Definite Integrals

The area under a curve can be approximated by adding up the areas of rectangles.


Example. Approximate the area under $y = \dfrac{1}{1 + x^3}$ from $x = 0$ to $x = 1$ using 20 equal subintervals and evaluating the function at the left-hand endpoints.

$$\sum_{n=0}^{19} f\left(n\cdot \dfrac{1}{20}\right) \cdot \dfrac{1}{20} = \sum_{n=0}^{19} \dfrac{1}{1 + \left(\dfrac{n}{20}\right)^3} \cdot \dfrac{1}{20} = 400 \sum_{n=0}^{19} \dfrac{1}{8000 + n^3}.$$

You can use a calculator to approximate this sum; it's around 0.84799.


Example. Note that the subintervals don't have to be the same size, and I don't have to choose the evaluation points systematically. These are both conveniences for the problems. For example, here is an approximation to the area under $y = x^2$ from $x = 0$ to $x = 6$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & interval & & x & & $f(x)$ & & $f(x)\cdot\Delta x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $[0,2]$ & & 1.0 & & 1.0 & & 2.0 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $[2,3]$ & & 2.8 & & 7.84 & & 7.84 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $[3,3.5]$ & & 3.0 & & 9.0 & & 4.5 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & $[3.5,6]$ & & 5.0 & & 25.0 & & 37.5 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & \cr & sum & & & & & & 51.84 & \cr height2pt & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

This gives an approximate area of 51.84. The actual area is 77.


In general, suppose I'm trying to find the area under $y = f(x)$ from $x = a$ to $x = b$ . I break the interval $[a,b]$ up into n subintervals of lengths $\Delta x_1$ , $\Delta x_2$ , ..., $\Delta x_n$ . In the k-th subinterval, I pick some point $x_k$ , and use $f(x_k)$ as the height of the k-th rectangle.

$$\hbox{\epsfysize=1.75in \epsffile{defint1.eps}}$$

The sum of the areas of the rectangles approximates the area under the curve:

$$\hbox{Area} \approx \sum_{k=1}^n f(x_k) \Delta x_k.$$

The more rectangles I take, the better the approximation. So it's reasonable to suppose that the exact area would be given by the limit of such sums, as n goes to infinity:

$$\hbox{Area} = \lim_{n \to \infty} \sum_{k=1}^n f(x_k) \Delta x_k.$$

The expression on the right --- a limit of a sum, or a Riemann sum --- is called the definite integral of $f(x)$ from a to b and is denoted as follows:

$$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{k=1}^n f(x_k) \Delta x_k.$$

It is possible to compute areas using the formula above, though it's not easy.


Example. Use the limit of a sum to compute the area under $y = 2x + 3$ from $x = 0$ to $x
   = 2$ .

I will need the following summation formulas:

$$\sum_{k=0}^n k = \dfrac{n(n - 1)}{2} \quad\hbox{and}\quad \sum_{k=0}^n c = nc.$$

Divide up the interval $[0,2]$ into n equal subintervals. Each has length $\Delta x = \dfrac{2}{n}$ . I'll evaluate the function at the left-hand endpoints. These are

$$0\cdot \dfrac{2}{n}, 1\cdot \dfrac{2}{n}, 2\cdot \dfrac{2}{n}, \ldots, (n - 1)\cdot \dfrac{2}{n}.$$

The k-th point is $\dfrac{2k}{n}$ , so the height of the k-th rectangle is

$$f\left(\dfrac{2k}{n}\right) = \dfrac{4k}{n} + 3.$$

I get the following expression for the area:

$$\lim_{n\to \infty} \sum_{k=0}^n \left(\dfrac{4k}{n} + 3\right)\cdot \dfrac{2}{n} = \lim_{n\to \infty} \dfrac{2}{n} \sum_{k=0}^n \left(\dfrac{4}{n}\cdot k + 3\right).$$

Apply the formulas from the beginning of the problem:

$$\lim_{n\to \infty} \dfrac{2}{n} \sum_{k=0}^n \left(\dfrac{4}{n}\cdot k + 3\right) = \lim_{n\to \infty} \dfrac{2}{n} \left(\dfrac{4}{n}\cdot \dfrac{n(n - 1)}{2} + 3n\right) = \lim_{n\to \infty} \dfrac{2}{n} \left(2(n - 1) + 3n\right) = \lim_{n\to \infty} \dfrac{10n - 4}{n} = 10.\quad\halmos$$


While this approach works, it's horrendously complicated. I'll discuss better ways to compute definite integrals shortly.

Since I'm taking a limit in computing the definite integral, it's possible for the limit (and hence, the definite integral) to be undefined. $f(x)$ is integrable on the interval $a \le x \le b$ if $\displaystyle
   \int_a^b f(x)\,dx$ is defined. The following fact says that many of the functions you'll use in calculus are integrable:

(A function on an interval $a \le x \le
   b$ is bounded if there is a number M such that $|f(x)| \le M$ for all x in the interval.)

For example,

$$f(x) = \cases{1 & if $x \le 0$ \cr x^2 & if $x > 0$ \cr}$$

is integrable on any interval.

$$\hbox{\epsfysize=1.5in \epsffile{defint2.eps}}$$

In particular, a continuous function is integrable.

On the other hand, $f(x) =
   \dfrac{1}{x^2}$ is not integrable on any interval containing 0.

$$\hbox{\epsfysize=1.5in \epsffile{defint3.eps}}$$

In some cases, you can use the fact that the definite integral represents the area under a curve to evaluate the integral geometrically.


Example. Compute $\displaystyle \int_0^4 (4 - x)\,dx$ .

$$\hbox{\epsfysize=1.5in \epsffile{defint4.eps}}$$

$$\int_0^4 (4 - x)\,dx = \dfrac{1}{2}\cdot 4\cdot 4 = 8.\quad\halmos$$


Example. Compute $\displaystyle \int_0^6 (4 - x)\,dx$ .

$$\hbox{\epsfysize=1.5in \epsffile{defint5.eps}}$$

The area consists of the piece in the last problem, together with a piece of area 2. But the second piece is below the x-axis, so it is taken as negative in the definite integral:

$$\int_0^6 (4 - x)\,dx = 8 - 2 = 6.\quad\halmos$$


Example. Compute $\displaystyle \int_{-1}^1 \sqrt{1 - x^2}\,dx$ .

$$\hbox{\epsfysize=1.25in \epsffile{defint6.eps}}$$

This is half the area of a circle of radius 1:

$$\int_{-1}^1 \sqrt{1 - x^2}\,dx = \dfrac{\pi}{2} \approx 1.57080.\quad\halmos$$


Here are some properties of definite integrals. I'll present them without proofs.

1. If k is a number, then

$$\int_a^b k\,dx = k(b - a).$$

This is another way of saying that the area of a rectangle is the base times the height.

$$\hbox{\epsfysize=1in \epsffile{defint7.eps}}$$


Example.

$$\int_5^{13} 7\,dx = 7\cdot (13 - 5) = 56.\quad\halmos$$


2. If f and g are integrable, then so is $f + g$ , and

$$\int_a^b \left(f(x) + g(x)\right)\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx.$$

This says that the integral of a sum is the sum of the integrals.

3. If f and g are integrable and $f(x)
   \ge g(x)$ for $a \le x \le b$ , then

$$\int_a^b f(x)\,dx \ge \int_a^b g(x)\,dx.$$

This says that bigger functions have bigger integrals.


Example. You can use the last rule to get estimates for integrals.

For example,

$$1 \ge \dfrac{x^2}{x^2 + 1} \quad\hbox{for all}\quad x.$$

Therefore,

$$\int_2^7 1\,dx \ge \int_2^7 \dfrac{x^2}{x^2 + 1}\,dx, \quad\quad\hbox{or}\quad\quad 5 \ge \int_2^7 \dfrac{x^2}{x^2 + 1}\,dx. \quad\halmos$$


4. If f is integrable, then

$$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx.$$

That is, switching the limits of integration multiplies the integral by -1.

5. If f is integrable on the interval $a
   \le x \le c$ and $a \le b \le c$ , then

$$\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx.$$

That is, integrating from a to b and then from b to c is the same as integrating all the way from a to c:

$$\hbox{\epsfysize=1.25in \epsffile{defint8.eps}}$$


Example.

$$\int_3^4 f(x)\,dx - \int_2^{-1} f(x)\,dx - \int_3^2 f(x)\,dx = \int_3^4 f(x)\,dx + \int_{-1}^2 f(x)\,dx + \int_2^3 f(x)\,dx = \int_{-1}^4 f(x)\,dx.\quad\halmos$$


6. ( The Mean Value Theorem for Integrals) There is a number c, $a \le c \le b$ , such that

$$\int_a^b f(x)\,dx = f(c)\cdot (b - a).$$

$f(c)$ represents the height of a rectangle on the integral $[a,b]$ which has the same area as the area under the graph of $f(x)$ .

$$\hbox{\epsfysize=1.25in \epsffile{defint9.eps}}$$


Example.

$$\int_0^2 \dfrac{dx}{x^2 + 1} = \dfrac{1}{c^2 + 1}\cdot (2 - 0) = \dfrac{2}{c^2 + 1}$$

for some c between 0 and 2.

Now

$$0 \le c \le 2 \quad\hbox{gives}\quad 0 \le c^2 \le 4.$$

Therefore,

$$1 \le c^2 + 1 \le 5, \quad \dfrac{1}{1} \ge \dfrac{1}{c^2 + 1} \ge \dfrac{1}{5}, \quad 2 \ge \dfrac{2}{c^2 + 1} \ge \dfrac{2}{5}.$$

So

$$2 \ge \int_0^2 \dfrac{dx}{x^2 + 1} \ge \dfrac{2}{5}.$$

I've gotten a rough estimate for the value of the integral.


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