# Riemann Sum Practice Problems with Answers and Solutions

> 10 Riemann sum practice problems with answers: left, right, midpoint and trapezoid sums, tables, over or under estimates and limits of sums.

Source: https://practicecalculus.com/practice-problems/riemann-sums
Updated: 2026-11-06

$$ \int_a^b f(x)\,dx \approx \sum_{i=1}^{n} f(x_i^*)\,\Delta x, \quad \Delta x = \frac{b-a}{n} $$

## At a glance

- **Problems:** 10, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 6
- **You need:** Function evaluation and sigma notation
- **Main idea:** Add up the areas of thin rectangles: height times width
- **Practice with new numbers:** [Riemann sums lesson](/learn/calculus-1/integrals/riemann-sums)

## Key takeaways

- Find $\Delta x = \frac{b-a}{n}$ first, then list the sample points before you evaluate anything.
- For an increasing function the left sum underestimates and the right sum overestimates. For a decreasing function it is the reverse.
- A limit of Riemann sums is a definite integral. Read off $\Delta x$ and the function to find the integral.

These **Riemann sum practice problems** cover left, right, midpoint and trapezoid sums, tables with uneven widths, over and under estimates, and turning a limit of sums into an integral. Try each one first. The answers come first, then full solutions.

The basic setup:

$$\Delta x = \frac{b-a}{n}, \qquad \text{sum} = \Delta x\,\big(f(x_1^*) + f(x_2^*) + \cdots + f(x_n^*)\big)$$

Left sums use the left end of each piece, right sums use the right end, and midpoint sums use the middle. For the ideas behind it, read the [Riemann sums guide](/blog/riemann-sums).

## Problems

**One function, four methods.** Use $f(x) = x^2$ on $[0, 2]$ with $n = 4$.

1. Find the left Riemann sum $L_4$.
2. Find the right Riemann sum $R_4$.
3. Find the midpoint sum $M_4$.
4. Find the trapezoid sum $T_4$, and compare all four with the exact value $\int_0^2 x^2\,dx = \frac{8}{3}$.

**Over or under**

5. Find the left sum $L_4$ for $f(x) = \frac{1}{x}$ on $[1, 3]$. Is it an overestimate or an underestimate of $\ln 3$?
6. A car's velocity is measured at a few times:

| $t$ (s) | 0 | 2 | 5 | 6 |
| --- | --- | --- | --- | --- |
| $v(t)$ (m/s) | 10 | 14 | 20 | 18 |

Estimate the distance traveled from $t = 0$ to $t = 6$ with a left sum, a right sum and a trapezoid sum.

7. For $f(x) = x^2$ on $[0, 2]$, explain why $M_4 < \frac{8}{3} < T_4$.

**Limits of sums**

8. Write $\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n}\left(1 + \frac{3i}{n}\right)^2$ as a definite integral and evaluate it.
9. Find the right sum $R_n$ for $f(x) = x$ on $[0, 2]$ in closed form, then take the limit as $n \to \infty$.
10. Estimate $\int_0^{\pi}\sin x\,dx$ with a midpoint sum, $n = 4$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $L_4 = 1.75$ |
| 2 | $R_4 = 3.75$ |
| 3 | $M_4 = 2.625$ |
| 4 | $T_4 = 2.75$; exact value $\approx 2.667$ |
| 5 | $L_4 = \frac{77}{60} \approx 1.283$, an overestimate |
| 6 | Left $82$ m, right $106$ m, trapezoid $94$ m |
| 7 | $x^2$ is concave up |
| 8 | $\int_1^4 x^2\,dx = 21$ |
| 9 | $R_n = \frac{2(n+1)}{n} \to 2$ |
| 10 | $\approx 2.052$ (exact value is $2$) |

## Step-by-step solutions

**1.** $\Delta x = \frac{2 - 0}{4} = 0.5$, so the points are $0, 0.5, 1, 1.5, 2$. The left sum uses the first four:

$$L_4 = 0.5\,(0 + 0.25 + 1 + 2.25) = 0.5(3.5) = 1.75$$

**2.** The right sum uses the last four points:

$$R_4 = 0.5\,(0.25 + 1 + 2.25 + 4) = 0.5(7.5) = 3.75$$

**3.** The midpoints are $0.25, 0.75, 1.25, 1.75$. Their squares are $0.0625, 0.5625, 1.5625, 3.0625$, which add to $5.25$:

$$M_4 = 0.5(5.25) = 2.625$$

**4.** With equal widths, the trapezoid sum is the average of the left and right sums:

$$T_4 = \frac{1.75 + 3.75}{2} = 2.75$$

The exact value $\frac{8}{3} \approx 2.667$ sits between $M_4$ and $T_4$. The midpoint sum is the closest.

**5.** $\Delta x = 0.5$ and the left points are $1, 1.5, 2, 2.5$:

$$L_4 = 0.5\left(1 + \tfrac{2}{3} + \tfrac{1}{2} + \tfrac{2}{5}\right) = 0.5\cdot\frac{77}{30} = \frac{77}{60} \approx 1.283$$

Since $\frac{1}{x}$ is decreasing, the left end of each piece is the highest point, so every rectangle sticks out above the curve. It is an overestimate of $\ln 3 \approx 1.099$.

**6.** The widths are $2$, $3$ and $1$. Multiply each width by the right height:

- Left: $2(10) + 3(14) + 1(20) = 82$ m
- Right: $2(14) + 3(20) + 1(18) = 106$ m
- Trapezoid: $2\cdot\frac{10+14}{2} + 3\cdot\frac{14+20}{2} + 1\cdot\frac{20+18}{2} = 24 + 51 + 19 = 94$ m

**7.** A concave up curve bends upward, so each trapezoid's top edge lies above the curve and $T_4$ is too big. Each midpoint rectangle has the same area as the trapezoid under the tangent line at that midpoint, and a concave up curve lies above its tangent lines, so $M_4$ is too small.

**8.** Match the pattern $\Delta x = \frac{3}{n}$ and $x_i = 1 + \frac{3i}{n}$. The interval starts at $1$ and has length $3$, so it is $[1, 4]$, and the function is $x^2$:

$$\int_1^4 x^2\,dx = \frac{4^3 - 1^3}{3} = \frac{63}{3} = 21$$

**9.** $\Delta x = \frac{2}{n}$ and $x_i = \frac{2i}{n}$. Use $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$:

$$R_n = \sum_{i=1}^{n}\frac{2}{n}\cdot\frac{2i}{n} = \frac{4}{n^2}\cdot\frac{n(n+1)}{2} = \frac{2(n+1)}{n} \to 2$$

This matches the triangle area $\frac{1}{2}(2)(2) = 2$.

**10.** $\Delta x = \frac{\pi}{4}$, and the midpoints are $\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}$. Their sines are about $0.3827, 0.9239, 0.9239, 0.3827$, which add to about $2.6131$:

$$M_4 \approx \frac{\pi}{4}(2.6131) \approx 2.052$$

Since $\sin x$ is concave down on $[0, \pi]$, the midpoint sum overestimates the exact value $2$.

## Common mistakes on these problems

- **Using $n + 1$ heights.** A left or right sum with $n = 4$ has exactly four rectangles. List the points and cross out the one you do not use.
- **Assuming equal widths in a table.** In problem 6 the gaps are $2$, $3$ and $1$. Using a single $\Delta x$ gives the wrong answer.
- **Mixing up the over and under rules.** Left and right depend on increasing or decreasing. Midpoint and trapezoid depend on concavity.
- **Misreading the interval in a limit.** In problem 8, $x_i = 1 + \frac{3i}{n}$ means the interval starts at $1$, not $0$.

## Keep practicing

For unlimited sums with new functions and instant feedback, work through the [Riemann sums lesson](/learn/calculus-1/integrals/riemann-sums) and the [trapezoid sums lesson](/learn/calculus-1/integrals/trapezoid-sums). Then see how the limit of these sums becomes an exact answer in the [definite integral problems](/practice-problems/definite-integrals) and the [Fundamental Theorem problems](/practice-problems/fundamental-theorem-of-calculus).

## Frequently asked questions

### How do you know if a Riemann sum is an overestimate or underestimate?

For left and right sums, look at whether the function is increasing or decreasing. For midpoint and trapezoid sums, look at concavity: if the graph is concave up, the trapezoid sum is too big and the midpoint sum is too small.

### What is the difference between the trapezoidal rule and a Riemann sum?

A Riemann sum uses rectangles. The trapezoidal rule uses trapezoids, which follow the slope of the curve. With equal widths, the trapezoid sum equals the average of the left and right sums.

### Do Riemann sums need equal widths?

No. With data in a table, the widths are whatever gaps the table gives. Multiply each height by its own width, as in problem 6.