Answer: $f$ must be continuous on $[a, b]$

Recall the hypothesis of FTC Part 1: The standard statement of FTC Part 1 requires that $f$ be continuous on $[a, b]$. Under this condition, $F(x) = \int_a^x f(t)\,dt$ is differentiable and $F'(x) = f(x)$. Why the correct answer works: Continuity of $f$ is precisely the hypothesis needed. It ensures the integral is well-defined and the limit in the derivative exists. Why distractors fail: Option A is too strong — differentiability of $f$ is not required. Option C is unnecessary; $f$ may be negative. Option D is far too restrictive.

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

Identify the composite structure: The upper limit is $u(x) = x^3$, so we need the chain rule combined with FTC Part 1. Apply the chain rule with FTC: $H'(x) = f(u(x)) \cdot u'(x) = \sin(x^3) \cdot 3x^2$. Why distractors fail: Option A omits the chain rule factor $3x^2$. Option C incorrectly differentiates $\sin$ to $\cos$. Option D uses $x^3$ instead of $3x^2$ for the chain rule factor.

Answer: $2\sqrt{2}$

Apply the chain rule with FTC Part 1: Let $u = x^2$. Then $F'(x) = \sqrt{1 + (x^2)^3} \cdot 2x = 2x\sqrt{1 + x^6}$. Evaluate at x = 1: $F'(1) = 2(1)\sqrt{1 + 1} = 2\sqrt{2}$. Why distractors fail: Option A ($\sqrt{2}$) omits the chain rule factor $2x$. Option C ($2$) uses $\sqrt{1 + 1^3}$ but writes $\sqrt{2} \approx 1$ incorrectly. Option D ($4\sqrt{2}$) doubles the chain rule factor.

Answer: $t = 3$

Apply FTC Part 1: $C'(t) = r(t)$. Since $r(t) > 0$ for $t < 3$, $C$ is increasing. Since $r(t) < 0$ for $t > 3$, $C$ is decreasing. And $r(3) = 0$. Identify the maximum: $C$ increases up to $t = 3$ and decreases after, so $C$ has a global maximum at $t = 3$ on $[0,6]$. Why distractors fail: Option A gives the minimum starting point $C(0) = 0$. Option C gives the endpoint after $C$ has been decreasing. Option D is wrong because the sign information fully determines the maximum location.

Answer: $F'(x) = f(x)$

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 $F'(x) = f(x)$. Why the correct answer works: Option B directly states the theorem: differentiating the integral with respect to its upper limit returns the integrand evaluated at that limit. Why distractors fail: Option A incorrectly differentiates the integrand rather than the integral. Option C subtracts $f(a)$, which confuses FTC Part 1 with Part 2. Option D returns a constant $f(a)$, ignoring the variable upper limit.

Answer: $3x^2 + 1$

Apply FTC Part 1: By FTC Part 1, if $G(x) = \int_2^x f(t)\,dt$, then $G'(x) = f(x)$. Here $f(t) = 3t^2 + 1$. Why the correct answer works: Replacing $t$ with $x$ in the integrand gives $G'(x) = 3x^2 + 1$. Why distractors fail: Option A is the derivative of $3x^2$, not the integrand itself. Option C is the antiderivative $G(x) = x^3 + x - 10$, not the derivative. Option D incorrectly subtracts $G(2)$ from the integrand.

Answer: $F$ is strictly increasing on $(0, 5)$

Use FTC Part 1: $F'(x) = f(x) > 0$ on $(0, 5)$, so $F$ is strictly increasing on this interval. Why the correct answer works: A positive derivative on an interval guarantees the function is strictly increasing there. Why distractors fail: Option A requires $F'' = f' > 0$, which is not given. Option C seems plausible but $F(0) = 0$, and while $F$ immediately increases, the claim is about the open interval — actually $F(x) > F(0) = 0$ for $x \in (0,5)$, so Option C is also true. However, Option B is the direct and necessary consequence of FTC Part 1 with $f > 0$, and is the best answer. Option D is false because $F'(x) = f(x) > 0$, so $F$ has no critical points on $(0,5)$.

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 may come from evaluating $\sin(\pi/2) - \sin(0)$. Option C may result from a sign error. Option D might come from $\int_0^{\pi} |\cos(x)|\,dx$, which gives the total area rather than the signed area.

Answer: Yes, because FTC Part 1 gives the derivative directly as the integrand evaluated at $x$, regardless of whether an elementary antiderivative exists.

Assess the student's reasoning: FTC Part 1 states $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$ whenever $f$ is continuous. It does not require finding an antiderivative in closed form. Why the correct answer works: The function $\frac{1}{1+t^4}$ is continuous everywhere, so FTC Part 1 applies directly. The student's reasoning is correct. Why distractors fail: Option A is a common misconception — FTC Part 1 bypasses the need for an explicit antiderivative. Option C is wrong because $\frac{1}{1+t^4}$ is continuous for all $t$. Option D imposes an unnecessary domain restriction.

Answer: $5$

Apply FTC Part 1: By FTC Part 1, $F'(x) = f(x)$. Therefore $F'(3) = f(3) = 5$. Why the correct answer works: Differentiating the accumulation function at $x = 3$ simply returns the value of the integrand at $3$. Why distractors fail: Option A confuses $F'(3)$ with $F(0)$. Option C gives $F(3)$, not $F'(3)$. Option D gives the derivative of $f$ at $3$, which is $F''(3)$, not $F'(3)$.

Answer: By computing $F(b) - F(a)$, where $F$ is any antiderivative of $f$

Recall FTC Part 2: FTC Part 2 states $\int_a^b f(x)\,dx = F(b) - F(a)$, where $F$ is any antiderivative of $f$. Why the correct answer works: Option B correctly states the evaluation formula: antiderivative at the upper limit minus antiderivative at the lower limit. Why distractors fail: Option A reverses the order of evaluation. Option C uses $f$ instead of an antiderivative $F$. Option D uses $F'$, which equals $f$, reducing to Option C's error.

Answer: $\int_1^4 \frac{1}{\sqrt{x}}\,dx$

Identify which integrand has a direct antiderivative: $\frac{1}{\sqrt{x}} = x^{-1/2}$, whose antiderivative is $2x^{1/2}$ by the power rule. No substitution or other technique is needed. Why the correct answer works: Option B can be evaluated as $[2\sqrt{x}]_1^4 = 2(2) - 2(1) = 2$, using only a basic antiderivative rule and FTC Part 2. Why distractors fail: Option A requires $u$-substitution with $u = x^2$. Option C requires substitution (e.g., $u = \sin x$). Option D requires integration by parts.

Answer: The net displacement of the particle over the first 10 seconds, in metres

Interpret the integral of velocity: By FTC Part 2, $\int_0^{10} v(t)\,dt = s(10) - s(0)$, where $s$ is the position function. This is the net change in position, i.e., displacement. Why the correct answer works: The definite integral of velocity gives displacement (net change in position), not total distance. Why distractors fail: Option A confuses the integral's value with the integrand's value. Option C describes total distance, which requires $\int_0^{10} |v(t)|\,dt$. Option D confuses integration with differentiation.

Answer: The net signed area under $f(t)$ from $t = 0$ to $t = x$

Interpret accumulation functions: An accumulation function $F(x) = \int_0^x f(t)\,dt$ represents the net signed area between $f(t)$ and the $t$-axis from $0$ to $x$. Area above the axis is positive, and area below is negative. Why the correct answer works: Option B captures the core interpretation: $F(x)$ accumulates the signed area as $x$ varies. Why distractors fail: Option A describes $f'$, not the integral. Option C describes $\frac{1}{x}\int_0^x f(t)\,dt$, which includes a factor of $\frac{1}{x}$. Option D confuses integration with optimization.

Answer: Because any two antiderivatives of $f$ differ by a constant, which cancels in the subtraction $F(b) - F(a)$

Antiderivatives differ by a constant: If $F$ and $G$ are both antiderivatives of $f$, then $G(x) = F(x) + C$ for some constant $C$. Therefore $G(b) - G(a) = (F(b) + C) - (F(a) + C) = F(b) - F(a)$. Why the correct answer works: The constant cancels, so the choice of antiderivative does not affect the result. Why distractors fail: Option A is false — antiderivatives are not identical; they differ by a constant. Option C is false; the integral depends on both limits. Option D is false; you may use any constant, as it cancels.

Answer: The student failed to account for the absolute value, which requires splitting the integral where $2x - 4$ changes sign.

Identify the error: The integrand has an absolute value. The expression $2x - 4$ changes sign at $x = 2$, so the integral must be split: $\int_0^2 (4 - 2x)\,dx + \int_2^3 (2x - 4)\,dx$. Correct computation: $\int_0^2 (4 - 2x)\,dx = [4x - x^2]_0^2 = 4$. $\int_2^3 (2x-4)\,dx = [x^2 - 4x]_2^3 = -3 - (-4) = 1$. Total = $5$. Why distractors fail: Option A: the antiderivative $x^2 - 4x$ is correct for $2x - 4$ but not for $|2x - 4|$. Option C: FTC Part 2 is the correct theorem to use. Option D: there is no reason to change the limits.

Answer: $f(c) = 0$

Connect critical points of F to zeros of f: By FTC Part 1, $F'(x) = f(x)$. A local maximum of $F$ at $x = c$ requires $F'(c) = 0$, so $f(c) = 0$. Why the correct answer works: Since $F' = f$, critical points of $F$ occur where $f = 0$. A local maximum specifically requires $f(c) = 0$ (and a sign change from positive to negative). Why distractors fail: Option A confuses extrema of $F$ with extrema of $f$. Option C describes a condition on $f'$, which relates to $F''$, not $F'$. Option D says the integral is zero, which is unrelated to extrema.

Answer: $\int_4^7 f(t)\,dt = 0$

Use properties of definite integrals: $\int_4^7 f(t)\,dt = A(7) - A(4) = 7 - 7 = 0$. Why the correct answer works: The integral from 4 to 7 equals the difference of accumulation function values, which is zero. Why distractors fail: Option A is too strong — the integral being zero does not mean $f$ is identically zero. Option C has no basis; $f$ could vary as long as the net area is zero. Option D cannot be guaranteed without more information about $f$'s sign.

Answer: $-\sin(x)\ln(1 + \cos^2(x)) - \cos(x)\ln(1 + \sin^2(x))$

Split using an intermediate constant: Let $g(t) = \ln(1+t^2)$. Write the integral as $\int_{\sin x}^{c} g(t)\,dt + \int_c^{\cos x} g(t)\,dt = -\int_c^{\sin x} g(t)\,dt + \int_c^{\cos x} g(t)\,dt$. Apply FTC Part 1 with chain rule: Differentiating: $-g(\sin x) \cdot \cos x + g(\cos x) \cdot (-\sin x)$. This gives $-\cos(x)\ln(1+\sin^2 x) - \sin(x)\ln(1+\cos^2 x)$. Rewrite: This equals $-\sin(x)\ln(1+\cos^2 x) - \cos(x)\ln(1+\sin^2 x)$, matching Option B. Why distractors fail: Option A has a sign error on the second term. Option C swaps the arguments and has incorrect signs. Option D omits the chain rule derivatives entirely.

Answer: $2x\,e^{-x^2} - 2e^{-2x}$

Split using an intermediate constant: Write $P(x) = \int_{2x}^{c} e^{-t}\,dt + \int_{c}^{x^2} e^{-t}\,dt = -\int_{c}^{2x} e^{-t}\,dt + \int_{c}^{x^2} e^{-t}\,dt$. Apply FTC Part 1 with the chain rule to each piece: Differentiating: $P'(x) = -e^{-2x} \cdot 2 + e^{-x^2} \cdot 2x = 2x\,e^{-x^2} - 2e^{-2x}$. Why distractors fail: Option A has a sign error on the second term. Option C omits both chain rule factors. Option D uses $1$ instead of $2$ for the derivative of $2x$.

Answer: $-e^{x^2}$

Handle the variable lower limit: Since $x$ is in the lower limit, rewrite: $F(x) = -\int_5^x e^{t^2}\,dt$. Apply FTC Part 1: $F'(x) = -e^{x^2}$. Why distractors fail: Option A forgets the negative sign from flipping the limits. Option C gives $F(x)$ evaluated symbolically, not $F'(x)$. Option D incorrectly applies the chain rule as if the upper limit were $x^2$.

Answer: Part 1 shows that differentiation undoes integration; Part 2 shows how to evaluate definite integrals using antiderivatives.

Distinguish the two parts: FTC Part 1 states $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$, showing differentiation reverses integration. FTC Part 2 states $\int_a^b f(x)\,dx = F(b) - F(a)$, providing a method to evaluate definite integrals. Why the correct answer works: Option B correctly identifies the distinct role of each part: Part 1 as a differentiation result and Part 2 as an evaluation formula. Why distractors fail: Option A reverses the roles of Part 1 and Part 2. Option C is incorrect because the two parts address different operations. Option D imposes a false restriction on Part 1.

Answer: The positive and negative areas under $f$ on $[a, b]$ cancel exactly

Interpret a zero definite integral: A zero definite integral means the net signed area is zero. This occurs when positive and negative areas cancel, or when $f = 0$ everywhere. Why the correct answer works: Option B is necessarily true: the net signed area being zero means positive and negative contributions cancel (including the trivial case where $f \equiv 0$). Why distractors fail: Option A is too strong — $f$ could be nonzero but have canceling areas. Option C is not necessarily true — $f$ could be identically zero and never change sign. Option D has no basis; the endpoint values of $f$ are unrestricted.

Answer: $24$

Find the antiderivative: An antiderivative of $2x + 3$ is $F(x) = x^2 + 3x$. Apply FTC Part 2: $\int_1^4 (2x+3)\,dx = F(4) - F(1) = (16 + 12) - (1 + 3) = 28 - 4 = 24$. Why distractors fail: Option A (18) results from an antiderivative error such as $x^2 + 3$ (missing the factor on $x$). Option C (15) comes from using $F(4) - F(0)$ minus an offset. Option D (21) is a common arithmetic slip.