Definite Integrals

MIT 18.01 Single Variable Calculus, lecture 16, explained in plain English

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Part 1 · 0:00–10:00

This part introduces definite integrals as a way to find the area under a curve between two points, using thin rectangles that get closer and closer to the true area.

Key ideas

A definite integral finds the area under a curve between a starting point a and an ending point b. The curve is y = f(x), and the bottom of the area is the x axis. 3:11
A curvy region cannot be split exactly into rectangles, so we use rectangles that are only almost rectangles. They overshoot in some places and miss in others. 4:17
To fix the errors, we make the rectangles thinner and thinner. The small amounts we miss or add go toward zero, so in the limit we get the exact area. 5:04
The lecturer will work one example, f(x) = x^2, starting at a = 0 and letting the end point b be any value. He picks the right end of each rectangle for its height. 7:27
Rectangles are easy to work with because their area is just base times height. The base length is always b divided by n, where n is the number of pieces. 9:19

Words to know

Definite integral: A calculus tool that gives the area under a curve between two set points, written with a start and an end on the integral sign.
Limit: The value that something gets closer and closer to as it changes, such as rectangles getting thinner and thinner.
Parabola: A U-shaped curve, like the graph of f(x) = x^2.
Integral: A calculus operation that adds up tiny pieces, used here to find areas under curves.

Example

Imagine trying to measure the area of a hill shape on a piece of paper by covering it with small square tiles. Some tiles stick out past the edge and some leave gaps, but if the tiles get smaller and smaller, the total gets closer to the true area. That is the same idea as using thinner and thinner rectangles.

3 checked practice questions for this part are in the app.

Part 2 · 10:00–20:00

This part shows how a row of thin rectangles can be added up to estimate the area under the curve y = x^2, and how a stack of stone blocks helps trap that total between two bounds.

Key ideas

Split the distance from 0 to b into n equal pieces, so each rectangle has base b/n. Each rectangle's height is the curve's value at its right edge, which is (kb/n)^2 for the kth rectangle. 10:00
Adding the rectangle areas gives (b/n)^3 times the sum 1^2 + 2^2 + ... + n^2. The lecturer pulls out the common (b/n)^3 factor to make the total easier to handle. 13:16
To understand the sum of squares, he builds a stepped stone pyramid with layers of n by n, (n-1) by (n-1), and so on down to 1. A smooth pyramid placed inside it gives a lower bound of 1/3 n^3. 18:21
A slightly larger smooth pyramid placed outside the stepped one gives an upper bound of 1/3 (n+1)^3. So the sum is trapped between these two values as n gets large. 19:07
The stacked layers are real solid blocks, so they can be treated as volumes, not just flat areas. The lecturer says he is making volumes out of small honest increments. 19:54

Words to know

limit: The value a quantity gets closer and closer to as a number, such as n, keeps growing larger and larger.
coefficient: A number multiplied in front of a variable or expression, such as the 2 in 2^2 or the 3 in 3^2 here.
volume: The amount of three-dimensional space a solid shape takes up.
base: The flat bottom face of a shape. In the volume formula, base times height is used, and for the pyramid the base is n by n.

Example

Imagine stacking boxes to make a staircase, then comparing it with a smooth ramp made of the same boxes. The ramp fits inside the staircase, and a bigger ramp covers it, so you know the staircase's total lies between those two. That is the same trick the lecturer uses to trap the sum.

3 checked practice questions for this part are in the app.

Part 3 · 20:00–30:00

This part shows how to find the exact area under the curve y = x² by trapping it between blocks, and introduces a short way to write long sums.

Key ideas

The area under x² from 0 to b is trapped between two stacks of blocks. The blocks are slightly too small and slightly too big, so the true area sits between them. 20:18
Dividing everything by n³ leaves a sum of squares divided by n³ that is squeezed between 1/3 and a number close to 1/3. As n gets very large, both edges reach 1/3. 21:28
The area under x² from 0 to b is 1/3 times b³. The sum of squares is hard to find exactly, so the method factors out the easy part and studies the hard part. 22:28
Summation notation, written with a capital sigma, is a short way to write a long list of added terms. It saves writing the same pattern again and again. 25:31
The second example is f(x) = x. The area under this line from 0 to b is a triangle with base b and height b, so its area is 1/2 times b squared. 29:19

Words to know

limit: the value something gets closer and closer to as a number grows larger and larger, such as n going to infinity.
closed form: a neat formula that gives the answer directly, without needing to add up a long list of terms.
summation notation: a short way to write adding up a list of terms, using a capital sigma and an index that counts through the terms.
distributive law: the rule that lets you pull a shared factor out of every term in a sum, such as writing b³/n³ times a sum.

Example

Think of estimating how much paint a wall needs by stacking small square tiles under the curve. If you use more and smaller tiles, your estimate gets closer to the true area. The lesson does the same thing with blocks: the more blocks, the tighter the trap around the true area.

3 checked practice questions for this part are in the app.

Part 4 · 30:00–40:00

This part shows that the area under a flat line is easy to find, that a pattern for powers of x helps find other areas, and that Riemann sums are a way to build up an area from thin rectangles.

Key ideas

When the function is the flat line f(x) = 1, the area from 0 to b is just the base b times the height 1, so the area is b. 30:42
The areas for x, x squared, and 1 all follow one pattern: for x to the power n, the area from 0 to b is b to the power n+1, divided by n+1. So for x cubed, the guess is b to the 4th, divided by 4. 33:01
To find an area in general, you cut the interval into n equal-width pieces, where each width is (b minus a) divided by n. Then you pick a height for each piece and add up the rectangles. 36:28
You may pick any height inside each piece, not only the right-hand end. Each choice gives a different staircase of rectangles, and all of them work. 37:10
When the rectangles get thinner and thinner, the sum becomes a definite integral, written with a curved S and dx. The dx is the limit of the small width delta x. 38:43

Words to know

definite integral: The exact area under a curve between two points, found by adding up very thin slices.
Riemann sum: A total made by adding up the areas of many rectangles that fill the space under a curve.
delta x: The width of one small slice or rectangle when you cut an interval into pieces.
limit: The value a sum gets closer and closer to as the pieces get thinner and thinner.

Example

Think of filling a long, flat tray with tiles. If every tile is the same height, you just multiply the length by the height to get the total. A curved area works the same way, except you cut it into thin strips, measure each strip as a rectangle, and add them up.

3 checked practice questions for this part are in the app.

Part 5 · 40:00–47:12

This part shows that an integral can be read as a running total, using a borrowing example where the amount you borrow and owe changes over a year.

Key ideas

An integral can be seen as a cumulative sum, which means adding up many small pieces. The lesson uses a bank loan to show this idea. 40:00
Units help you check a formula. The rate is in dollars per year, and multiplying by a time in years leaves dollars, which is the amount actually borrowed. 41:09
Splitting the year into 365 days, each day's borrowing is about the rate times a tiny time step, delta t = 1/365. Adding all these days up is very close to the integral from 0 to 1 of f(t) dt. 41:59
With continuous compounding, a loan of P grows to P times e^(rt) after time t. The lesson gives 0.05 per year as an example interest rate. 45:04
To find what you owe at year end, each day's borrowing grows for the time left in the year, which is 1 minus t. The total becomes an integral with e^(r(1-t)) inside, and the lesson says this is the kind of integral a borrowing strategy must deal with. 46:15

Words to know

cumulative sum: A running total made by adding up many small amounts, one after another, until you reach the end.
delta t: A very small step in time. Here it is 1/365, one day measured as a fraction of a year.
principal: The original amount of money you borrow or invest, before any interest is added.
compounded continuously: Interest that is added all the time, not just once a month or once a year, so the balance keeps growing smoothly.

Example

Think of filling a bathtub with a tap whose flow changes over the day. Each small slice of time adds a little water, and the total water is the running sum of all those slices. In the lesson, each day's borrowing is one slice, and adding the slices gives the total borrowed.

3 checked practice questions for this part are in the app.

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Notes written by AI (Claude) from the lecture's official transcript, so check anything important against the lecture. MIT OpenCourseWare · CC BY-NC-SA 4.0 · not affiliated with or endorsed by MIT. These notes are shared under the same license.

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