AP Calculus AB Practice Quiz: The Fundamental Theorem of Calculus and Accumulation Functions

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

Let $g$ be a function defined by $g(x) = \int_{2}^{x} \\cos(t^2) dt$. Which of the following is an expression for $g'(x)$?

A)$\\cos(x^2)$B)$\\sin(x^2)$C)$2x \\cos(x^2)$D)$\\cos(x^2) - \\cos(4)$

All Questions (9)

Let $g$ be a function defined by $g(x) = \int_{2}^{x} \\cos(t^2) dt$. Which of the following is an expression for $g'(x)$?

A) $\\cos(x^2)$

B) $\\sin(x^2)$

C) $2x \\cos(x^2)$

D) $\\cos(x^2) - \\cos(4)$

Correct Answer: A

According to the Fundamental Theorem of Calculus, Part 1, if $g(x) = \int_{a}^{x} f(t) dt$, then $g'(x) = f(x)$. In this case, $f(t) = \\cos(t^2)$, so $g'(x) = \\cos(x^2)$. [cite: 2643]

The rate at which a pollutant is entering a lake is modeled by the function $R(t)$, where $t$ is the time in hours since midnight. Which of the following expressions represents the total amount of pollutant that entered the lake between 8 AM ($t=8$) and noon ($t=12$)?

A) $R(12) - R(8)$

B) $\\frac{R(12) - R(8)}{12 - 8}$

C) $\int_{8}^{12} R(t) dt$

D) $\\frac{d}{dt} \int_{8}^{12} R(t) dt$

Correct Answer: C

An accumulation function can be represented by a definite integral. To find the total accumulation of a quantity given its rate of change, $R(t)$, over an interval $[a, b]$, we calculate the definite integral of the rate function over that interval. Therefore, the total amount of pollutant is the integral of the rate $R(t)$ from $t=8$ to $t=12$. [cite: 2641, 2647]

Let $f$ be a continuous function. A new function $H$ is defined by $H(x) = \int_{-3}^{x} (t^2 + 5) dt$. What is the value of $H'(-1)$?

A) -2

B) 6

C) 8

D) 10

Correct Answer: B

By the Fundamental Theorem of Calculus, Part 1, $H'(x) = \\frac{d}{dx} \int_{-3}^{x} (t^2 + 5) dt = x^2 + 5$. To find $H'(-1)$, we substitute $x=-1$ into the expression for $H'(x)$: $H'(-1) = (-1)^2 + 5 = 1 + 5 = 6$. [cite: 2643]

The graph of a continuous function $f$ is shown above, consisting of two line segments and a semicircle. Let the function $g$ be defined by $g(x) = \int_{0}^{x} f(t) dt$. What is the value of $g(4)$?

A) $2 - 2\\pi$

B) $2 - \\pi$

C) $4 - \\pi$

D) $4 - 2\\pi$

Correct Answer: B

The value of $g(4)$ is the net area under the curve of $f$ from $t=0$ to $t=4$. The area from $t=0$ to $t=2$ is a triangle with base 2 and height 2, so its area is $\\frac{1}{2}(2)(2) = 2$. The area from $t=2$ to $t=4$ is a semicircle below the x-axis with radius 1. Its area is $\\frac{1}{2}\\pi(1)^2 = \\frac{\\pi}{2}$. Since it is below the axis, we subtract this area. However, the graph shows a semicircle of radius 2 from x=2 to x=6. The question asks for g(4), which is the area from 0 to 4. The area from 2 to 4 is a quarter circle of radius 2 below the x-axis. Area = $\\frac{1}{4}\\pi(2)^2 = \\pi$. So, $g(4) = (\\text{Area from 0 to 2}) - (\\text{Area from 2 to 4}) = 2 - \\pi$. [cite: 2641, 2642, 2648]

Let $f$ be the function whose graph is shown above. If $g(x) = \int_{-2}^{x} f(t) dt$, on which of the following intervals is $g$ increasing?

A) $(-2, 0)$

B) $(0, 3)$

C) $(3, 5)$

D) $(-2, 3)$

Correct Answer: D

The function $g(x)$ is increasing when its derivative, $g'(x)$, is positive. According to the Fundamental Theorem of Calculus, $g'(x) = f(x)$. Therefore, $g$ is increasing when $f(x) > 0$. Based on the provided graph, the function $f(x)$ is positive on the interval $(-2, 3)$. [cite: 2643]

If $F(x) = \int_{c}^{x} f(t) dt$, where $f$ is a continuous function and $c$ is a constant, which of the following statements is always true?

A) $F(x) = f'(x)$

B) $F'(x) = f(x)$

C) $F(c) = f(c)$

D) $F'(x) = f(x) - f(c)$

Correct Answer: B

This question is a direct conceptual check of the Fundamental Theorem of Calculus, Part 1. The theorem states that for a continuous function $f$, the derivative of its accumulation function $F(x) = \int_{c}^{x} f(t) dt$ is the function $f(x)$ itself. Thus, $F'(x) = f(x)$. [cite: 2643]

Let $f$ be a continuous function. What is $\\frac{d}{dx} \int_{1}^{x^3} f(t) dt$?

A) $f(x^3)$

B) $f'(x^3)$

C) $3x^2 f(x^3)$

D) $f(x^3) - f(1)$

Correct Answer: C

This problem requires the Fundamental Theorem of Calculus combined with the Chain Rule. Let $u = x^3$. Then we are finding $\\frac{d}{dx} \int_{1}^{u} f(t) dt$. By the Chain Rule, this is $(\\frac{d}{du} \int_{1}^{u} f(t) dt) \\cdot \\frac{du}{dx}$. The first part is $f(u)$ by FTC, and the second part is $3x^2$. Substituting $u=x^3$ back gives $f(x^3) \\cdot 3x^2$. [cite: 2643]

The function $f$ is continuous and its graph is shown above. If $h(x)$ is a function defined by $h(x) = \int_{1}^{x} f(t) dt$, what is the value of $h(5)$?

A) -1

B) 0

C) 1

D) 2

Correct Answer: A

The value of $h(5)$ is the net signed area under the graph of $f$ from $t=1$ to $t=5$. The area from $t=1$ to $t=3$ is a triangle above the axis with base 2 and height 1, so its area is $\\frac{1}{2}(2)(1) = 1$. The area from $t=3$ to $t=5$ is a triangle below the axis with base 2 and height 2, so its area is $\\frac{1}{2}(2)(2) = 2$. The net area is the area above minus the area below: $1 - 2 = -1$. [cite: 2641, 2642, 2648]

The graph of the continuous function $f$ is shown above. Let $G(x) = \int_{0}^{x} f(t) dt$. For what value of $x$ does $G$ have a local maximum?

A) 1

B) 2

C) 4

D) 5

Correct Answer: C

A function $G(x)$ has a local maximum where its derivative, $G'(x)$, changes from positive to negative. By the Fundamental Theorem of Calculus, $G'(x) = f(x)$. We need to find where the graph of $f(x)$ changes from positive (above the x-axis) to negative (below the x-axis). This occurs at $x=4$. [cite: 2643]