Calculus 1 Quiz: Ftc And Accumulation Functions
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Ftc And Accumulation FunctionsQuestion 1 of 7

What is the equation of the line tangent to the graph of F(x)=∫1x2et−1 dtF(x) = \int_{1}^{x^2} e^{t-1} \, dt at the point where x=1x=1?

y=2x−2y = 2x-2
y=x−1y = x-1
y=2ex−2ey = 2ex-2e
y=0y=0
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Calculus 1 Quiz

Calculus 1 Quiz: Ftc And Accumulation Functions

Practice Ftc And Accumulation Functions in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Ftc And Accumulation Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

What is the equation of the line tangent to the graph of F(x)=∫1x2et−1 dtF(x) = \int_{1}^{x^2} e^{t-1} \, dt at the point where x=1x=1?

  1. y=2x−2y = 2x-2 (correct answer)
  2. y=x−1y = x-1
  3. y=2ex−2ey = 2ex-2e
  4. y=0y=0
Explanation: The equation of a tangent line is y−F(1)=F′(1)(x−1)y - F(1) = F'(1)(x-1). First, we find the point of tangency by calculating F(1)=∫112et−1 dt=∫11et−1 dt=0F(1) = \int_{1}^{1^2} e^{t-1} \, dt = \int_{1}^{1} e^{t-1} \, dt = 0. So the point is (1,0)(1, 0). Next, we find the slope by calculating F′(x)F'(x) using the Fundamental Theorem of Calculus with the Chain Rule: F′(x)=ex2−1⋅ddx(x2)=ex2−1⋅2xF'(x) = e^{x^2-1} \cdot \frac{d}{dx}(x^2) = e^{x^2-1} \cdot 2x. Now we evaluate the slope at x=1x=1: F′(1)=e12−1⋅2(1)=e0⋅2=1⋅2=2F'(1) = e^{1^2-1} \cdot 2(1) = e^0 \cdot 2 = 1 \cdot 2 = 2. Plugging the point (1,0)(1,0) and slope m=2m=2 into the point-slope formula gives y−0=2(x−1)y - 0 = 2(x-1), which simplifies to y=2x−2y = 2x - 2.

Question 2

Let ff be a continuous function. If F(x)=∫2x0f(t) dtF(x) = \int_{2x}^{0} f(t) \, dt and f(0)=3f(0)=3, what is the value of F′(0)F'(0)?

  1. −6-6 (correct answer)
  2. −3-3
  3. 3
  4. 6
Explanation: To find F′(x)F'(x), we first rewrite the integral as F(x)=−∫02xf(t) dtF(x) = -\int_{0}^{2x} f(t) \, dt. Now we apply the Fundamental Theorem of Calculus, Part 1, along with the Chain Rule. The derivative is F′(x)=−f(2x)⋅ddx(2x)=−f(2x)⋅2=−2f(2x)F'(x) = -f(2x) \cdot \frac{d}{dx}(2x) = -f(2x) \cdot 2 = -2f(2x). To find the value of F′(0)F'(0), we substitute x=0x=0 into this expression: F′(0)=−2f(2⋅0)=−2f(0)F'(0) = -2f(2 \cdot 0) = -2f(0). We are given that f(0)=3f(0)=3. Therefore, F′(0)=−2(3)=−6F'(0) = -2(3) = -6.

Question 3

Let g(x)=∫−4x(∣t+2∣−1) dtg(x) = \int_{-4}^x (|t+2| - 1) \, dt. Find the value of g(2)g(2).

  1. 0
  2. 4 (correct answer)
  3. 6
  4. 8
Explanation: To evaluate the integral, we must split it at the point where the expression inside the absolute value changes sign, which is at t=−2t=-2. For t<−2t < -2, ∣t+2∣=−(t+2)|t+2| = -(t+2). For t≥−2t \ge -2, ∣t+2∣=t+2|t+2| = t+2. So, we split the integral: g(2)=∫−4−2(−(t+2)−1) dt+∫−22((t+2)−1) dtg(2) = \int_{-4}^{-2} (-(t+2) - 1) \, dt + \int_{-2}^2 ((t+2) - 1) \, dt. This becomes g(2)=∫−4−2(−t−3) dt+∫−22(t+1) dtg(2) = \int_{-4}^{-2} (-t-3) \, dt + \int_{-2}^2 (t+1) \, dt. Evaluating the integrals: [−t22−3t]−4−2+[t22+t]−22[-\frac{t^2}{2} - 3t]_{-4}^{-2} + [\frac{t^2}{2} + t]_{-2}^2. The first part is (−42−3(−2))−(−162−3(−4))=(−2+6)−(−8+12)=4−4=0(-\frac{4}{2} - 3(-2)) - (-\frac{16}{2} - 3(-4)) = (-2+6) - (-8+12) = 4-4=0. The second part is (42+2)−(42−2)=(2+2)−(2−2)=4−0=4(\frac{4}{2} + 2) - (\frac{4}{2} - 2) = (2+2) - (2-2) = 4-0=4. The total value is 0+4=40+4=4.

Question 4

Let G(x)=∫x21cos⁡(t2) dtG(x) = \int_{x^2}^{1} \cos(t^2) \, dt. Find G′(x)G'(x).

  1. 2xcos⁡(x4)2x \cos(x^4)
  2. cos⁡(1)−cos⁡(x4)\cos(1) - \cos(x^4)
  3. −2xcos⁡(x4)-2x \cos(x^4) (correct answer)
  4. −cos⁡(x4)-\cos(x^4)
Explanation: To find the derivative of the integral, we first reverse the limits of integration, which changes the sign: G(x)=−∫1x2cos⁡(t2) dtG(x) = -\int_{1}^{x^2} \cos(t^2) \, dt. Now, we apply the Fundamental Theorem of Calculus, Part 1, along with the Chain Rule. Let f(t)=cos⁡(t2)f(t) = \cos(t^2) and u(x)=x2u(x) = x^2. Then G(x)=−∫1u(x)f(t) dtG(x) = -\int_1^{u(x)} f(t) \, dt. The derivative is G′(x)=−f(u(x))⋅u′(x)G'(x) = -f(u(x)) \cdot u'(x). Substituting the functions, we get G′(x)=−cos⁡((x2)2)⋅(2x)=−2xcos⁡(x4)G'(x) = -\cos((x^2)^2) \cdot (2x) = -2x \cos(x^4).

Question 5

If H(x)=∫sin⁡xx31+t4 dtH(x) = \int_{\sin x}^{x^3} \sqrt{1+t^4} \, dt, what is H′(x)H'(x)?

  1. 3x21+x12+(cos⁡x)1+sin⁡4x3x^2 \sqrt{1+x^{12}} + (\cos x) \sqrt{1+\sin^4 x}
  2. 3x21+x12−(cos⁡x)1+sin⁡4x3x^2 \sqrt{1+x^{12}} - (\cos x) \sqrt{1+\sin^4 x} (correct answer)
  3. 1+x12−1+sin⁡4x\sqrt{1+x^{12}} - \sqrt{1+\sin^4 x}
  4. 1+(x3)4−1+(sin⁡x)4\sqrt{1+(x^3)^4} - \sqrt{1+(\sin x)^4}
Explanation: We use the general form of the Fundamental Theorem of Calculus, Part 1, for an integral with variable limits: ddx∫h(x)g(x)f(t) dt=f(g(x))g′(x)−f(h(x))h′(x)\frac{d}{dx} \int_{h(x)}^{g(x)} f(t) \, dt = f(g(x))g'(x) - f(h(x))h'(x). Here, f(t)=1+t4f(t) = \sqrt{1+t^4}, g(x)=x3g(x) = x^3, and h(x)=sin⁡xh(x) = \sin x. Their derivatives are g′(x)=3x2g'(x) = 3x^2 and h′(x)=cos⁡xh'(x) = \cos x. Applying the formula: H′(x)=f(x3)⋅(3x2)−f(sin⁡x)⋅(cos⁡x)=1+(x3)4⋅(3x2)−1+(sin⁡x)4⋅(cos⁡x)=3x21+x12−(cos⁡x)1+sin⁡4xH'(x) = f(x^3) \cdot (3x^2) - f(\sin x) \cdot (\cos x) = \sqrt{1+(x^3)^4} \cdot (3x^2) - \sqrt{1+(\sin x)^4} \cdot (\cos x) = 3x^2\sqrt{1+x^{12}} - (\cos x)\sqrt{1+\sin^4 x}.

Question 6

Let ff be a continuous function. Define two functions G(x)=∫0xf(t) dtG(x) = \int_0^x f(t) \, dt and H(x)=∫3xf(t) dtH(x) = \int_3^x f(t) \, dt. If G(3)=5G(3)=5, what is the value of H(0)H(0)?

  1. 55
  2. 00
  3. −5-5 (correct answer)
  4. Cannot be determined from the information given.
Explanation: We are given G(3)=∫03f(t) dt=5G(3) = \int_0^3 f(t) \, dt = 5. We need to find H(0)=∫30f(t) dtH(0) = \int_3^0 f(t) \, dt. Using the properties of definite integrals, reversing the limits of integration negates the value of the integral. Therefore, H(0)=∫30f(t) dt=−∫03f(t) dtH(0) = \int_3^0 f(t) \, dt = -\int_0^3 f(t) \, dt. Since we know ∫03f(t) dt=5\int_0^3 f(t) \, dt = 5, we have H(0)=−5H(0) = -5.

Question 7

Let ff be a continuous function whose average value on the interval [1,5][1, 5] is 10. If G(x)=∫1xf(t) dtG(x) = \int_1^x f(t) \, dt, what is the value of G(5)G(5)?

  1. 10
  2. Cannot be determined from the information given.
  3. 2.5
  4. 40 (correct answer)
Explanation: This question tests your understanding of the relationship between average value of a function and definite integrals. When you see "average value" paired with an integral expression like G(x)=∫1xf(t) dtG(x) = \int_1^x f(t) \, dt, think about how these concepts connect through the Fundamental Theorem of Calculus. The average value of a continuous function ff on interval [a,b][a,b] is defined as 1b−a∫abf(t) dt\frac{1}{b-a}\int_a^b f(t) \, dt. Since ff has average value 10 on [1,5][1,5], we have: 15−1∫15f(t) dt=10\frac{1}{5-1}\int_1^5 f(t) \, dt = 10 This gives us 14∫15f(t) dt=10\frac{1}{4}\int_1^5 f(t) \, dt = 10, so ∫15f(t) dt=40\int_1^5 f(t) \, dt = 40. Since G(x)=∫1xf(t) dtG(x) = \int_1^x f(t) \, dt, we get G(5)=∫15f(t) dt=40G(5) = \int_1^5 f(t) \, dt = 40. Choice (A) 10 represents the common error of confusing the average value with the definite integral itself. The average value is 10, but that's not what G(5)G(5) represents. Choice (B) "Cannot be determined" might seem tempting since we don't know the explicit form of f(t)f(t), but the average value gives us exactly the information we need to find the definite integral. Choice (C) 2.5 appears to come from incorrectly dividing 10 by 4, perhaps confusing the relationship between average value and the interval length. Remember: average value problems often give you enough information to find definite integrals even when the function's formula isn't provided. Focus on the average value formula and work backwards to find what you need.