Integrals with Riemann Sums for MATH 141

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Computing integrals via Riemann sums is conceptual in MATH 141. May appear as one midterm question.

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Lesson

Understanding Riemann Sums

Before we had the Fundamental Theorem of Calculus, mathematicians needed a way to actually define what an integral means. You can't just say "find the area under a curve" without explaining how to calculate it. Riemann sums give us that definition — they show that an integral is really just a limit of adding up rectangle areas.

Even though we now have easier methods (like antiderivatives), understanding Riemann sums is important because:

They explain why integrals work
Some limits can only be evaluated by recognizing them as Riemann sums
Numerical integration (what computers do) is based on this idea

The Core Idea

To find the area under a curve from $x = a$ to $x = b$:

Divide the interval $[a, b]$ into $n$ equal pieces
Build rectangles on each piece (using the function value at some point in each piece)
Add up all the rectangle areas
Take the limit as $n \to \infty$ (more and more thinner rectangles)

Key Components

Width of Each Rectangle

If you divide $[a, b]$ into $n$ equal parts:

$$\Delta x = \frac{b - a}{n}$$

Sample Points

The $i$-th subinterval goes from $x_{i-1}$ to $x_i$, where:

$$x_i = a + i \cdot \Delta x$$

You can use different points in each subinterval:

Type	Sample Point $x_i^*$	Description
Right endpoint	$x_i = a + i \cdot \Delta x$	Uses right edge of each rectangle
Left endpoint	$x_{i-1} = a + (i-1) \cdot \Delta x$	Uses left edge of each rectangle
Midpoint	$\frac{x_{i-1} + x_i}{2}$	Uses center of each rectangle

The Riemann Sum

We write $x_i^*$ (read "x-i-star") to represent the sample point in the $i$-th subinterval. The star is a placeholder — it could be the left endpoint, right endpoint, or midpoint. Your problem will tell you which one to use. Until then, we use the star to keep the formula general.

$$R_n = \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x$$

This is the sum of all rectangle areas: height × width for each rectangle.

Play with this interactive graph to see Riemann sums in action! Adjust:

$a$ and $b$: the left and right endpoints of the interval
$n$: the number of rectangles (try increasing it to see the approximation improve)
$c$: the sample point type (0 = left endpoint, 0.5 = midpoint, 1 = right endpoint)

Notice how each rectangle's area (height × width) contributes to the total sum, and how the approximation gets better as $n$ increases.

The Definite Integral

$$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x$$

The integral is the limit of Riemann sums as the number of rectangles approaches infinity.

Useful Summation Formulas

When computing Riemann sums, you'll need these:

$$\sum_{i=1}^{n} 1 = n$$

$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

$$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$

$$\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2$$

Example 1: Setting Up a Riemann Sum

Write a right-endpoint Riemann sum for $\int_0^3 x^2 \, dx$ with $n$ rectangles.

Step 1: Find $\Delta x$

$$\Delta x = \frac{b - a}{n} = \frac{3 - 0}{n} = \frac{3}{n}$$

Step 2: Find the right endpoints

$$x_i = a + i \cdot \Delta x = 0 + i \cdot \frac{3}{n} = \frac{3i}{n}$$

Step 3: Write $f(x_i)$

$$f(x_i) = x_i^2 = \left(\frac{3i}{n}\right)^2 = \frac{9i^2}{n^2}$$

Step 4: Write the Riemann sum

$$R_n = \sum_{i=1}^{n} f(x_i) \cdot \Delta x = \sum_{i=1}^{n} \frac{9i^2}{n^2} \cdot \frac{3}{n} = \sum_{i=1}^{n} \frac{27i^2}{n^3}$$

$$\boxed{R_n = \frac{27}{n^3} \sum_{i=1}^{n} i^2}$$

Example 2: Evaluating the Integral via Riemann Sum

Compute $\int_0^2 x^2 \, dx$ using the definition (right endpoints).

Step 1: Set up

$\Delta x = \frac{2}{n}$, and $x_i = \frac{2i}{n}$

Step 2: Write the Riemann sum

$$R_n = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n} = \sum_{i=1}^{n} \frac{4i^2}{n^2} \cdot \frac{2}{n} = \frac{8}{n^3} \sum_{i=1}^{n} i^2$$

Step 3: Apply the summation formula

$$R_n = \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{8n(n+1)(2n+1)}{6n^3}$$

Step 4: Simplify

$$R_n = \frac{4(n+1)(2n+1)}{3n^2} = \frac{4(2n^2 + 3n + 1)}{3n^2} = \frac{8n^2 + 12n + 4}{3n^2}$$

$$= \frac{8}{3} + \frac{4}{n} + \frac{4}{3n^2}$$

Step 5: Take the limit

$$\int_0^2 x^2 \, dx = \lim_{n \to \infty} R_n = \lim_{n \to \infty} \left(\frac{8}{3} + \frac{4}{n} + \frac{4}{3n^2}\right) = \frac{8}{3}$$

$$\boxed{\frac{8}{3}}$$

Example 3: Evaluating a Limit Using Riemann Sums

Evaluate $\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2$.

Step 1: Recognize this as a Riemann sum

Compare to $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \cdot \Delta x$

Here: $\Delta x = \frac{1}{n}$ and $x_i = \frac{i}{n}$

So $f(x_i) = \left(\frac{i}{n}\right)^2 = x_i^2$, meaning $f(x) = x^2$

Step 2: Identify the interval

With $x_i = \frac{i}{n}$ for $i = 1, 2, \ldots, n$:

The values range from $\frac{1}{n} \to 0$ to $\frac{n}{n} = 1$

This is a right Riemann sum on $[0, 1]$.

Step 3: Write as an integral and evaluate

$$\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 = \int_0^1 x^2 \, dx$$

$$= \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \boxed{\frac{1}{3}}$$

Example 4: Left vs Right Endpoints

For $f(x) = x$ on $[0, 4]$ with $n = 4$ rectangles, compute both left and right Riemann sums.

Step 1: Find $\Delta x$ and partition points

$\Delta x = \frac{4-0}{4} = 1$

The partition points are: $x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4$

Step 2: Compute the left Riemann sum

Use left endpoints: $x_0, x_1, x_2, x_3$

$$L_4 = f(0) \cdot 1 + f(1) \cdot 1 + f(2) \cdot 1 + f(3) \cdot 1$$ $$= 0 + 1 + 2 + 3 = 6$$

Step 3: Compute the right Riemann sum

Use right endpoints: $x_1, x_2, x_3, x_4$

$$R_4 = f(1) \cdot 1 + f(2) \cdot 1 + f(3) \cdot 1 + f(4) \cdot 1$$ $$= 1 + 2 + 3 + 4 = 10$$

Step 4: Compare to actual area

$$\int_0^4 x \, dx = \frac{x^2}{2}\Big|_0^4 = 8$$

The left sum underestimates ($6 < 8$) and right sum overestimates ($10 > 8$) because $f(x) = x$ is increasing.

$$\boxed{L_4 = 6, \quad R_4 = 10}$$

Example 5: Midpoint Riemann Sum

Estimate $\int_0^2 x^3 \, dx$ using a midpoint Riemann sum with $n = 4$.

Step 1: Find $\Delta x$ and subintervals

$\Delta x = \frac{2-0}{4} = 0.5$

Subintervals: $[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]$

Step 2: Find midpoints

$x_1^* = 0.25, \quad x_2^* = 0.75, \quad x_3^* = 1.25, \quad x_4^* = 1.75$

Step 3: Evaluate and sum

$$M_4 = \Delta x \cdot [f(0.25) + f(0.75) + f(1.25) + f(1.75)]$$ $$= 0.5 \cdot [(0.25)^3 + (0.75)^3 + (1.25)^3 + (1.75)^3]$$ $$= 0.5 \cdot [0.015625 + 0.421875 + 1.953125 + 5.359375]$$ $$= 0.5 \cdot 7.75 = \boxed{3.875}$$

Actual value: $\int_0^2 x^3 \, dx = \frac{x^4}{4}\Big|_0^2 = 4$

The midpoint sum (3.875) is quite close to the actual value (4).

Example 6: True or False

True or False: As $n \to \infty$, left and right Riemann sums always converge to the same value for a continuous function.

True. For any continuous function on a closed interval $[a, b]$, the definite integral exists and equals the limit of any Riemann sum (left, right, midpoint, or any other choice of sample points).

The difference between left and right sums shrinks as $n$ increases, and both converge to $\int_a^b f(x) \, dx$.

Example 7: True or False

True or False: If $f(x)$ is increasing on $[a, b]$, then the left Riemann sum underestimates the integral.

True. For an increasing function, the left endpoint of each subinterval is the minimum value on that subinterval. So each rectangle's height is less than (or equal to) the actual function values, giving an underestimate.

Similarly, the right Riemann sum would overestimate because it uses the maximum value on each subinterval.

Example 8: True or False

True or False: The expression $\lim_{n \to \infty} \sum_{i=1}^{n} f(a + i\Delta x) \cdot \Delta x$ where $\Delta x = \frac{b-a}{n}$ equals $\int_a^b f(x) \, dx$.

True. This is exactly the definition of the definite integral using right endpoints:

$\Delta x = \frac{b-a}{n}$ is the width of each rectangle
$x_i = a + i\Delta x$ gives the right endpoints
$f(x_i) \cdot \Delta x$ is each rectangle's area
The sum adds all rectangles
The limit as $n \to \infty$ defines the integral

Common Mistakes and Misunderstandings

❌ Mistake: Confusing $x_i$ with $i$

Wrong: Using $f(i)$ instead of $f(x_i)$ in the Riemann sum.

Why it's wrong: The variable $i$ is just a counter (1, 2, 3, ...). The actual $x$-value is $x_i = a + i \cdot \Delta x$.

Correct: Always substitute $x_i = a + i \cdot \frac{b-a}{n}$ into the function.

❌ Mistake: Forgetting that $\Delta x$ depends on $n$

Wrong: Treating $\Delta x$ as a constant when taking the limit.

Why it's wrong: $\Delta x = \frac{b-a}{n}$ contains $n$, so it affects how terms simplify when you take $\lim_{n \to \infty}$.

Correct: Always write out $\Delta x = \frac{b-a}{n}$ and simplify the entire expression before taking the limit.

❌ Mistake: Wrong limits when converting a limit to an integral

Wrong: Assuming every Riemann sum limit has bounds $[0, 1]$.

Why it's wrong: The bounds depend on how $\Delta x$ and $x_i$ are defined. Look at the structure carefully.

Correct: Identify $\Delta x$ and $x_i$ from the given limit. The interval is $[a, b]$ where $a$ is the smallest $x_i$ value and $b = a + n \cdot \Delta x$.

Formulas & Reference

Riemann Sum

R_n = \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x

The sum of rectangle areas approximating the area under a curve. Divide [a,b] into n equal pieces, build rectangles using function values at sample points, and add up the areas.

Variables:

$n$:: number of rectangles
$f(x_i^*)$:: height of the i-th rectangle (function value at sample point)
$\Delta x$:: width of each rectangle, equal to (b-a)/n
$x_i^*$:: sample point in the i-th subinterval (left, right, or midpoint)

Definite Integral as Limit of Riemann Sums

\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x

The definite integral is defined as the limit of Riemann sums as the number of rectangles approaches infinity. This is the formal definition of the integral.