In Section 5.1, we discovered how to find the exact area under a curve by taking the limit of a sum of rectangles. We saw that the area $A$ under a continuous, positive function $f(x)$ from $x=a$ to $x=b$ is:
$$ A = \lim_{n \to \infty} R_n = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x $$
(We could also have used the left-hand sum $L_n$).
This single, powerful concept is the heart of integral calculus. In this section, we will formalize this limit, give it a name and a new notation, and explore its properties.
So far, we've used left endpoints ($L_n$) or right endpoints ($R_n$) to set the height of our rectangles. But we could have chosen any point inside the $i$-th subinterval $[x_{i-1}, x_i]$.
We call these general points sample points, denoted $x_i^*$ (read "x-i-star"). A sum based on these general points is called a Riemann Sum, named after the 19th-century German mathematician Bernhard Riemann.
This leads to our formal definition of the integral, which is one of the most important definitions in all of calculus.
If $f$ is a function defined for $x$ in the interval $[a, b]$, the definite integral of $f$ from $a$ to $b$ is the number:
$$ \int_a^b f(x) \,dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x $$
...provided that this limit exists and gives the same value for all possible choices of sample points $x_i^*$. If it does exist, we say that $f$ is integrable on $[a, b]$.
This new notation is full of meaning. Let's dissect it:
The entire expression $\int_a^b f(x) \,dx$ is a single number. It is not a function. It represents the value of the limit of the Riemann Sum.
If $f$ is continuous on $[a, b]$, or if $f$ has only a finite number of jump discontinuities, then $f$ is integrable on $[a, b]$.
This is great news! Nearly all the functions we work with in this course (polynomials, roots, trig functions, exponentials) are continuous, so we are guaranteed they can be integrated.
When $f(x)$ is positive and continuous, we know from Section 5.1 that the integral $\int_a^b f(x) \,dx$ is the exact area under the curve from $a$ to $b$.
But what if $f(x)$ is not always positive?
If $f(x)$ dips below the $x$-axis, then the value of $f(x_i^*)$ will be negative. In that case, the term $f(x_i^*) \Delta x$ (height $\times$ width) will be a negative number. When we sum them up, the integral will add the areas above the axis as positive values and subtract the areas below the axis as negative values.
The definite integral $\int_a^b f(x) \,dx$ represents the net area of the region bounded by $y=f(x)$, the $x$-axis, $x=a$, and $x=b$.
Net Area = (Area Above $x$-axis) - (Area Below $x$-axis)
This graph illustrates the idea. The function $f(x)$ has a region $A_1$ above the axis and a region $A_2$ below the axis.
The definite integral would be calculated as $\int_a^b f(x) \,dx = A_1 - A_2$.
This geometric interpretation gives us a new way to evaluate integrals if the shapes are simple (like rectangles, triangles, or circles).
Evaluate $\int_0^3 (x-1) \,dx$ by interpreting it as net area.
The function is $y = x-1$, which is a straight line. We are looking for the net area from $x=0$ to $x=3$.
We can see two regions:
The integral is the net area:
$$ \int_0^3 (x-1) \,dx = A_2 - A_1 = 2 - 0.5 = 1.5 $$
Evaluate $\int_{-2}^2 \sqrt{4-x^2} \,dx$.
We need to recognize the function $y = \sqrt{4-x^2}$. If we square both sides, we get $y^2 = 4-x^2$, which rearranges to $x^2 + y^2 = 4$.
This is the equation of a circle centered at $(0,0)$ with radius $r=2$. Since the original function was $y = +\sqrt{...}$, we are only looking at the top half of the circle (a semicircle).
The integral from $-2$ to $2$ is asking for the entire area of this top semicircle.
The area of a full circle is $\pi r^2$. The area of our semicircle is $\frac{1}{2}\pi r^2$.
$$ \int_{-2}^2 \sqrt{4-x^2} \,dx = \frac{1}{2} \pi (2)^2 = 2\pi $$
Evaluate $\int_0^4 (2x - 4) \,dx$ by interpreting it as net area.
Calculating the limit of a Riemann Sum is hard (we'll find an amazing shortcut in Section 5.3!). For now, we can approximate integrals numerically by calculating the sum for a fixed, finite $n$.
We've already seen $L_n$ and $R_n$. A third, and often more accurate, method is the Midpoint Rule. Instead of using the left or right edge for the height, we use the value of the function at the midpoint of the interval.
We use the midpoint of each subinterval as the sample point $x_i^*$.
The midpoint of the $i$-th subinterval $[x_{i-1}, x_i]$ is $\bar{x}_i = \frac{x_{i-1} + x_i}{2}$.
$$ M_n = \sum_{i=1}^n f(\bar{x}_i) \Delta x $$
The Midpoint Rule is often more accurate because the part of the rectangle that overshoots the curve is often cancelled out by the part that undershoots it.
Note to students: red is upper limit, blue is lower limit, green is the midpoint, click on $R_n$ , $L_n$ and $M_p$ to see the approximations.
Use the Midpoint Rule with $n=4$ to approximate $\int_1^2 \frac{1}{x} \,dx$.
1. Find $\Delta x$: $\Delta x = \frac{b-a}{n} = \frac{2-1}{4} = \frac{1}{4} = 0.25$.
2. Find Subintervals: $[1, 1.25], [1.25, 1.5], [1.5, 1.75], [1.75, 2]$.
3. Find Midpoints ($\bar{x}_i$):
4. Sum: $M_4 = \Delta x [f(\bar{x}_1) + f(\bar{x}_2) + f(\bar{x}_3) + f(\bar{x}_4)]$
$$ M_4 = 0.25 \left[ \frac{1}{1.125} + \frac{1}{1.375} + \frac{1}{1.625} + \frac{1}{1.875} \right] $$
$$ M_4 \approx 0.25 [ 0.8888... + 0.7272... + 0.6153... + 0.5333... ] $$
$$ M_4 \approx 0.25 [ 2.7648... ] \approx 0.6912 $$
(The true value is $\ln(2) \approx 0.6931$, so this is a very good approximation!)
Set up the Midpoint Rule approximation for $\int_0^5 x^2 \,dx$ using $n=5$.
1. Find $\Delta x$: $\Delta x = \frac{5-0}{5} = 1$.
2. Find Subintervals: $[0, 1], [1, 2], [2, 3], [3, 4], [4, 5]$.
3. Find Midpoints ($\bar{x}_i$):
4. Sum: $M_5 = \Delta x [f(0.5) + f(1.5) + f(2.5) + f(3.5) + f(4.5)]$
$$ M_5 = 1 \cdot [ (0.5)^2 + (1.5)^2 + (2.5)^2 + (3.5)^2 + (4.5)^2 ] $$
$$ M_5 = 0.25 + 2.25 + 6.25 + 12.25 + 20.25 = 41.25 $$
What are the midpoints ($\bar{x}_i$) you would use to calculate $M_4$ for the integral $\int_0^\pi \sin(x) \,dx$?
Just like limits, derivatives, and sums, definite integrals have properties that make them much easier to work with. These properties let us split complex integrals into simpler ones.
| Property Name | Algebraic Statement | Geometric / Meaning |
|---|---|---|
| Basic Properties | ||
| Integral of a Constant | $\int_a^b c \,dx = c(b-a)$ | The area is a rectangle with height $c$ and width $(b-a)$. |
| Sum/Difference | $\int_a^b [f(x) \pm g(x)] \,dx = \int_a^b f(x) \,dx \pm \int_a^b g(x) \,dx$ | The area of a sum is the sum of the areas. |
| Constant Multiple | $\int_a^b c f(x) \,dx = c \int_a^b f(x) \,dx$ | Stretching the function vertically by $c$ stretches the area by $c$. |
| Interval Properties | ||
| Adjacent Intervals | $\int_a^c f(x) \,dx + \int_c^b f(x) \,dx = \int_a^b f(x) \,dx$ | We can split an area into two side-by-side pieces. |
| Reversed Limits | $\int_b^a f(x) \,dx = -\int_a^b f(x) \,dx$ | Integrating "backwards" negates the result. |
| Zero Width | $\int_a^a f(x) \,dx = 0$ | Area of a region with zero width is zero. |
| Comparison Properties | ||
| Positivity | If $f(x) \ge 0$ for $a \le x \le b$, then $\int_a^b f(x) \,dx \ge 0$. | If the function is above the axis, the area is positive. |
| Order Property | If $f(x) \ge g(x)$ for $a \le x \le b$, then $\int_a^b f(x) \,dx \ge \int_a^b g(x) \,dx$. | A "larger" function has a larger area under it. |
| Boundedness | If $m \le f(x) \le M$ for $a \le x \le b$, then $m(b-a) \le \int_a^b f(x) \,dx \le M(b-a)$. |
The area is trapped between a small rectangle (height $m$) and a big rectangle (height $M$). |
Let's practice using these properties.
Given that $\int_0^2 f(x) \,dx = 5$ and $\int_0^2 g(x) \,dx = -3$, find $\int_0^2 [2f(x) - 4g(x)] \,dx$.
We use the Sum and Constant Multiple properties to break the integral apart:
$$ \int_0^2 [2f(x) - 4g(x)] \,dx = \int_0^2 2f(x) \,dx - \int_0^2 4g(x) \,dx $$
$$ = 2 \int_0^2 f(x) \,dx - 4 \int_0^2 g(x) \,dx $$
Now we just substitute the given values:
$$ = 2(5) - 4(-3) = 10 + 12 = 22 $$
Show that $\int_0^1 \sqrt{1+x^2} \,dx \le \sqrt{2}$.
We can use the Boundedness or Order property. Notice that on the interval $[0, 1]$, the maximum value of $x^2$ is $1$.
Therefore, $1+x^2 \le 1+1 = 2$.
Taking the square root (which is an increasing function), we get $\sqrt{1+x^2} \le \sqrt{2}$.
So, our function is always less than the constant function $M = \sqrt{2}$. By the Comparison Property:
$$ \int_0^1 \sqrt{1+x^2} \,dx \le \int_0^1 \sqrt{2} \,dx $$
$$ \int_0^1 \sqrt{1+x^2} \,dx \le \sqrt{2}(1-0) = \sqrt{2} $$
Evaluate $\int_2^5 4 \,dx$. Which property does this use?
In this section, we gave a formal name and notation to the limit of a Riemann Sum: the Definite Integral. We learned that this integral represents the net area (area above minus area below) and that we can evaluate simple integrals using geometry. We also learned how to approximate integrals with the Midpoint Rule and, most importantly, we learned the properties that allow us to manipulate integrals algebraically.
After this lecture, you should be able to: