AP Calculus AB Practice Quiz: Riemann Sums, Summation Notation, and Definite Integral Notation

Correct Answer: C

Based on the provided content, a Riemann sum is defined as 'the sum of products, each of which is the value of the function at a point in a subinterval multiplied by the length of that subinterval of the partition.' [cite: 2635]

Correct Answer: C

The provided content states that the definite integral, denoted by $\int_{a}^{b} f(x) dx$, 'is the limit of Riemann sums as the widths of the subintervals approach 0.' [cite: 2637]

Correct Answer: A

The limit of a Riemann sum $\lim_{n \to \\infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$ over the interval $[a, b]$ is represented by the definite integral $\int_{a}^{b} f(x) dx$. In this case, the function is $f(x) = x^3$ and the interval is $[2, 5]$. [cite: 2636, 2638]

Correct Answer: B

This is the limit of a right Riemann sum. The interval starts at $a=3$. The width of each subinterval is $\Delta x = \\frac{2}{n}$, so the total width of the interval is $b-a = n \\cdot \\frac{2}{n} = 2$. Thus, $b = a+2 = 3+2=5$. The sample points are $x_i = 3 + i\Delta x = 3 + \\frac{2i}{n}$. The function is $f(x_i) = \\cos(x_i)$, so $f(x) = \\cos(x)$. The corresponding definite integral is $\int_{3}^{5} \\cos(x) dx$. [cite: 2636]

Correct Answer: B

The provided content explicitly states that we can 'interpret the limiting case of the Riemann sum as a definite integral' and that 'The limit of an approximating Riemann sum can be interpreted as a definite integral.' [cite: 2633, 2634]

Correct Answer: B

For the integral $\int_{1}^{4} \\sqrt{x} dx$, the interval is $[a, b] = [1, 4]$. The width of each subinterval is $\Delta x = \\frac{b-a}{n} = \\frac{4-1}{n} = \\frac{3}{n}$. For a right Riemann sum, the sample points are $x_i = a + i\Delta x = 1 + \\frac{3i}{n}$. The function is $f(x) = \\sqrt{x}$, so $f(x_i) = \\sqrt{1 + \\frac{3i}{n}}$. The Riemann sum is $\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \\sqrt{1 + \\frac{3i}{n}} \\frac{3}{n}$. The definite integral is the limit of this sum. [cite: 2638]

Correct Answer: D

The content states that 'The definite integral of a continuous function f over the interval [a, b] ... is the limit of Riemann sums as the widths of the subintervals approach 0.' This limit represents the exact value. [cite: 2637]

Correct Answer: A

We can identify the components of the Riemann sum. $\Delta x = \\frac{1}{n}$. If we choose the interval to start at $a=0$, then $b-a = n \\cdot \Delta x = n \\cdot \\frac{1}{n} = 1$, so $b=1$. The sample points are $x_i = a + i\Delta x = 0 + \\frac{i}{n} = \\frac{i}{n}$. The function is $f(x_i) = \\frac{1}{1+x_i}$, so $f(x) = \\frac{1}{1+x}$. The integral is $\int_{0}^{1} \\frac{1}{1+x} dx$. (Note: Choosing $a=1$ would lead to $\int_1^2 \\frac{1}{x} dx$, which is equivalent but not an option). [cite: 2636]

Correct Answer: C

The content explicitly states: 'A definite integral can be translated into the limit of a related Riemann sum, and the limit of a Riemann sum can be written as a definite integral.' [cite: 2638]

Correct Answer: A

For the integral $\int_{0}^{\\pi} \\sin(x) dx$, we have $[a, b] = [0, \\pi]$. The subinterval width is $\Delta x = \\frac{b-a}{n} = \\frac{\\pi-0}{n} = \\frac{\\pi}{n}$. The right endpoints are $x_i = a + i\Delta x = 0 + i\\frac{\\pi}{n} = \\frac{i\\pi}{n}$. The function is $f(x) = \\sin(x)$, so $f(x_i) = \\sin(\\frac{i\\pi}{n})$. The corresponding limit of the Riemann sum is $\lim_{n \to \\infty} \sum_{i=1}^{n} f(x_i) \Delta x = \lim_{n \to \\infty} \sum_{i=1}^{n} \\sin(\\frac{i\\pi}{n}) \\frac{\\pi}{n}$. [cite: 2638]

Correct Answer: B

The expression for the sample points in a right Riemann sum is $x_i = a + i\Delta x$. Comparing this to the term in the exponent, $5 + 2i/n$, we can identify the starting point of the interval as $a=5$. The width of the subintervals is $\Delta x = \\frac{2}{n}$. The total width of the interval is $b-a = n \\cdot \Delta x = n \\cdot \\frac{2}{n} = 2$. Therefore, the endpoint is $b = a+2 = 5+2=7$. The interval is $[5, 7]$. [cite: 2636]

Correct Answer: D

This limit can be interpreted in two common ways. Interpretation 1: Let $a=2$. Then $\Delta x = 1/n$, so $b=a+1=3$. The sample points are $x_i = 2+i/n$, and the function is $f(x)=x^4$. This gives $\int_{2}^{3} x^4 dx$. Interpretation 2: Let $a=0$. Then $\Delta x = 1/n$, so $b=1$. The sample points are $x_i = i/n$, and the function is $f(x)=(2+x)^4$. This gives $\int_{0}^{1} (2+x)^4 dx$. Both definite integrals are valid representations of the limit and are equivalent. [cite: 2636, 2638]

Correct Answer: B

The provided content explicitly states that 'A Riemann sum, which requires a partition of an interval I, is the sum of products...'. This indicates that a partition is a prerequisite for a Riemann sum. [cite: 2635]

Correct Answer: C

The definite integral is over the interval $[a, b] = [4, 6]$. For a Riemann sum with $n$ equal subintervals, the width of each subinterval is $\Delta x = \\frac{b-a}{n}$. In this case, $\Delta x = \\frac{6-4}{n} = \\frac{2}{n}$. [cite: 2635, 2636]

Correct Answer: A

First, manipulate the expression to fit the form $f(x_i)\Delta x$. Divide the numerator and denominator by $n^2$: $\\frac{n/n^2}{(n^2+i^2)/n^2} = \\frac{1/n}{1+(i/n)^2}$. The sum becomes $\lim_{n \to \\infty} \sum_{i=1}^{n} \\frac{1}{1+(i/n)^2} \\cdot \\frac{1}{n}$. We can now identify $\Delta x = 1/n$ and $x_i = i/n$. This corresponds to an integral over $[0, 1]$ with the function $f(x) = \\frac{1}{1+x^2}$. The definite integral is $\int_{0}^{1} \\frac{1}{1+x^2} dx$. [cite: 2636, 2638]