AP Calculus AB Quiz: Fundamental Theorem Of Calculus Accumulation Functions
20 questions · exam conditions
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Fundamental Theorem Of Calculus Accumulation FunctionsQuestion 1 of 20

Suppose K(x)=∫0x(2t+cos⁡t) dtK(x)=\int_{0}^{x}\big(2^{t}+\cos t\big)\,dt. What is K′(x)K'(x)?

2x+cos⁡x2^{x}+\cos x
2xln⁡2−sin⁡x2^{x}\ln 2-\sin x
∫0x(2t+cos⁡t) dt\int_{0}^{x}\big(2^{t}+\cos t\big)\,dt
2t+cos⁡t2^{t}+\cos t
[2tln⁡2+sin⁡t]0x\left[\dfrac{2^{t}}{\ln 2}+\sin t\right]_{0}^{x}
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AP Calculus AB Quiz

AP Calculus AB Quiz: Fundamental Theorem Of Calculus Accumulation Functions

Practice Fundamental Theorem Of Calculus Accumulation Functions in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Fundamental Theorem Of Calculus Accumulation Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Suppose K(x)=∫0x(2t+cos⁡t) dtK(x)=\int_{0}^{x}\big(2^{t}+\cos t\big)\,dt. What is K′(x)K'(x)?

  1. 2x+cos⁡x2^{x}+\cos x (correct answer)
  2. 2xln⁡2−sin⁡x2^{x}\ln 2-\sin x
  3. ∫0x(2t+cos⁡t) dt\int_{0}^{x}\big(2^{t}+\cos t\big)\,dt
  4. 2t+cos⁡t2^{t}+\cos t
  5. [2tln⁡2+sin⁡t]0x\left[\dfrac{2^{t}}{\ln 2}+\sin t\right]_{0}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function K(x) is defined as the integral from 0 to x of (2ᵗ + cos t) dt, which accumulates the area under the curve of f(t) = 2ᵗ + cos t from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative K'(x) equals the integrand evaluated at x, which is 2ˣ + cos x. This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 0 to x, but that actually computes K(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

Question 2

Let C(x)=∫−5x∣t∣ dtC(x)=\int_{-5}^{x} \big| t \big| \, dt. What is C′(x)C'(x)?

  1. ∣x∣|x| (correct answer)
  2. x∣x∣\dfrac{x}{|x|}
  3. ∫−5x∣t∣ dt\int_{-5}^{x}|t|\,dt
  4. ∣t∣|t|
  5. [t∣t∣2]−5x\left[\dfrac{t|t|}{2}\right]_{-5}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function C(x) is defined as the integral from -5 to x of ∣t∣|t| dt, which accumulates the area under the curve of f(t) = ∣t∣|t| from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative C'(x) equals the integrand evaluated at x, which is ∣x∣|x|. This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from -5 to x, but that actually computes C(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

Question 3

Define H(x)=∫−1x(t−1)5 dtH(x)=\int_{-1}^{x}(t-1)^5\,dt. What is H′(x)H'(x)?

  1. (x−1)5(x-1)^5 (correct answer)
  2. 5(x−1)45(x-1)^4
  3. ∫−1x(t−1)5 dt\int_{-1}^{x}(t-1)^5\,dt
  4. (t−1)5(t-1)^5
  5. [(t−1)66]−1x\left[\dfrac{(t-1)^6}{6}\right]_{-1}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function H(x) is defined as the integral from -1 to x of (t - 1)⁵ dt, which accumulates the area under the curve of f(t) = (t - 1)⁵ from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative H'(x) equals the integrand evaluated at x, which is (x - 1)⁵. This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from -1 to x, but that actually computes H(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

Question 4

Suppose S(x)=∫3x(cos⁡t+1t2) dtS(x)=\int_{3}^{x}\big(\cos t+\dfrac{1}{t^2}\big)\,dt. What is S′(x)S'(x)?

  1. cos⁡x+1x2\cos x+\dfrac{1}{x^2} (correct answer)
  2. −sin⁡x−2x3-\sin x-\dfrac{2}{x^3}
  3. ∫3x(cos⁡t+1t2) dt\int_{3}^{x}\big(\cos t+\dfrac{1}{t^2}\big)\,dt
  4. cos⁡t+1t2\cos t+\dfrac{1}{t^2}
  5. [sin⁡t−1t]3x\left[\sin t-\dfrac{1}{t}\right]_{3}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function S(x) is defined as the integral from 3 to x of (cos t + 1/t21/t^2) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative S'(x) is simply the integrand evaluated at t = x, so S'(x) = cos x + 1/x^2. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 5

Let P(x)=∫−2x1t+3 dtP(x)=\int_{-2}^{x}\dfrac{1}{\sqrt{t+3}}\,dt for x>−3x>-3. What is P′(x)P'(x)?

  1. 1x+3\dfrac{1}{\sqrt{x+3}} (correct answer)
  2. −12(x+3)3/2\dfrac{-1}{2(x+3)^{3/2}}
  3. ∫−2x1t+3 dt\int_{-2}^{x}\dfrac{1}{\sqrt{t+3}}\,dt
  4. 1t+3\dfrac{1}{\sqrt{t+3}}
  5. [2t+3]−2x\left[2\sqrt{t+3}\right]_{-2}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function P(x) is defined as the integral from -2 to x of 1/sqrt(t+3) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative P'(x) is simply the integrand evaluated at t = x, so P'(x) = 1/sqrt(x+3). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 6

Let b(x)=∫0x(sec⁡t) dtb(x)=\int_{0}^{x}\big(\sec t\big)\,dt on an interval where sec⁡t\sec t is defined. What is b′(x)b'(x)?

  1. sec⁡x\sec x (correct answer)
  2. sec⁡xtan⁡x\sec x\tan x
  3. ∫0xsec⁡t dt\int_{0}^{x}\sec t\,dt
  4. sec⁡t\sec t
  5. [ln⁡∣sec⁡t+tan⁡t∣]0x\left[\ln|\sec t+\tan t|\right]_{0}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function b(x) is defined as the ∫0xsec⁡t dt\int_0^x \sec t \, dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative b'(x) is simply the integrand evaluated at t = x, so b′(x)=sec⁡xb'(x) = \sec x. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 7

Let X(x)=∫−2x1(t2+1)1/2 dtX(x)=\int_{-2}^{x}\dfrac{1}{(t^2+1)^{1/2}}\,dt. What is X′(x)X'(x)?

  1. 1x2+1\dfrac{1}{\sqrt{x^2+1}} (correct answer)
  2. −x(x2+1)3/2\dfrac{-x}{(x^2+1)^{3/2}}
  3. ∫−2x1(t2+1)1/2 dt\int_{-2}^{x}\dfrac{1}{(t^2+1)^{1/2}}\,dt
  4. 1t2+1\dfrac{1}{\sqrt{t^2+1}}
  5. [ln⁡∣t+t2+1∣]−2x\left[\ln\left|t+\sqrt{t^2+1}\right|\right]_{-2}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function X(x) is defined as the integral from -2 to x of 1t2+1\frac{1}{\sqrt{t^2+1}} dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative X'(x) is simply the integrand evaluated at t = x, so X'(x) = 1x2+1\frac{1}{\sqrt{x^2+1}}. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 8

Let g(x)=∫5x11+t2 dtg(x)=\int_{5}^{x}\frac{1}{1+t^2}\,dt. What is g′(x)g'(x)?

  1. ∫5x11+t2 dt\displaystyle \int_{5}^{x}\frac{1}{1+t^2}\,dt
  2. 11+t2∣t=5\displaystyle \frac{1}{1+t^2}\big|_{t=5}
  3. ∫5x−2t(1+t2)2 dt\displaystyle \int_{5}^{x}\frac{-2t}{(1+t^2)^2}\,dt
  4. 11+x2\dfrac{1}{1+x^2} (correct answer)
  5. −2x(1+x2)2\dfrac{-2x}{(1+x^2)^2}
Explanation: This is a direct application of the Fundamental Theorem of Calculus Part 1, which tells us that the derivative of an accumulation function equals the integrand evaluated at the variable limit. For g(x)=∫5x11+t2 dtg(x) = \int_{5}^{x}\frac{1}{1+t^2}\,dt, we get g′(x)=11+x2g'(x) = \frac{1}{1+x^2} by substituting x for t in the integrand. The process doesn't require finding an antiderivative or performing any integration. Choice E (−2x(1+x2)2\frac{-2x}{(1+x^2)^2}) represents the derivative of the integrand function, which would be incorrect since FTC Part 1 calls for evaluation, not differentiation. To master FTC Part 1, remember that ddx∫axf(t) dt=f(x)\frac{d}{dx}\int_{a}^{x}f(t)\,dt = f(x) when a is constant.

Question 9

Let G(x)=∫5x11+t2 dtG(x)=\int_{5}^{x}\dfrac{1}{1+t^2}\,dt. What is G′(x)G'(x)?​​

  1. ∫5x11+t2 dt\displaystyle \int_{5}^{x}\dfrac{1}{1+t^2}\,dt
  2. 11+x2\dfrac{1}{1+x^2} (correct answer)
  3. −11+x2\dfrac{-1}{1+x^2}
  4. 11+52\dfrac{1}{1+5^2}
  5. arctan⁡(x)−arctan⁡(5)\arctan(x)-\arctan(5)
Explanation: This is a direct application of the Fundamental Theorem of Calculus Part 1, which states that the derivative of ∫axf(t) dt\int_{a}^{x}f(t)\,dt equals f(x)f(x). Given G(x)=∫5x11+t2 dtG(x) = \int_{5}^{x}\frac{1}{1+t^2}\,dt, we find G′(x)G'(x) by evaluating the integrand at the upper limit xx. This gives us G′(x)=11+x2G'(x) = \frac{1}{1+x^2}, obtained by replacing tt with xx in the integrand. Choice E (arctan⁡(x)−arctan⁡(5)\arctan(x)-\arctan(5)) might tempt students who recognize that ∫11+t2 dt=arctan⁡(t)+C\int\frac{1}{1+t^2}\,dt = \arctan(t)+C, but this is the antiderivative, not the derivative we seek. Remember: when differentiating an accumulation function, simply evaluate the integrand at the variable upper limit.

Question 10

If T(x)=∫−3x(e2t−cos⁡t)dtT(x)=\int_{-3}^{x}\left(e^{2t}-\cos t\right)dt, what is T′(x)T'(x)?

  1. 2e2x+sin⁡x2e^{2x}+\sin x
  2. ∫−3x(e2t−cos⁡t)dt\displaystyle \int_{-3}^{x}\left(e^{2t}-\cos t\right)dt
  3. e2x−cos⁡xe^{2x}-\cos x (correct answer)
  4. ∫−3x(2e2t+sin⁡t)dt\displaystyle \int_{-3}^{x}\left(2e^{2t}+\sin t\right)dt
  5. (e2t−cos⁡t)∣t=−3\displaystyle (e^{2t}-\cos t)\big|_{t=-3}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1, where the derivative of an integral with variable upper limit equals the integrand evaluated at that limit. Given T(x)=∫−3x(e2t−cos⁡t) dtT(x) = \int_{-3}^{x}(e^{2t} - \cos t)\,dt, we get T′(x)=e2x−cos⁡xT'(x) = e^{2x} - \cos x by replacing t with x in the integrand. The process is purely evaluation; no integration or differentiation of the integrand is performed. Choice A (2e2x+sin⁡x2e^{2x} + \sin x) shows the derivatives of e2xe^{2x} and −cos⁡x-\cos x, which would result from incorrectly differentiating after applying FTC Part 1. To master this concept, remember that FTC Part 1 converts "derivative of integral" problems into simple substitution exercises.

Question 11

Let A(x)=∫2x(t3−4t+1) dtA(x)=\int_{2}^{x}(t^3-4t+1)\,dt. What is A′(x)A'(x)?

  1. ∫2x(t3−4t+1) dt\displaystyle \int_{2}^{x}(t^3-4t+1)\,dt
  2. x3−4x+1x^3-4x+1 (correct answer)
  3. (t3−4t+1)∣t=x\displaystyle (t^3-4t+1)\big|_{t=x}
  4. 3x2−43x^2-4
  5. ∫2x(3t2−4) dt\displaystyle \int_{2}^{x}(3t^2-4)\,dt
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1, which states that the derivative of an accumulation function equals the integrand evaluated at the upper limit. Since A(x)=∫2x(t3−4t+1) dtA(x) = \int_{2}^{x}(t^3-4t+1)\,dt, we differentiate with respect to x to get A′(x)=(t3−4t+1)∣t=x=x3−4x+1A'(x) = (t^3-4t+1)|_{t=x} = x^3-4x+1. The integrand (t3−4t+1)(t^3-4t+1) is simply evaluated at t=xt=x, replacing all instances of t with x. Choice D (3x2−43x^2-4) might tempt students who incorrectly try to differentiate the integrand itself rather than evaluate it at x. To recognize FTC Part 1 problems, look for the derivative of a definite integral where the variable appears in one or both limits of integration.

Question 12

Define M(x)=∫6xt1+t dtM(x)=\int_{6}^{x}\dfrac{\sqrt{t}}{1+t}\,dt for x≥0x\ge 0. What is M′(x)M'(x)?

  1. x1+x\dfrac{\sqrt{x}}{1+x} (correct answer)
  2. 12x(1+x)−x(1+x)2\dfrac{\frac{1}{2\sqrt{x}}(1+x)-\sqrt{x}}{(1+x)^2}
  3. ∫6xt1+t dt\int_{6}^{x}\dfrac{\sqrt{t}}{1+t}\,dt
  4. t1+t\dfrac{\sqrt{t}}{1+t}
  5. [∫t1+t dt]6x\left[\int \dfrac{\sqrt{t}}{1+t}\,dt\right]_{6}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function M(x) is defined as the integral from 6 to x of t1+t dt\dfrac{\sqrt{t}}{1+t}\, dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative M'(x) is simply the integrand evaluated at t = x, so M'(x) = x1+x\dfrac{\sqrt{x}}{1+x}. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 13

Define Y(x)=∫0xsin⁡tt2+1 dtY(x)=\int_{0}^{x}\dfrac{\sin t}{t^2+1}\,dt. What is Y′(x)Y'(x)?

  1. sin⁡xx2+1\dfrac{\sin x}{x^2+1} (correct answer)
  2. cos⁡x(x2+1)−2xsin⁡x(x2+1)2\dfrac{\cos x(x^2+1)-2x\sin x}{(x^2+1)^2}
  3. ∫0xsin⁡tt2+1 dt\int_{0}^{x}\dfrac{\sin t}{t^2+1}\,dt
  4. sin⁡tt2+1\dfrac{\sin t}{t^2+1}
  5. [∫sin⁡tt2+1 dt]0x\left[\int \dfrac{\sin t}{t^2+1}\,dt\right]_{0}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function Y(x) is defined as the integral from 0 to x of (sin t)/(t2+1t^2+1) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative Y'(x) is simply the integrand evaluated at t = x, so Y'(x) = (sin x)/(x2+1x^2+1). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 14

Let V(x)=∫2x(t+t3) dtV(x)=\int_{2}^{x}\big(\sqrt{t}+\sqrt[3]{t}\big)\,dt for x≥0x\ge 0. What is V′(x)V'(x)?

  1. x+x3\sqrt{x}+\sqrt[3]{x} (correct answer)
  2. 12x+13x2/3\dfrac{1}{2\sqrt{x}}+\dfrac{1}{3x^{2/3}}
  3. ∫2x(t+t3) dt\int_{2}^{x}\big(\sqrt{t}+\sqrt[3]{t}\big)\,dt
  4. t+t3\sqrt{t}+\sqrt[3]{t}
  5. [23t3/2+34t4/3]2x\left[\dfrac{2}{3}t^{3/2}+\dfrac{3}{4}t^{4/3}\right]_{2}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function V(x) is defined as the integral from 2 to x of (t+t1/3\sqrt{t} + t^{1/3}) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative V'(x) is simply the integrand evaluated at t = x, so V'(x) = x+x1/3\sqrt{x} + x^{1/3}. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 15

Let G(x)=∫4x1tln⁡t dtG(x)=\int_{4}^{x}\dfrac{1}{t\ln t}\,dt for x>1x>1. What is G′(x)G'(x)?

  1. 1xln⁡x\dfrac{1}{x\ln x} (correct answer)
  2. −ln⁡x−1x2(ln⁡x)2\dfrac{-\ln x-1}{x^2(\ln x)^2}
  3. ∫4x1tln⁡t dt\int_{4}^{x}\dfrac{1}{t\ln t}\,dt
  4. 1tln⁡t\dfrac{1}{t\ln t}
  5. [ln⁡∣ln⁡t∣]4x\left[\ln|\ln t|\right]_{4}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function G(x) is defined as the integral from 4 to x of 1/(t ln t) dt, which accumulates the area under the curve of f(t) = 1/(t ln t) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative G'(x) equals the integrand evaluated at x, which is 1/(x ln x). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the evaluation of an antiderivative from 4 to x, but that actually computes G(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

Question 16

Suppose e(x)=∫−2xtt2+5 dte(x)=\int_{-2}^{x}\dfrac{t}{\sqrt{t^2+5}}\,dt. What is e′(x)e'(x)?

  1. xx2+5\dfrac{x}{\sqrt{x^2+5}} (correct answer)
  2. 5(x2+5)3/2\dfrac{5}{(x^2+5)^{3/2}}
  3. ∫−2xtt2+5 dt\int_{-2}^{x}\dfrac{t}{\sqrt{t^2+5}}\,dt
  4. tt2+5\dfrac{t}{\sqrt{t^2+5}}
  5. [t2+5]−2x\left[\sqrt{t^2+5}\right]_{-2}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function e(x) is defined as the integral from -2 to x of t/sqrt(t2+5t^2+5) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative e'(x) is simply the integrand evaluated at t = x, so e'(x) = x/sqrt(x2+5x^2+5). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 17

Suppose W(x)=∫1x11+t dtW(x)=\int_{1}^{x}\dfrac{1}{1+\sqrt{t}}\,dt for x≥0x\ge 0. What is W′(x)W'(x)?

  1. 11+x\dfrac{1}{1+\sqrt{x}} (correct answer)
  2. −12x(1+x)2\dfrac{-1}{2\sqrt{x}(1+\sqrt{x})^2}
  3. ∫1x11+t dt\int_{1}^{x}\dfrac{1}{1+\sqrt{t}}\,dt
  4. 11+t\dfrac{1}{1+\sqrt{t}}
  5. [2t−2ln⁡(1+t)]1x\left[2\sqrt{t}-2\ln(1+\sqrt{t})\right]_{1}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function W(x) is defined as the integral from 1 to x of 1/(1+sqrt(t)) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative W'(x) is simply the integrand evaluated at t = x, so W'(x) = 1/(1+sqrt(x)). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 18

Suppose O(x)=∫0x(π+t) dtO(x)=\int_{0}^{x}\big(\pi+t\big)\,dt. What is O′(x)O'(x)?

  1. π+x\pi+x (correct answer)
  2. 11
  3. ∫0x(π+t) dt\int_{0}^{x}(\pi+t)\,dt
  4. π+t\pi+t
  5. [πt+t22]0x\left[\pi t+\dfrac{t^2}{2}\right]_{0}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function O(x) is defined as the integral from 0 to x of (π + t) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative O'(x) is simply the integrand evaluated at t = x, so O'(x) = π + x. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 19

Let N(x)=∫1x1(1+t2)2 dtN(x)=\int_{1}^{x}\dfrac{1}{(1+t^2)^2}\,dt. What is N′(x)N'(x)?

  1. 1(1+x2)2\dfrac{1}{(1+x^2)^2} (correct answer)
  2. −4x(1+x2)3\dfrac{-4x}{(1+x^2)^3}
  3. ∫1x1(1+t2)2 dt\int_{1}^{x}\dfrac{1}{(1+t^2)^2}\,dt
  4. 1(1+t2)2\dfrac{1}{(1+t^2)^2}
  5. [∫1(1+t2)2 dt]1x\left[\int \dfrac{1}{(1+t^2)^2}\,dt\right]_{1}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function N(x) is defined as the integral from 1 to x of 1/(1+t^2)^2 dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative N'(x) is simply the integrand evaluated at t = x, so N'(x) = 1/(1+x^2)^2. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 20

Define g(x)=∫0x11+cos⁡t dtg(x)=\int_{0}^{x}\dfrac{1}{1+\cos t}\,dt where defined. What is g′(x)g'(x)?

  1. 11+cos⁡x\dfrac{1}{1+\cos x} (correct answer)
  2. sin⁡x(1+cos⁡x)2\dfrac{\sin x}{(1+\cos x)^2}
  3. ∫0x11+cos⁡t dt\int_{0}^{x}\dfrac{1}{1+\cos t}\,dt
  4. 11+cos⁡t\dfrac{1}{1+\cos t}
  5. [∫11+cos⁡t dt]0x\left[\int \dfrac{1}{1+\cos t}\,dt\right]_{0}^{x}
Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function g(x) is defined as the integral from 0 to x of 1/(1+cos t) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative g'(x) is simply the integrand evaluated at t = x, so g'(x) = 1/(1+cos x). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.