Welcome to the second major part of calculus. Until now, we have been focused on the "Problem of Tangents," which led us to the derivative. Now, we begin our study of the "Problem of Areas," which will lead us to the integral.
It may not be obvious why finding the "area under a curve" is useful. We will motivate this new, powerful idea by starting with a familiar concept: distance.
Let's begin with a question that requires no calculus at all.
If you drive a car at a constant velocity of 60 mph for 2 hours, how far have you traveled?
This is a simple calculation: $$ \text{Distance} = \text{Velocity} \times \text{Time} $$ $$ D = 60 \text{ mi/hr} \times 2 \text{ hr} = 120 \text{ miles} $$
Now, let's look at this graphically. The graph of your velocity $v(t) = 60$ is a horizontal line. The time interval is from $t=0$ to $t=2$.
Look closely at the graph. The "120" has a clear geometric meaning: it is the area of the shaded rectangle.
$$ \text{Area} = \text{Height} \times \text{Width} = 60 \times 2 = 120 $$
This gives us our first "big idea" of integral calculus:
Distance Traveled = Area Under the Velocity Curve
The constant velocity case was easy. But what happens if the velocity is not constant? This is the "Distance Problem" that calculus must solve.
For example, what if the velocity is given by $v(t) = t^2$ and we want to find the distance traveled from $t=0$ to $t=2$?
We can no longer just multiply velocity by time, because the velocity is always changing. But our "big idea" from Part 1 still holds: the distance we want to find is the area under the velocity curve $v(t) = t^2$ from $t=0$ to $t=2$.
We have a problem. We don't have a simple geometric formula for the area of this "curvy" shape (called a parabolic spandrel). We can't find the area, so we can't find the distance.
This leads us to the central "Area Problem" of calculus: How do we find the area under a general function $f(x)$ on an interval $[a, b]$?
If we can solve this general Area Problem, we will be able to solve the Distance Problem (and many others!).
We don't know the area of the curvy shape, but we do know how to find the area of rectangles. Our strategy will be to "chop and sum":
Let's try this for $f(x) = x^2$ on the interval $[0, 2]$. We'll start with $n=4$ rectangles.
First, we "chop" the interval $[0, 2]$ into 4 pieces. The width of each piece is:
$$ \Delta x = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5 $$
This gives us four subintervals: $[0, 0.5]$, $[0.5, 1]$, $[1, 1.5]$, and $[1.5, 2]$.
Now, how do we choose the height of the rectangle for each interval? We have two simple choices:
This graph shows our $f(x)=x^2$ with $n=4$ rectangles. You can toggle between the Right-Hand Sum ($R_n$) and Left-Hand Sum ($L_n$) to see the difference.
Find the area approximation for $f(x) = x^2$ on $[0, 2]$ using $n=4$ rectangles and right endpoints.
The width is $\Delta x = 0.5$. The subintervals are $[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]$.
For $R_4$, we use the right endpoints for the heights: $x_1=0.5, x_2=1, x_3=1.5, x_4=2$.
The total area, $R_4$, is the sum of the areas of the 4 rectangles:
$$ R_4 = (\text{Area 1}) + (\text{Area 2}) + (\text{Area 3}) + (\text{Area 4}) $$
$$ R_4 = (\Delta x \cdot f(0.5)) + (\Delta x \cdot f(1)) + (\Delta x \cdot f(1.5)) + (\Delta x \cdot f(2)) $$
We can factor out the common width $\Delta x = 0.5$:
$$ R_4 = 0.5 \cdot [f(0.5) + f(1) + f(1.5) + f(2)] $$
$$ R_4 = 0.5 \cdot [ (0.5)^2 + (1)^2 + (1.5)^2 + (2)^2 ] $$
$$ R_4 = 0.5 \cdot [ 0.25 + 1 + 2.25 + 4 ] $$
$$ R_4 = 0.5 \cdot [ 7.5 ] = 3.75 $$
From the graph, we can clearly see this is an overestimate of the true area.
Find the area approximation for $f(x) = x^2$ on $[0, 2]$ using $n=4$ rectangles and left endpoints.
The width is still $\Delta x = 0.5$. The subintervals are $[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]$.
For $L_4$, we use the left endpoints for the heights: $x_0=0, x_1=0.5, x_2=1, x_3=1.5$.
$$ L_4 = \Delta x \cdot [f(0) + f(0.5) + f(1) + f(1.5)] $$
$$ L_4 = 0.5 \cdot [ (0)^2 + (0.5)^2 + (1)^2 + (1.5)^2 ] $$
$$ L_4 = 0.5 \cdot [ 0 + 0.25 + 1 + 2.25 ] $$
$$ L_4 = 0.5 \cdot [ 3.5 ] = 1.75 $$
From the graph, we can clearly see this is an underestimate of the true area.
So, we know the true area $A$ is somewhere between our two estimates: $1.75 < A < 3.75$. This is a wide range, so our approximation isn't very good yet.
Calculate the right-hand sum $R_2$ for $f(x) = x^2$ on the interval $[0, 2]$.
How can we get a better approximation of the area?
The answer is to use more rectangles. If we use $n=8$ or $n=20$, the width $\Delta x$ of each rectangle gets smaller, and they "fit" the curve much better. The error (the white space between the rectangles and the curve) gets much smaller.
Go back to the graph in Part 3 and move the slider for $n$ from $n=4$ to $n=20$. Watch how both $L_n$ and $R_n$ "hug" the curve more tightly. The underestimate $L_n$ grows, and the overestimate $R_n$ shrinks.
This leads us to the logical conclusion...
To get a better and better approximation, we just let $n$ get bigger and bigger. To get the exact area, we let $n$ go to infinity.
The exact area $A$ is the limit of the sum of the areas of the approximating rectangles.
As we use more rectangles, our sum for the area gets very long and tiring to write. For example, $R_n$ is:
$$ R_n = f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + \dots + f(x_n)\Delta x $$
Mathematicians developed a shorthand called Sigma Notation (using the Greek letter $\Sigma$) to express this efficiently. The expression above is written as:
$$ \sum_{i=1}^n f(x_i) \Delta x $$
Let's break down its anatomy:
We read this as: "The sum, as $i$ goes from 1 to $n$, of $f(x_i)$ times $\Delta x$."
Now we can write our formal definition using this powerful new notation.
The area $A$ of the region under the graph of a continuous function $f(x)$ from $a$ to $b$ is the limit of the sum of the areas of approximating rectangles:
$$ A = \lim_{n \to \infty} R_n = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x $$
(We could also use the left-hand sum, $L_n$. As $n \to \infty$, both limits converge to the same value, $A$.)
This sum is called a Riemann Sum, and this limit is the foundation of the integral. We have successfully connected the "Area Problem" to our first big idea in calculus: the limit.
To set up these limit definitions, we just need a general set of formulas for $\Delta x$ and the endpoints $x_i$.
For a function $f(x)$ on an interval $[a, b]$ with $n$ equal subintervals:
The sum is $R_n = \sum_{i=1}^n f(a + i\Delta x) \Delta x$
The sum is $L_n = \sum_{i=1}^n f(a + (i-1)\Delta x) \Delta x$
Let's practice setting these up. We will not evaluate the limit (that's for Section 5.2!), but just write the expression for the area.
Write an expression for the exact area under $f(x) = \sqrt{x}$ from $x=1$ to $x=4$ using the limit of a right-hand sum.
We are just building the formula $A = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x$. We need the pieces:
1. Interval: $[a, b] = [1, 4]$. So, $a=1$ and $b=4$.
2. Width: $\Delta x = \frac{b-a}{n} = \frac{4-1}{n} = \frac{3}{n}$.
3. Right Endpoints ($x_i$): $x_i = a + i\Delta x = 1 + i \cdot \left(\frac{3}{n}\right) = 1 + \frac{3i}{n}$.
4. Function Value ($f(x_i)$): $f(x) = \sqrt{x}$, so $f(x_i) = \sqrt{1 + \frac{3i}{n}}$.
5. Assemble the Limit:
$$ A = \lim_{n \to \infty} \sum_{i=1}^n \left( \sqrt{1 + \frac{3i}{n}} \right) \cdot \left( \frac{3}{n} \right) $$
Write an expression for the exact area under $f(x) = \cos(x)$ from $x=0$ to $x=\pi/2$ using the limit of a left-hand sum.
We need $A = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x$.
1. Interval: $[a, b] = [0, \pi/2]$. So, $a=0$ and $b=\pi/2$.
2. Width: $\Delta x = \frac{b-a}{n} = \frac{\pi/2 - 0}{n} = \frac{\pi}{2n}$.
3. Left Endpoints ($x_i$): $x_i = a + (i-1)\Delta x = 0 + (i-1) \cdot \left(\frac{\pi}{2n}\right) = \frac{(i-1)\pi}{2n}$.
4. Function Value ($f(x_i)$): $f(x) = \cos(x)$, so $f(x_i) = \cos\left(\frac{(i-1)\pi}{2n}\right)$.
5. Assemble the Limit:
$$ A = \lim_{n \to \infty} \sum_{i=1}^n \cos\left(\frac{(i-1)\pi}{2n}\right) \cdot \left( \frac{\pi}{2n} \right) $$
Consider the function $f(x) = \frac{1}{x}$ on the interval $[1, 3]$. For any $n$, will the right-hand sum $R_n$ be an overestimate or an underestimate of the true area?
In this section, we built the entire foundation for integral calculus. We started with a simple idea (Distance = Velocity $\times$ Time) and saw that it was an "Area Problem." We then solved the Area Problem by approximating the area with rectangles and refining that approximation with a limit.
After this lecture, you should be able to: