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6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

1 min readjune 18, 2024

Emery

Emery

Emery

Emery

6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

So far in Unit 6, we’ve learned how to approximate the area under the curve. But what if we wanted to find its exact value? For that, we’ll need to learn about integrals and how they are related to the Riemann sum. We’re bringing limits back! 🧠


Σ Summation Notation

Before we get into integrals, we first need to learn how to make taking a Riemann sum faster. To do this, we’ll introduce summation notation so that we can apply algebra instead of manually computing the Riemann sum.

Let’s consider the function f(x)=x2f(x)=x^2 over the interval 0 to 5:

Untitled

Image courtesy of Emery

Now, let’s use a left Riemann sum with 5 subintervals to approximate this area.

Untitled

Image courtesy of Emery

To convert this to summation notation, we need to create a function that will give us the area of each rectangle. For now, let’s name this function AA so that A(i)A(i) gives us the area of the iith rectangle. With this function, we can write the Riemann sum as:

i=04A(i)\sum_{i=0}^{4}A(i)

Now that we have our summation notation, we need to find an expression for A(i)A(i). We know that the area of the rectangle will be bhb\cdot h, where bb is the width of the base and hh is the height.

The width of our base will be constant, and we can find it by dividing the entire interval by the number of subintervals we want:

505=1\frac{5-0}{5}=1

The height is a little more tricky to notate, but it is just the value of the function ff at each of the left endpoints. To find this next (or iith) point, let’s first find the xx-value, which we’ll denote as xix_i. To do this, we start at 0 and repeatedly add 1, the length of our subinterval.

PNG image 23.png

Image courtesy of Emery

We can write the formula for this xx-value as 1i1i or just ii. We can obtain the yy-value, or height, by simply plugging in this formula for the xx-value to our original formula like so: yi=(i)2y_i=(i)^2.

We can put these two things together in the area formula as:

A(i)=1i2A(i)=1\cdot i^2

which simplifies to

A(i)=i2A(i)=i^2

Now, we simply plug that back into our summation notation:

i=15i2\sum_{i=1}^{5}i^2

✏️ General Form of Riemann Sum

The general form for the Riemann sum in summation notation is as follows:

i=0n1Δxf(xi)\sum_{i=0}^{n-1} {\Delta x}\cdot{f({x_i})}

for left-endpoint, and

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

for right-endpoint.

Let’s break down what each symbol means!

  • nn: This denotes the number of subintervals, and is usually given in the problem
  • aa: The lower bound of the interval.
  • bb: The upper bound of the interval.
  • Δx\Delta x: This is the width of each subinterval, and can be calculated by using the formula ban\frac{b-a}{n}.
  • xix_i: This denotes the right endpoint for each rectangle, such that xi=a+Δxix_i=a+\Delta x\cdot i.
  • f(xi)f(x_i): The original function with xix_i plugged in.
  • \sum: Tells us to add up the values from 0n10\rightarrow n-1 for left Riemann sums or 1n1\rightarrow n for right Riemann sums.

Whew! That’s a lot of pieces to put together. 🧩


∫ Connecting to Definite Integrals

We learned in the last lesson that our Riemann sum becomes more precise if we use more rectangles or a greater value for nn. But what if we could have infinite rectangles, with infinitely tiny widths?

view.gif

GIF courtesy of Maplesoft

We see that as nn\rightarrow \infty, our approximation for the area under the curve gets closer and closer to perfect. This means that we can connect the Riemann sum to the definite integral like so:

limni=1nΔxf(xi)=abf(x) dx\lim_{n\to \infty}\sum_{i=1}^{n} {\Delta x}\cdot{f({x_i})}=\int_a^bf(x)\ dx

Where Δx=ban\Delta x=\frac{b-a}{n} and xi=a+Δxix_i=a+\Delta x\cdot i.

The truly amazing thing is that this limit tells us the exact value for the definite integral—not just an approximation! 👌

📌 Connecting to Definite Integrals Walkthrough

Let’s connect all the dots and practice converting from summation to definite integral notation with this example problem!

Given the function f(x)=x2+3f(x)=x^2+3 over the interval [0,5][0,5], give the summation notation for the right Riemann sum using 10 subintervals and calculate its value. Then, write the definite integral as the limit of the right Riemann sum as nn approaches infinity.

1) Solving Summation Notation

The first part of the problem asks us to give and solve the summation notation with a fixed, finite nn. Let’s first write the general form without our area function defined:

i=110Δxf(xi)\sum_{i=1}^{10}\Delta x\cdot f(x_i)

Now, we need to define our terms, starting with Δx\Delta x, which is defined as ban=5010=0.5\frac{b-a}{n}=\frac{5-0}{10}=0.5. Then, we need to define xix_i so that we can plug it into the original equation. To get the iith right endpoint, we use the formula a+Δxia+\Delta x\cdot i. In our case, a=0a= 0 and Δx=0.5\Delta x=0.5, so xi=0.5ix_i=0.5i. Now, we plug that back into our original equation to find that f(xi)=0.5i2+3f(x_i)=0.5i^2+3. Putting it all together, we find that the right Riemann sum using 10 subintervals is:

i=1100.5(0.5i2+3)\sum_{i=1}^{10}0.5\cdot (0.5i^2+3)

The next part of the problem asks us to calculate this sum. This sum essentially tells us to plug in values from 1 to 10 and sum them all up, like so:

0.5(0.5(12)+3)+0.5(0.5(22)+3)+0.5(0.5(32)+3)+0.5(0.5(42)+3)+0.5(0.5(52)+3)+0.5(0.5(62)+3)+0.5(0.5(72)+3)+0.5(0.5(82)+3)+0.5(0.5(92)+3)+0.5(0.5(102)+3)0.5(0.5(1^2)+3)+0.5(0.5(2^2)+3)+0.5(0.5(3^2)+3)+0.5(0.5(4^2)+3)+0.5(0.5(5^2)+3)+0.5(0.5(6^2)+3)+0.5(0.5(7^2)+3)+0.5(0.5(8^2)+3)+0.5(0.5(9^2)+3)+0.5(0.5(10^2)+3)

This can be rewritten as

0.5(3.5+5+7.5+11+15.5+21+27.5+35+43.5+53)=111.250.5(3.5+5+7.5+11+15.5+21+27.5+35+43.5+53)=\boxed{111.25}

Lots of math, but not too bad, right?

2) Writing the Definite Integral

The final part of the problem asks us to write the definite integral as the limit of the right Riemann sum as nn approaches infinity. First, let’s write the definite integral:

05x2+3 dx\int_0^5x^2+3\ dx

Now, let’s remind ourselves of the general form for equation definite integrals to Riemann sums:

limni=1nΔxf(xi)=abf(x) dx\lim_{n\to \infty}\sum_{i=1}^{n} {\Delta x}\cdot{f({x_i})}=\int_a^bf(x)\ dx

This is where things get a little tricky—since nn is the variable of our limit, we need to define our other terms in terms of nn.

  1. Let’s start by first defining Δx\Delta x as 50n=5n\frac{5-0}{n}=\frac{5}{n}.
  2. Then, let’s define xix_i in terms of nn, too. xi=a+Δxi=0+5ni=5nix_i=a+\Delta x\cdot i=0+\frac{5}{n}\cdot i=\frac{5}{n}i.
  3. Now let’s plug that back into f(x)f(x) to find that f(xi)=(5ni)2+3f(x_i)=(\frac{5}{n}i)^2+3.

Finally, let’s put that all back into our equation to get:

05x2+3 dx=limni=1n5n[(5ni2)+3]\int_0^5x^2+3\ dx=\lim_{n\to \infty}\sum_{i=1}^{n} \frac{5}{n}\cdot \Bigg[ \Big( \frac{5}{n} i^2\Big)+3\Bigg]

These problems take a lot of practice, but once you commit the process for finding Δx\Delta x and xix_i to memory, they become very doable. Good luck! 🍀


✏️ Riemann Sum Practice

Try to tackle these three problems yourself! When you’re done, you can check out the solutions and how to walk through them.

❓Practice Problems

  1. For the function f(x)=4x2+2x1f(x)=4x^2+2x-1, determine the definite integral 13f(x) dx\int_1^3f(x)\ dx as the limit of a Riemann sum.
  2. Express the definite integral 022ex+1 dx\int_0^2 2e^x+1 \ dx dx as the limit of a Riemann sum.
  3. The velocity function v(t)v(t) is given by v(t)=2t+1v(t)=2t+1 for the time interval [1,4][1,4]. Express the definite integral 14v(t) dt\int_1^4v(t)\ dt as the limit of a Riemann sum.

Answers below! ⬇️

📝 Solutions

Practice Problem 1 Solution

Start by defining Δx\Delta x in terms of nn: 31n=2n\frac{3-1}{n}=\frac{2}{n}. Then, do the same for xix_i: xi=a+Δxi=1+2nix_i=a+\Delta x\cdot i=1+\frac{2}{n}i. Then, plug this in to f(x)f(x) to get:

f(xi)=4(1+2ni)2+2(1+2ni)1f(x_i)=4(1+\frac{2}{n}i)^2+2(1+\frac{2}{n}i)-1

Finally, plug this all into the general limit equation to find that:

13f(x) dx=limni=1n2n(4(1+2ni)2+2(1+2ni)1)\int_1^3f(x)\ dx=\lim_{n\to \infty}\sum_{i=1}^n \frac{2}{n}\cdot \Bigg(4(1+\frac{2}{n}i)^2+2(1+\frac{2}{n}i)-1\Bigg)

The next question follows a very similar process.

Practice Problem 2 Solution

You can again start by defining Δx\Delta x in terms of nn: 20n=2n\frac{2-0}{n}=\frac{2}{n}. Then, do the same for xix_i: xi=a+Δxi=0+2ni=2nix_i=a+\Delta x\cdot i=0+\frac{2}{n}i=\frac{2}{n}i. Then, plug this in to f(x)f(x) to get:

f(xi)=2e2i/n+1f(x_i)=2e^{2i/n}+1

Finally, plug this all into the general limit equation to find that:

02f(x) dx=limni=1n2n(2e2i/n+1)\int_0^2f(x)\ dx=\lim_{n\to \infty}\sum_{i=1}^n \frac{2}{n}\cdot (2e^{2i/n}+1)

Practice Problem 3 Solution

Last question! Start by defining Δt\Delta t in terms of nn: 41n=3n\frac{4-1}{n}=\frac{3}{n}. Then, do the same for tit_i: ti=a+Δti=1+3nit_i=a+\Delta t\cdot i=1+\frac{3}{n}i. Then, plug this in to v(t)v(t) to get:

v(ti)=2(1+3ni)+1=2+6ni+1=6ni+3v(t_i)=2\Big(1+\frac{3}{n}i\Big)+1=2+\frac{6}{n}i+1=\frac{6}{n}i+3

Finally, plug this all into the general limit equation to find that:

14v(t) dt=limni=1n3n(6ni+3)\int_1^4v(t)\ dt=\lim_{n\to \infty}\sum_{i=1}^n \frac{3}{n}\cdot \Big(\frac{6}{n}i+3\Big)

💫 Closing

Now that you’ve completed this guide, you should feel comfortable expressing Riemann sums using summation notation, understand the connection between definite integrals and Riemann sums, and be able to rewrite a definite integral as the limit of a Riemann sum! 🏃

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6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

1 min readjune 18, 2024

Emery

Emery

Emery

Emery

6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

So far in Unit 6, we’ve learned how to approximate the area under the curve. But what if we wanted to find its exact value? For that, we’ll need to learn about integrals and how they are related to the Riemann sum. We’re bringing limits back! 🧠


Σ Summation Notation

Before we get into integrals, we first need to learn how to make taking a Riemann sum faster. To do this, we’ll introduce summation notation so that we can apply algebra instead of manually computing the Riemann sum.

Let’s consider the function f(x)=x2f(x)=x^2 over the interval 0 to 5:

Untitled

Image courtesy of Emery

Now, let’s use a left Riemann sum with 5 subintervals to approximate this area.

Untitled

Image courtesy of Emery

To convert this to summation notation, we need to create a function that will give us the area of each rectangle. For now, let’s name this function AA so that A(i)A(i) gives us the area of the iith rectangle. With this function, we can write the Riemann sum as:

i=04A(i)\sum_{i=0}^{4}A(i)

Now that we have our summation notation, we need to find an expression for A(i)A(i). We know that the area of the rectangle will be bhb\cdot h, where bb is the width of the base and hh is the height.

The width of our base will be constant, and we can find it by dividing the entire interval by the number of subintervals we want:

505=1\frac{5-0}{5}=1

The height is a little more tricky to notate, but it is just the value of the function ff at each of the left endpoints. To find this next (or iith) point, let’s first find the xx-value, which we’ll denote as xix_i. To do this, we start at 0 and repeatedly add 1, the length of our subinterval.

PNG image 23.png

Image courtesy of Emery

We can write the formula for this xx-value as 1i1i or just ii. We can obtain the yy-value, or height, by simply plugging in this formula for the xx-value to our original formula like so: yi=(i)2y_i=(i)^2.

We can put these two things together in the area formula as:

A(i)=1i2A(i)=1\cdot i^2

which simplifies to

A(i)=i2A(i)=i^2

Now, we simply plug that back into our summation notation:

i=15i2\sum_{i=1}^{5}i^2

✏️ General Form of Riemann Sum

The general form for the Riemann sum in summation notation is as follows:

i=0n1Δxf(xi)\sum_{i=0}^{n-1} {\Delta x}\cdot{f({x_i})}

for left-endpoint, and

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

for right-endpoint.

Let’s break down what each symbol means!

  • nn: This denotes the number of subintervals, and is usually given in the problem
  • aa: The lower bound of the interval.
  • bb: The upper bound of the interval.
  • Δx\Delta x: This is the width of each subinterval, and can be calculated by using the formula ban\frac{b-a}{n}.
  • xix_i: This denotes the right endpoint for each rectangle, such that xi=a+Δxix_i=a+\Delta x\cdot i.
  • f(xi)f(x_i): The original function with xix_i plugged in.
  • \sum: Tells us to add up the values from 0n10\rightarrow n-1 for left Riemann sums or 1n1\rightarrow n for right Riemann sums.

Whew! That’s a lot of pieces to put together. 🧩


∫ Connecting to Definite Integrals

We learned in the last lesson that our Riemann sum becomes more precise if we use more rectangles or a greater value for nn. But what if we could have infinite rectangles, with infinitely tiny widths?

view.gif

GIF courtesy of Maplesoft

We see that as nn\rightarrow \infty, our approximation for the area under the curve gets closer and closer to perfect. This means that we can connect the Riemann sum to the definite integral like so:

limni=1nΔxf(xi)=abf(x) dx\lim_{n\to \infty}\sum_{i=1}^{n} {\Delta x}\cdot{f({x_i})}=\int_a^bf(x)\ dx

Where Δx=ban\Delta x=\frac{b-a}{n} and xi=a+Δxix_i=a+\Delta x\cdot i.

The truly amazing thing is that this limit tells us the exact value for the definite integral—not just an approximation! 👌

📌 Connecting to Definite Integrals Walkthrough

Let’s connect all the dots and practice converting from summation to definite integral notation with this example problem!

Given the function f(x)=x2+3f(x)=x^2+3 over the interval [0,5][0,5], give the summation notation for the right Riemann sum using 10 subintervals and calculate its value. Then, write the definite integral as the limit of the right Riemann sum as nn approaches infinity.

1) Solving Summation Notation

The first part of the problem asks us to give and solve the summation notation with a fixed, finite nn. Let’s first write the general form without our area function defined:

i=110Δxf(xi)\sum_{i=1}^{10}\Delta x\cdot f(x_i)

Now, we need to define our terms, starting with Δx\Delta x, which is defined as ban=5010=0.5\frac{b-a}{n}=\frac{5-0}{10}=0.5. Then, we need to define xix_i so that we can plug it into the original equation. To get the iith right endpoint, we use the formula a+Δxia+\Delta x\cdot i. In our case, a=0a= 0 and Δx=0.5\Delta x=0.5, so xi=0.5ix_i=0.5i. Now, we plug that back into our original equation to find that f(xi)=0.5i2+3f(x_i)=0.5i^2+3. Putting it all together, we find that the right Riemann sum using 10 subintervals is:

i=1100.5(0.5i2+3)\sum_{i=1}^{10}0.5\cdot (0.5i^2+3)

The next part of the problem asks us to calculate this sum. This sum essentially tells us to plug in values from 1 to 10 and sum them all up, like so:

0.5(0.5(12)+3)+0.5(0.5(22)+3)+0.5(0.5(32)+3)+0.5(0.5(42)+3)+0.5(0.5(52)+3)+0.5(0.5(62)+3)+0.5(0.5(72)+3)+0.5(0.5(82)+3)+0.5(0.5(92)+3)+0.5(0.5(102)+3)0.5(0.5(1^2)+3)+0.5(0.5(2^2)+3)+0.5(0.5(3^2)+3)+0.5(0.5(4^2)+3)+0.5(0.5(5^2)+3)+0.5(0.5(6^2)+3)+0.5(0.5(7^2)+3)+0.5(0.5(8^2)+3)+0.5(0.5(9^2)+3)+0.5(0.5(10^2)+3)

This can be rewritten as

0.5(3.5+5+7.5+11+15.5+21+27.5+35+43.5+53)=111.250.5(3.5+5+7.5+11+15.5+21+27.5+35+43.5+53)=\boxed{111.25}

Lots of math, but not too bad, right?

2) Writing the Definite Integral

The final part of the problem asks us to write the definite integral as the limit of the right Riemann sum as nn approaches infinity. First, let’s write the definite integral:

05x2+3 dx\int_0^5x^2+3\ dx

Now, let’s remind ourselves of the general form for equation definite integrals to Riemann sums:

limni=1nΔxf(xi)=abf(x) dx\lim_{n\to \infty}\sum_{i=1}^{n} {\Delta x}\cdot{f({x_i})}=\int_a^bf(x)\ dx

This is where things get a little tricky—since nn is the variable of our limit, we need to define our other terms in terms of nn.

  1. Let’s start by first defining Δx\Delta x as 50n=5n\frac{5-0}{n}=\frac{5}{n}.
  2. Then, let’s define xix_i in terms of nn, too. xi=a+Δxi=0+5ni=5nix_i=a+\Delta x\cdot i=0+\frac{5}{n}\cdot i=\frac{5}{n}i.
  3. Now let’s plug that back into f(x)f(x) to find that f(xi)=(5ni)2+3f(x_i)=(\frac{5}{n}i)^2+3.

Finally, let’s put that all back into our equation to get:

05x2+3 dx=limni=1n5n[(5ni2)+3]\int_0^5x^2+3\ dx=\lim_{n\to \infty}\sum_{i=1}^{n} \frac{5}{n}\cdot \Bigg[ \Big( \frac{5}{n} i^2\Big)+3\Bigg]

These problems take a lot of practice, but once you commit the process for finding Δx\Delta x and xix_i to memory, they become very doable. Good luck! 🍀


✏️ Riemann Sum Practice

Try to tackle these three problems yourself! When you’re done, you can check out the solutions and how to walk through them.

❓Practice Problems

  1. For the function f(x)=4x2+2x1f(x)=4x^2+2x-1, determine the definite integral 13f(x) dx\int_1^3f(x)\ dx as the limit of a Riemann sum.
  2. Express the definite integral 022ex+1 dx\int_0^2 2e^x+1 \ dx dx as the limit of a Riemann sum.
  3. The velocity function v(t)v(t) is given by v(t)=2t+1v(t)=2t+1 for the time interval [1,4][1,4]. Express the definite integral 14v(t) dt\int_1^4v(t)\ dt as the limit of a Riemann sum.

Answers below! ⬇️

📝 Solutions

Practice Problem 1 Solution

Start by defining Δx\Delta x in terms of nn: 31n=2n\frac{3-1}{n}=\frac{2}{n}. Then, do the same for xix_i: xi=a+Δxi=1+2nix_i=a+\Delta x\cdot i=1+\frac{2}{n}i. Then, plug this in to f(x)f(x) to get:

f(xi)=4(1+2ni)2+2(1+2ni)1f(x_i)=4(1+\frac{2}{n}i)^2+2(1+\frac{2}{n}i)-1

Finally, plug this all into the general limit equation to find that:

13f(x) dx=limni=1n2n(4(1+2ni)2+2(1+2ni)1)\int_1^3f(x)\ dx=\lim_{n\to \infty}\sum_{i=1}^n \frac{2}{n}\cdot \Bigg(4(1+\frac{2}{n}i)^2+2(1+\frac{2}{n}i)-1\Bigg)

The next question follows a very similar process.

Practice Problem 2 Solution

You can again start by defining Δx\Delta x in terms of nn: 20n=2n\frac{2-0}{n}=\frac{2}{n}. Then, do the same for xix_i: xi=a+Δxi=0+2ni=2nix_i=a+\Delta x\cdot i=0+\frac{2}{n}i=\frac{2}{n}i. Then, plug this in to f(x)f(x) to get:

f(xi)=2e2i/n+1f(x_i)=2e^{2i/n}+1

Finally, plug this all into the general limit equation to find that:

02f(x) dx=limni=1n2n(2e2i/n+1)\int_0^2f(x)\ dx=\lim_{n\to \infty}\sum_{i=1}^n \frac{2}{n}\cdot (2e^{2i/n}+1)

Practice Problem 3 Solution

Last question! Start by defining Δt\Delta t in terms of nn: 41n=3n\frac{4-1}{n}=\frac{3}{n}. Then, do the same for tit_i: ti=a+Δti=1+3nit_i=a+\Delta t\cdot i=1+\frac{3}{n}i. Then, plug this in to v(t)v(t) to get:

v(ti)=2(1+3ni)+1=2+6ni+1=6ni+3v(t_i)=2\Big(1+\frac{3}{n}i\Big)+1=2+\frac{6}{n}i+1=\frac{6}{n}i+3

Finally, plug this all into the general limit equation to find that:

14v(t) dt=limni=1n3n(6ni+3)\int_1^4v(t)\ dt=\lim_{n\to \infty}\sum_{i=1}^n \frac{3}{n}\cdot \Big(\frac{6}{n}i+3\Big)

💫 Closing

Now that you’ve completed this guide, you should feel comfortable expressing Riemann sums using summation notation, understand the connection between definite integrals and Riemann sums, and be able to rewrite a definite integral as the limit of a Riemann sum! 🏃