AP Calculus BC Practice Quiz: The Fundamental Theorem of Calculus and Definite Integrals
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 10 questions to check your progress.
Question 1 of 10
All Questions (10)
A) 4
B) 6
C) 8
D) 10
Correct Answer: C
Based on the Fundamental Theorem of Calculus [cite: 2665], if $f(x) = 2x$, then an antiderivative is $F(x) = x^2$ [cite: 2663]. The value of the definite integral is $F(3) - F(1) = 3^2 - 1^2 = 9 - 1 = 8$ [cite: 2662].
A) $g(x) = f'(x)$
B) $g'(x) = f(x)$
C) $\int g(x) dx = f(x)$
D) $g(x) = f(x)$
Correct Answer: B
This question tests the definition of an antiderivative [cite: 2663]. An antiderivative of a function $f$ is a function $g$ whose derivative is $f$. Therefore, $g'(x) = f(x)$.
A) $\cos(x^2)$
B) $2x\cos(x^2)$
C) $\sin(x^2)$
D) $-\cos(x^2)$
Correct Answer: C
According to the Fundamental Theorem of Calculus [cite: 2664], if a function is defined as $F(x) = \int_{a}^{x} f(t) dt$, then $F'(x) = f(x)$. In this case, $f(t) = \sin(t^2)$, so $F'(x) = \sin(x^2)$.
A) $F(x)$ is the derivative of $f(x)$.
B) $f(x)$ is the derivative of $F(x)$.
C) $F(x)$ and $f(x)$ are unrelated.
D) $F(x) = f(x)$.
Correct Answer: B
The Fundamental Theorem of Calculus [cite: 2665] states that to evaluate $\int_{a}^{b} f(x) dx$, we use an antiderivative $F$ of $f$. By definition [cite: 2663], an antiderivative is a function whose derivative is the original function, so $F'(x) = f(x)$, which means $f(x)$ is the derivative of $F(x)$.
A) -1
B) 0
C) 1
D) 2
Correct Answer: C
To evaluate the definite integral analytically [cite: 2662], we use the Fundamental Theorem of Calculus [cite: 2665]. An antiderivative of $f(x) = \cos(x)$ is $F(x) = \sin(x)$. The value of the integral is $F(\pi/2) - F(0) = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.
A) $6x + 4$
B) $x^3 + 2x^2$
C) $3x^3 + 4x^2$
D) $x^2 + x$
Correct Answer: B
An antiderivative of a function $f$ is a function $g$ whose derivative is $f$ [cite: 2663]. We need to find a function $g(x)$ such that $g'(x) = 3x^2 + 4x$. The derivative of $g(x) = x^3 + 2x^2$ is $g'(x) = \frac{d}{dx}(x^3 + 2x^2) = 3x^2 + 4x$, which matches $f(x)$.
A) $F(2) - F(7)$
B) $F(7) - F(2)$
C) $f(7) - f(2)$
D) $f'(7) - f'(2)$
Correct Answer: B
This is a direct application of the evaluation part of the Fundamental Theorem of Calculus [cite: 2665]. If $F$ is an antiderivative of $f$, then $\int_{a}^{b} f(x) dx = F(b) - F(a)$. Here, $a=2$ and $b=7$, so the integral is equal to $F(7) - F(2)$.
A) $G(x) = \ln(x) + C$
B) $G'(x) = -\frac{1}{x^2}$
C) $G(1) = 0$
D) $G(x)$ is decreasing for $x > 0$.
Correct Answer: C
The function is defined by $G(x) = \int_{a}^{x} f(t) dt$ where $a=1$ [cite: 2664]. Evaluating this function at the lower limit of integration gives $G(1) = \int_{1}^{1} \frac{1}{t} \,dt$. The integral of any continuous function from a point to itself is 0. Therefore, $G(1) = 0$.
A) 1
B) 2
C) 3/2
D) 4
Correct Answer: B
We must first evaluate the definite integral analytically using the Fundamental Theorem of Calculus [cite: 2662, 2665]. The integrand is $f(x) = x^{-2}$. An antiderivative is $F(x) = -x^{-1} = -\frac{1}{x}$. The integral is $F(k) - F(1) = (-\frac{1}{k}) - (-\frac{1}{1}) = 1 - \frac{1}{k}$. We set this equal to $\frac{1}{2}$: $1 - \frac{1}{k} = \frac{1}{2}$. Solving for $k$, we get $\frac{1}{k} = 1 - \frac{1}{2} = \frac{1}{2}$, so $k=2$.
A) $f$ must be differentiable on $[a, b]$.
B) $f$ must be a polynomial function.
C) $F$ must be a positive function.
D) $f$ must be continuous on $[a, b]$.
Correct Answer: D
The statement of the Fundamental Theorem of Calculus [cite: 2665] begins with the condition, 'If $f$ is continuous on the interval $[a, b]$...'. This is a required condition for the theorem to apply. The other options are not necessary conditions.