AP Calculus AB Practice Quiz: The Fundamental Theorem of Calculus and Definite Integrals

Correct Answer: C

According to the Fundamental Theorem of Calculus [cite: 2665], if $F$ is an antiderivative of $f$, then $\int_{a}^{b} f(x) dx = F(b) - F(a)$. First, find an antiderivative of $f(x) = 2x$. An antiderivative is $F(x) = x^2$ [cite: 2663]. Then, evaluate $F(3) - F(1) = 3^2 - 1^2 = 9 - 1 = 8$ [cite: 2662].

Correct Answer: B

The Fundamental Theorem of Calculus states that if a function is defined by $F(x) = \int_{a}^{x} f(t) dt$, then $F(x)$ is an antiderivative of $f(x)$ [cite: 2664]. This means that the derivative of $F(x)$ is the original function $f(x)$, with the variable $t$ replaced by $x$. Therefore, $F'(x) = x^2 - \\cos(x)$.

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 whose derivative is $e^x + 3$. The derivative of $e^x$ is $e^x$, and the derivative of $3x$ is $3$. Therefore, the derivative of $g(x) = e^x + 3x$ is $f(x) = e^x + 3$, making it an antiderivative.

Correct Answer: C

To evaluate the definite integral, we use the Fundamental Theorem of Calculus: $\int_{a}^{b} f(x) dx = F(b) - F(a)$ [cite: 2665]. First, find an antiderivative of $f(x) = \\sin(x)$. An antiderivative is $F(x) = -\\cos(x)$ [cite: 2663]. Then, we evaluate $F(\\pi/2) - F(0) = -\\cos(\\pi/2) - (-\\cos(0)) = -0 - (-1) = 1$ [cite: 2662].

Correct Answer: B

This question is a direct application of the definition from the Fundamental Theorem of Calculus, Part 2. The theorem states that if $f$ is continuous on the interval $[a, b]$ and $F$ is an antiderivative of $f$, then $\int_{a}^{b} f(x) dx = F(b) - F(a)$ [cite: 2665].

Correct Answer: C

Using the Fundamental Theorem of Calculus [cite: 2665], we first need an antiderivative of $f(x) = \\frac{5}{x}$ [cite: 2663]. The antiderivative is $F(x) = 5 \\ln|x|$. Now, we evaluate $F(b) - F(a)$: $F(e) - F(1) = 5 \\ln(e) - 5 \\ln(1) = 5(1) - 5(0) = 5$ [cite: 2662].

Correct Answer: B

The function is defined as $g(x) = \int_{-4}^{x} \\sqrt{t^3+8} \, dt$ [cite: 2664]. To find $g(-4)$, we substitute $x=-4$ into the definition: $g(-4) = \int_{-4}^{-4} \\sqrt{t^3+8} \, dt$. The definite integral of any continuous function from a point $a$ to itself is always zero. Thus, the value is 0.

Correct Answer: A

According to the Fundamental Theorem of Calculus [cite: 2665], we must first find an antiderivative of $f(x) = 3x^2 - 4$. An antiderivative is $F(x) = x^3 - 4x$ [cite: 2663]. Next, we evaluate $F(2) - F(-1)$. $F(2) = 2^3 - 4(2) = 8 - 8 = 0$. $F(-1) = (-1)^3 - 4(-1) = -1 + 4 = 3$. Therefore, the integral is $F(2) - F(-1) = 0 - 3 = -3$ [cite: 2662].

Correct Answer: C

The question asks for the value of $\int_{1}^{e} f(t) \, dt$. The problem defines the function $G(x) = \int_{1}^{x} f(t) \, dt$ [cite: 2664]. Therefore, the value of the integral is simply the value of the function $G(x)$ evaluated at the upper limit of integration, $x=e$. We are given that $G(x) = x^2 \\ln(x)$. So, we calculate $G(e) = e^2 \\ln(e) = e^2(1) = e^2$. This problem connects the definition of a function as an integral with its evaluation [cite: 2665].

Correct Answer: A

The properties of definite integrals state that $\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx$. We can write $\int_{0}^{10} f(x) \, dx = \int_{0}^{8} f(x) \, dx + \int_{8}^{10} f(x) \, dx$. Substituting the given values: $17 = 12 + \int_{8}^{10} f(x) \, dx$. Solving for the unknown integral gives $\int_{8}^{10} f(x) \, dx = 17 - 12 = 5$. This is an application of the evaluation of definite integrals [cite: 2662].