 Indefinite Integration 1.1. Simple Indefinite Integrals

Indefinite integration, also known as antidifferentiation, is the reversing of the process of differentiation. Given a function f, one finds a function F such that F' = f.

Finding an antiderivative is an important process in calculus. It is used as a method to obtain the area under a curve and to obtain many physical and electrical equations that scientists and engineers use everyday. For example, the equation for the current through a capacitor is , where I is current in Amperes, C is capacitance in Farads, V is voltage in Volts and t is time in seconds. To obtain an unknown (like V), one would have to use integration to obtain a voltage at a certain time interval.

While a true integral exists between a given boundary, taking the indefinite integral is simply reversing differentiation in much the same way division reverses multiplication. Instead of having a set of boundary values, one only finds an equation that would produce the integral due to differentiation without having to use the values to get a definite answer.

Suppose we have the equation f(x) = 3x2. We wish to find an equation F(x) so that F'(x) = 3x2. One method that could be used is the power rule from differentiation in reverse to obtain F(x) = x3. However, this is not the only answer. Remember, when one differentiates a constant, the result is zero (0). Therefore, the function could be any of the following:

F(x) = x3 - 16

F(x) = x3 + 1234567

F(x) = x3 + p

As seen, if one differentiates each one of the equations, the result becomes the same: F(x) = 3x2. Clearly, there are an infinite amount of results that one could obtain, and they all differ by the constant--the constant of integration (C). If F is the antiderivative of f, then (F + C) is the antiderivative of f. This is summarized into the following equation:

[F(x) + C]' = F'(x) + 0 = f(x).

THEOREM 1: Antiderivatives differ by a constant

If F is an antiderivative of the continuous function f, then any other antiderivative must have the form

G(x) = F(x) + C.

This says that two derivatives of the same function differ by the value of the constant of integration, which could be zero. Proof: If F and G are both antiderivatives of f, then F' = f and G' and f and the Constant Difference Theorem states that G(x) - F(x) = C, so G(x) = F(x) + C.

Before attempting some examples, it is necessary to define some basic integration rules that will allow one to take the antiderivative of a differential function. The following chart lists the basic rules. More complex methods will follow.

THEOREM 2: Basic Integration Rules

 Procedural Rules Differentiation Formulas Integration Formulas Constant Multiple ò cf(u) du = cò f(u) du Sum Rule ò [f(u) + g(u)]du = ò f(u)du + ò g(u)du Difference Rule ò [f(u) - g(u)]du = ò f(u)du - ò g(u)du Constant Rule ò 0du = c Power Rule n ¹ -1 Trigonometric Rules ò sin(u) du = -cos(u) + C ò cos(u) du = sin(u) + C ò sec2 u du = tan u + C eu Rule n is a constant ò neu du = enu + C ; n is a cosntant ln(u) Rule

Proof: Each of these formulas can be derived by reversing the corresponding derivative formula. For example, to obtain the Power Rule, then

so that is an antiderivative of un and

for n ¹ -1.

The next few examples demonstrate how to apply the rules given in Theorem 2.

EXAMPLE 1: Find ò x dx.

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EXAMPLE 2: Find ò cos(x) dx

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EXAMPLE 3: Find ò (4x2 + 1)dx

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EXAMPLE 4: Find ò 3e3x dx.

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1.2. Indefinite Integrals Using the Substitution Method

Often, integrals are too complex to simply use a rule. One method for solving complex integrals is the method of substitution, where one substitutes a variable for part of the integral, integrates the function with the new variable and then plugs the original value in place of the variable. This is the integration version of the chain rule. Recall that according to the chain rule, the derivative of (x2 + 3x + 5)3 is

Thus, 3(x2 + 3x + 5)2(2x + 3)dx = (x2 + 3x + 5)3 + C

Note that the product is of the form , where g(u) = 3u2 and u = x2 + 3x + 5. The first example will go over this problem in detail.

Theorem 3: Integration by substitution

Let f, g and u be differentiable functions of x so that

Then

where G is the antiderivative of g.

Proof: If G is an antiderivative of g, then G'(u) = g(u) and, through use of the chain rule

Integrating both sides of the equation yields

Example 1: Find ò 3(x2 + 3x + 5)2(2x + 3)dx.

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Example 2: Find

Example 3: Find ò (sin3(x) cos2(x))dx.

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1.3. Integration by Parts

With some complex integrals, like xex dx both methods of inspection and the method of simple substitution can not be used. Simple substitution involves substituting for only one value and its derivative. When one uses the product rule of differentiation, the result is a function that is too complex to solve by the aforementioned means. A new method must be derived to handle these types of problems.

The product rule of differentiation is as follows:

If we substitute u for f(x) and v for g(x), we have

We can reverse the product rule to be a statement about antiderivatives

uv = ò v du + ò u dv.

This equation can then be manipulated to produce the formula for integration by parts

ò u dv = uv - ò v du.

There are several steps one must go through in order to properly use the formula:

Step 1: Let u = f(x) and dv = g(x) dx, where f(x) g(x) dx is the original integrand. Good choices to make are integrals dv = g(x) dx, which are easy to integrate.

Step 2: Compute du = f'(x) dx and v = g(x) dx.

Step 3: Substitute u, v, du and dv into the formula

ò u dv = uv - ò v du.

Step 4: Calculate uv - ò v du. If v du becomes too difficult to integrate, try choosing different values for u and dv.

Step 5: Check your solution by differentiating it.

Example 1: Find ò xex dx.

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Example 2: Find ò ln(x) dx.

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Example 3: Find ò arctan(x) dx.

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