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

Correct Answer: B

The 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: D

The content defines a Riemann sum 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: B

This limit represents a Riemann sum. The width of each subinterval is $\Delta x = \frac{2}{n}$. The sample points are $x_i = 1 + \frac{2i}{n}$. This corresponds to the interval $[a, b]$ where $a=1$ and $b-a=2$, so $b=3$. The function is $f(x) = x^3$. Therefore, the limit is equivalent to the definite integral $\int_{1}^{3} x^3 dx$. [cite: 2636, 2638]

Correct Answer: C

The provided content explicitly states that the limit of an approximating Riemann sum can be interpreted as a definite integral. The given expression is the formal definition of a definite integral as the limit of a Riemann sum. [cite: 2633, 2634]

Correct Answer: C

For the integral $\int_{2}^{5} x^2 dx$, the interval is $[2, 5]$, so the total width is $b-a = 5-2=3$. For $n$ subintervals, the width of each is $\Delta x = \frac{3}{n}$. The right endpoint of the $i$-th subinterval is $x_i = a + i\Delta x = 2 + i(\frac{3}{n})$. The function is $f(x) = x^2$, so $f(x_i) = (2 + \frac{3i}{n})^2$. The Riemann sum is $\sum_{i=1}^{n} f(x_i)\Delta x$, and its limit is the expression in option C. [cite: 2638]

Correct Answer: B

The term inside the function is of the form $a + i\Delta x$. Here, it is $3 + \frac{4i}{n}$. This means the starting point of the interval is $a=3$. The width of each subinterval is $\Delta x = \frac{4}{n}$. Since $\Delta x = \frac{b-a}{n}$, we have $\frac{4}{n} = \frac{b-3}{n}$, which implies $b-3=4$, so $b=7$. The interval is $[3, 7]$. [cite: 2636]

Correct Answer: A

In this Riemann sum, the width of the subintervals is $\Delta x = \frac{1}{n}$. The sample points are $x_i = a + i\Delta x$. Since the term inside the function is $\frac{i}{n}$, we can infer $a=0$, making $x_i = \frac{i}{n}$. The function being evaluated is $f(x_i) = \sqrt{\frac{i}{n}}$, so $f(x) = \sqrt{x}$. The interval is $[0, 1]$ since $a=0$ and $b-a = n \cdot \Delta x = n \cdot \frac{1}{n} = 1$. Thus, the integral is $\int_{0}^{1} \sqrt{x} dx$. [cite: 2636, 2638]

Correct Answer: C

The definite integral is the limit of the sum $\sum f(c_i) \Delta x$. As the number of subintervals $n$ approaches infinity, the width of each subinterval $\Delta x$ approaches 0. The notation 'dx' in the integral represents this infinitesimal width of the subintervals. [cite: 2637]

Correct Answer: C

To interpret this as a Riemann sum, we need to factor it into the form $\sum f(x_i) \Delta x$. We can rewrite the term as $\frac{i^4}{n^5} = (\frac{i}{n})^4 \cdot \frac{1}{n}$. Here, we can identify $\Delta x = \frac{1}{n}$ and $x_i = \frac{i}{n}$. This implies an interval starting at $a=0$ and with a total width of 1, so the interval is $[0, 1]$. The function is $f(x_i) = (\frac{i}{n})^4$, so $f(x) = x^4$. The corresponding integral is $\int_{0}^{1} x^4 dx$. [cite: 2636, 2638]

Correct Answer: A

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

Correct Answer: C

The content explicitly states that '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.' This highlights the direct, convertible relationship between the two concepts. [cite: 2638]

Correct Answer: B

We identify the components of the Riemann sum. The width of the subintervals is $\Delta x = \frac{5}{n}$. The term inside the function is $x_i = 2 + \frac{5i}{n}$. This is in the form $a + i\Delta x$, so the starting point is $a=2$. The total width of the interval is $b-a = n \cdot \Delta x = n \cdot \frac{5}{n} = 5$. Therefore, the endpoint is $b = a+5 = 2+5 = 7$. The function is $f(x_i) = e^{x_i}$, so $f(x) = e^x$. The resulting integral is $\int_{2}^{7} e^x dx$. [cite: 2636, 2638]

Correct Answer: C

A Riemann sum requires a partition of an interval $I$. If the number of subintervals, $n$, increases over a fixed interval $[a, b]$, the width of each subinterval, $\Delta x = (b-a)/n$, must decrease. The definite integral is the limit as these widths approach zero. [cite: 2635, 2637]

Correct Answer: B

Let's analyze the Riemann sum. The width is $\Delta x = \frac{3}{n}$. The sample point evaluation is on $1 + \frac{3i}{n}$. If we let the function be $f(x) = \ln(x)$, then $x_i = 1 + \frac{3i}{n}$. This means $a=1$ and $\Delta x = \frac{3}{n}$, so $b-a=3 \implies b=4$. This gives the integral $\int_{1}^{4} \ln(x) dx$. Alternatively, if we let the function be $f(x) = \ln(1+x)$, then $x_i = \frac{3i}{n}$. This means $a=0$ and $\Delta x = \frac{3}{n}$, so $b-a=3 \implies b=3$. This gives the integral $\int_{0}^{3} \ln(1+x) dx$. Both are mathematically equivalent. Since option B is present, it is the correct choice based on a common way of interpreting these limits. [cite: 2636, 2638]

Correct Answer: C

The content states that we can 'interpret the limiting case of the Riemann sum as a definite integral.' A definite integral of a non-negative function represents the exact area under the curve. The Riemann sum itself is a finite sum of rectangle areas that approximates this value. [cite: 2633]