$; the absolute value requires splitting the integral at $x = 2$ before applying FTC Part 2.

Identify the issue with absolute value: $|2x - 4| = \begin{cases} 4 - 2x & x < 2 \\ 2x - 4 & x \geq 2 \end{cases}$. The function $x^2 - 4x$ is an antiderivative of $2x - 4$, not $|2x - 4|$. Correct approach: Split: $\int_0^2 (4 - 2x)\,dx + \int_2^3 (2x - 4)\,dx = [4x - x^2]_0^2 + [x^2 - 4x]_2^3 = 4 + 1 = 5$. Why distractors fail: Option A ignores that the antiderivative must match the integrand including the absolute value. Option B incorrectly claims FTC doesn't apply at all; it does apply on each subinterval. Option D uses the wrong antiderivative and gets $9$.

Answer: $F(x)$ is the antiderivative of $f$, so $F'(x) = f(x)$

Recall FTC Part 1: FTC Part 1 states that if $f$ is continuous on $[a, b]$ and $F(x) = \int_a^x f(t)\, dt$, then $F$ is differentiable and $F'(x) = f(x)$. Why Option A is correct: This directly restates the theorem: differentiating the accumulation function $F(x)$ returns the integrand $f(x)$. Why distractors fail: Option B confuses FTC Part 1 with Part 2 and also misapplies it (Part 2 uses an antiderivative $F$, not $f$ itself). Option C incorrectly defines $F(x)$ as area 'above' the curve; the integral gives net signed area. Option D confuses the derivative of the integral of $f$ with the integral of the derivative of $f$, which is the Net Change Theorem.

Answer: $15$

Find the antiderivative: An antiderivative of $2x$ is $F(x) = x^2$. Apply FTC Part 2: $\int_1^4 2x\,dx = F(4) - F(1) = 4^2 - 1^2 = 16 - 1 = 15$. Why distractors fail: Option A ($8$) might come from $2 \cdot 4 = 8$, evaluating the integrand rather than the antiderivative. Option C ($16$) forgets to subtract $F(1) = 1$. Option D ($6$) may result from an arithmetic error such as $4 - 1 = 3$ then $2 \cdot 3 = 6$.

Answer: Yes, $F(a) = 0$ is correct because the integral from $a$ to $a$ is zero, but this only means $F$ passes through zero at $x = a$, not that it 'crosses' the axis.

Verify $F(a) = 0$: $F(a) = \int_a^a f(t)\,dt = 0$ for any integrable $f$. This is correct. Distinguish 'passes through' from 'crosses': 'Crossing' typically means the function changes sign. $F$ touches zero at $x = a$ but may stay non-negative or non-positive nearby, depending on $f$. Why distractors fail: Option A makes an unsupported claim about $F(b)$. Option C is wrong; $\int_a^a f(t)\,dt = 0$ is well-defined. Option D is wrong; $F(a) = 0$ regardless of the value of $f(a)$.

Answer: $f(x)\int_0^x g(t)\,dt + g(x)\int_0^x f(t)\,dt$

Identify the structure: This is the derivative of a product of two functions: $U(x) = \int_0^x f(t)\,dt$ and $V(x) = \int_0^x g(t)\,dt$. Apply the product rule and FTC Part 1: $\frac{d}{dx}[U \cdot V] = U'V + UV'$. By FTC Part 1, $U' = f(x)$ and $V' = g(x)$. So the result is $f(x)\int_0^x g(t)\,dt + g(x)\int_0^x f(t)\,dt$. Why distractors fail: Option A only gives $U'V'$, ignoring the product rule. Option C incorrectly combines the integrands into one integral. Option D mixes derivatives of $f$ and $g$ without justification.

Answer: Evaluate the definite integral $\int_a^b f(x)\, dx$ exactly

Connect the two parts: FTC Part 1 tells us that $F(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$. FTC Part 2 then uses any antiderivative to evaluate the definite integral as $F(b) - F(a)$. Why Option B is correct: Part 2 leverages the antiderivative (whose existence is guaranteed by Part 1) to compute the exact value of a definite integral. Why distractors fail: Option A relates to the second derivative, not to FTC. Option C reverses the direction: FTC starts with $f$ and builds the antiderivative, not the derivative. Option D is false — continuity of $f$ does not imply differentiability of $f$.

Answer: The integral of a rate of change gives the total accumulated change in the original quantity over the interval.

Interpret the Net Change Theorem: When you integrate the rate of change $f'(x)$ from $a$ to $b$, you recover the net change $f(b) - f(a)$. Why Option A is correct: The integral of a derivative (rate of change) yields the total accumulated change in the original function over $[a, b]$. Why distractors fail: Option B describes the average value formula $\frac{1}{b-a}\int_a^b f(x)\,dx$, not the net change. Option C confuses integration with evaluating a derivative at a point. Option D confuses net change with the Extreme Value Theorem.

Answer: $4$

Verify the antiderivative: $\int x^{-1/2}\,dx = 2x^{1/2} + C$. This is correct since $\frac{d}{dx}(2\sqrt{x}) = \frac{1}{\sqrt{x}}$. Apply FTC Part 2: $[2\sqrt{x}]_1^9 = 2\sqrt{9} - 2\sqrt{1} = 6 - 2 = 4$. Why distractors fail: Option A ($2$) might come from $2\sqrt{1} = 2$ alone. Option C ($6$) forgets to subtract $F(1)$. Option D ($8$) could result from an error like $2(4) = 8$, using $\sqrt{9} \approx 4$.

Answer: The net displacement of the vehicle from $t = 0$ to $t = 10$ seconds

Apply the Net Change Theorem: Since velocity is the derivative of position, $v(t) = s'(t)$, the Net Change Theorem gives $\int_0^{10} v(t)\,dt = s(10) - s(0)$, which is the net change in position (displacement). Why Option C is correct: The integral of velocity gives displacement — the net change in position, accounting for direction. Why distractors fail: Option A confuses the integral of velocity with the derivative of velocity (acceleration). Option B describes $\frac{1}{10}\int_0^{10} v(t)\,dt$, not the integral itself. Option D describes $\int_0^{10} |v(t)|\,dt$ (total distance requires the absolute value).

Answer: The student forgot to subtract $F(1) = 2(1)^3 = 2$ from $F(3) = 54$, so the correct answer is $52$.

Check the antiderivative: The antiderivative of $6x^2$ is $2x^3$. This part is correct. Identify the omission: FTC Part 2 requires $F(b) - F(a)$. The student computed $F(3) = 54$ but did not subtract $F(1) = 2$. The correct answer is $54 - 2 = 52$. Why distractors fail: Option A incorrectly claims the antiderivative is wrong. Option C uses an incorrect antiderivative $6x^3$. Option D claims no error, but the student clearly omitted $F(1)$.

Answer: $3x^2 \sin(x^3)$

Identify the structure: The upper limit is $g(x) = x^3$, so this requires FTC Part 1 combined with the chain rule. Apply the formula: $\frac{d}{dx}\int_1^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x) = \sin(x^3) \cdot 3x^2$. Why distractors fail: Option A applies FTC Part 1 but omits the chain rule factor $g'(x) = 3x^2$. Options C and D use $\cos$ instead of $\sin$, incorrectly differentiating the integrand rather than just evaluating it at $g(x)$.

Answer: $F(b) - F(a)$

State FTC Part 2: FTC Part 2 says $\int_a^b f(x)\, dx = F(b) - F(a)$, where $F$ is any antiderivative of $f$. Why Option B is correct: It matches the exact formula: evaluate the antiderivative at the upper limit minus the lower limit. Why distractors fail: Option A reverses the subtraction order. Option C uses $F'$ (which is $f$), confusing the antiderivative with the original function. Option D uses $f$ rather than $F$; this would be the Net Change Theorem applied to $f'$, not to $f$.

Answer: $0$

Find the antiderivative: An antiderivative of $\cos(x)$ is $\sin(x)$. Apply FTC Part 2: $\int_0^{\pi} \cos(x)\,dx = \sin(\pi) - \sin(0) = 0 - 0 = 0$. Why distractors fail: Option A ($1$) might come from $\sin(\pi/2) = 1$, confusing the upper limit. Option B ($-1$) could arise from $\cos(\pi) = -1$, evaluating the wrong function. Option D ($2$) might come from $\int_0^{\pi} |\cos(x)|\,dx$, the total unsigned area.

Answer: The student should have written $\ln(x^2) \cdot 2x$ to account for the chain rule.

Identify the issue: The upper limit is $g(x) = x^2$, not $x$. FTC Part 1 with the chain rule gives $f(g(x)) \cdot g'(x) = \ln(x^2) \cdot 2x$. Why Option A is correct: The student applied FTC Part 1 but forgot to multiply by $g'(x) = 2x$. Why distractors fail: Option B differentiates $\ln(t)$ to get $1/t$, but FTC Part 1 returns the integrand, not its derivative. Option C ignores the chain rule requirement. Option D is unnecessarily complicated and the whole point of FTC Part 1 is to avoid evaluating the integral first.

Answer: $f(t) = (t - 3)(t - 7)$

Use FTC Part 1: $A'(x) = f(x)$. For a local max at $x = 3$, we need $f(3) = 0$ with $f$ changing from positive to negative. For a local min at $x = 7$, we need $f(7) = 0$ with $f$ changing from negative to positive. Check Option A: $f(t) = (t-3)(t-7)$. Zeros at $t = 3, 7$. For $t < 3$: both factors negative → product positive. For $3 < t < 7$: first positive, second negative → product negative. For $t > 7$: both positive → product positive. So $f$ goes $+ \to - \to +$, giving max at 3 and min at 7. ✓ Why distractors fail: Option B is $-(t-3)(t-7)$, which gives sign pattern $- \to + \to -$, producing a min at 3 and max at 7 — the reverse of what is needed. Option C: $(t-3)^2(t-7)$ — at $t = 3$, the squared factor means $f$ does not change sign, so there is no extremum at $x = 3$. Option D: $(t+3)(t-7)$ has zeros at $t = -3$ and $t = 7$, not at $t = 3$, so there is no critical point at $x = 3$.

Answer: It replaces the limit of a sum with a single subtraction of antiderivative values at the endpoints.

Compare methods: Without FTC Part 2, evaluating $\int_a^b f(x)\,dx$ requires computing $\lim_{n \to \infty} \sum_{i=1}^n f(x_i^*)\Delta x$, which is often laborious. Why Option A is correct: FTC Part 2 says: find any antiderivative $F$, then compute $F(b) - F(a)$. This replaces the infinite summation process with simple evaluation and subtraction. Why distractors fail: Option B is wrong because you still need to find the antiderivative and evaluate it — it doesn't eliminate computation. Option C describes numerical methods, which is the opposite of what FTC Part 2 does. Option D is simply incorrect; $\frac{f(a)+f(b)}{2}$ is related to the Trapezoidal Rule, not to FTC.

Answer: $e^9$

Apply FTC Part 1: Since $G(x) = \int_0^x e^{t^2}\, dt$, FTC Part 1 gives $G'(x) = e^{x^2}$. Evaluate at $x = 3$: $G'(3) = e^{3^2} = e^9$. Why distractors fail: Option B multiplies by $2 \cdot 3$, which would be the chain rule factor if the upper limit were $x^2$ rather than $x$. Option C uses $e^6$ from an error like $e^{2 \cdot 3}$. Option D ($0$) has no justification — $G'(3)$ is clearly positive.

Answer: $e^{x^2} \cdot 2x - e^{\sin x} \cdot \cos x$

Split the integral at a constant: Write $\int_{\sin x}^{x^2} e^t\,dt = \int_0^{x^2} e^t\,dt - \int_0^{\sin x} e^t\,dt$. Differentiate each part using FTC Part 1 with the chain rule: $\frac{d}{dx}\int_0^{x^2} e^t\,dt = e^{x^2} \cdot 2x$ and $\frac{d}{dx}\int_0^{\sin x} e^t\,dt = e^{\sin x} \cdot \cos x$. Combine: The result is $e^{x^2} \cdot 2x - e^{\sin x} \cdot \cos x$. Why distractors fail: Option B omits both chain rule factors. Option C uses $+$ instead of $-$ for the lower limit contribution. Option D ignores the lower limit entirely.