]> Haese & Harris - Chapter 25 Notes

25: Integration

A: Antidifferentiation

Differentiation takes a function and returns the derivative. Antidifferentiation takes the derivative and returns the original function!

The antiderivative of $f (x)$ is any function $F (x)$ where $F^{'} (x) = f (x)$ .

For example: since $y = x^{2}$ has a derivative of $y = 2 x$ , the antiderivative of $y = 2 x$ is $y = x^{2}$ .

Note the word any…a single function has only one derivative, but a single derivative has lots of functions that it could have come from! For example—if I tell you that the derivative is $y = 2 x$ , then the original function could be $y = x^{2}$ or $y = x^{2} + 2$ or $y = x^{2} - 5$ …

Thus: differentiation gives a single answer, but antidifferentiation has an infinite number of answers!

B: Integration

The word antidifferentiation is really long (and gives the spell checker nightmares)—so let's use some notation (and use shorter words).

If we are given a derivative, and want to find an antiderivative, then we'll integrate the function. Integration and antidifferentiation are the same thing.

Notation: if $F^{'} (x) = f (x)$ , then $\int f (x) d x = F (x) + C$ .

The symbol ∫ is called the integral (symbol), and $f (x)$ is called the integrand. The $d x$ tells us which variable is independent—we integrate with respect to that variable. The $+ C$ on the end is called the constant of integration—remember that there are an infinite number of answers. However, the differences among those answers can only vary by a constant (since the derivative of a constant is zero). If we are given some information about a desired function, then we can solve for the constant of integration, and determine one answer to the integral.

Constants can be factored out of derivatives—thus, constants can be factored out of integrals. $\int k \cdot f (x) d x = k \int f (x) d x$ .

The derivative of a sum/difference of two functions can be written as the sum/difference of two derivatives—the same is true of integrals: $\int [f (x) \pm g (x)] d x = \int f (x) d x \pm \int g (x) d x$ .

Now for some (fairly) simple integrals:

$\int x^{n} d x = \frac{1}{n + 1} x^{n + 1} + C$

$e^{x} d x = e^{x} + C$

$\int \frac{1}{x} d x = ln | x | + C$

C: Integrating Power and Exponential Functions

Since the derivatives of $y = e^{x}$ and $y = e^{a x + b}$ differ by a constant, it turns out that their integrals differ by a constant also! The same is true for power functions and logarithmic functions. We'll investigate this more closely in class—for now, here are the results:

$\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + C$

$\int {(a x + b)}^{n} d x = \frac{1}{a} \cdot \frac{1}{n + 1} {(a x + b)}^{n + 1} + C$

$\int \frac{1}{(a x + b)} d x = \frac{1}{a} \cdot ln | (a x + b) | + C$

D: Integration by Substitution

If we take the derivative of a composition of functions, then we must employ the chain rule—resulting in an answer that involves a product. Working this backwards through integration is harder, but possible!

The derivative of $f [g (x)]$ is $f^{'} [g (x)] \cdot g^{'} (x)$ ; so the integral of $f^{'} [g (x)] \cdot g^{'} (x)$ must be $f [g (x)] + C$ .

In practice, seeing this can be hard; thus, we attack it from a different angle—a change of variables. Let part of the integrand become u, then find $d u$ —can the integral with respect to x now be changed to an integral with respect to u?

As an example, consider $\int 2 (x^{2} + 3 x) (2 x + 3) d x$ . I can see that part of that integrand is the derivative of another part. Let u be the part that is the "original" function: $u = x^{2} + 3 x$ . Now, take the derivative: $\frac{d u}{d x} = 2 x + 3$ . We will want to write this in differential form: $d u = (2 x + 3) d x$ . Now, start swapping: can you replace all instances of x with some instance of u?

$\int 2$	$(x^{2} + 3 x)$	$(2 x + 3) d x$

$\int 2$	$u$	$d u$

So $\int 2 (x^{2} + 3 x) (2 x + 3) d x = \int (2 u) d u$ . This can be done! $\int (2 u) d u = u^{2} + C$ . However, it's only polite to put things back like you found them—we should go back and exchange u for x: $u^{2} + C = {(x^{2} + 3 x)}^{2} + C$ .

The trick is in picking u—the only teacher for that is experience!

E: Definite Integrals

I'm not going to go into any depth on this just yet—suffice to say that I find it exceedingly weird to talk about definite integrals at this point.

Well, here goes: $\int_{a}^{b} f (x) d x = F (x)]_{a}^{b} = F (b) - F (a)$ . The change is that you're going to be given two limits, and they must be plugged in and subtracted to obtain a number.

It is worth noting that $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ , and that $\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ for any value of c.

As to what all this means—we'll save that for later.

Page last validated 2010-08-15