Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 20

We discuss now the main rules to find a primitive function, and to compute definite integrals. They rest on rules for derivation.



Integration by parts

Theorem

Let

denote continuously differentiable functions. Then

Proof  

Due to the product rule the function is a primitive function for . Therefore,


In using integration by parts, two things are to be considered. Firstly, the function to be integrated is usually not in the form , but just as a product (if there is no product, then this rule will probably not help, however, sometimes the trivial product might help). Then for one factor, we have to find a primitive function, and we have to differentiate the other factor. If is a primitive function of , then the formula reads

Secondly, integration by parts only helps when the integral on the right, i.e. , can be integrated.


Example

We determine a primitive function for the natural logarithm , with integration by parts We write , and we integrate the constant function , and we differentiate the logarithm. Then

So a primitive function is .


Example

A primitive function for the sine function is . In order to find a primitive function for , we use integration by parts to get a recursive relation to a power with a smaller exponent. To make this more precise, we work over an interval, the primitive function shall start at and have the value there. For , with integration by parts we get

Multiplication with and rearranging yields

In particular, for , we have



Integration of inverse function

Theorem

Let denote a bijective differentiable function, and let denote a primitive function for . Then

is a primitive function for the inverse function .

Proof  

Differentiating, using Lemma 14.7 and Theorem 14.8 , yields


The graph with its inverse function, and the areas relevant for the computation of the integral of the inverse function.

For this statement, there exists also an easy geometric explanation. When is a strictly increasing continuous function (and therefore induces a bijection between and ), then the following relation between the areas holds:

or, equivalently,

For the primitive function of with starting point , we have, if denotes a primitive function for , the relation

where is a constant of integration.


Example

We compute a primitive function for , using Theorem 20.4 . A primitive function of tangent is

Hence,

is a primitive function for .



Substitution

Theorem

Suppose that denotes a real interval, and let

denote a continuous function. Let

be a continuously differentiable function. Then

holds.

Proof  

Since is continuous and is continuously differentiable, both integrals exist. Let denote a primitive function for , which exists, due to Theorem 18.17 . Because of the chain rule the composite function

has the derivative . Therefore,



Example

Typical examples, where one sees immediately that one can apply substitution are

with the primitive function

or

with the primitive function

Often, the indefinite integral is not in a form where one can apply the preceding rule directly. Then the following variant is more appropriate.


Corollary

Suppose that

is a continuous function, and let

denote a bijective continuously differentiable function. Then

holds.

Proof  

Because of Theorem 20.6 , we have



Remark

The substitution is applied in the following way: suppose that the integral

has to be computed. Then one needs an idea that the integral gets simpler by the substitution

(taking into account the derivative and that the inverse function has to be determined). Setting and , we have the situation

In certain cases, some standard substitutions help.

In order to make a substitution, three operations have to be done.

  1. Replace by .
  2. Replace by .
  3. Replace the integration bounds and by and .

To remember the second step, think of

which in the framework "differential forms“, has a meaning.


Example

The upper curve of the unit circle is the set

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For a given , , there exists exactly one fulfilling this condition, namely . Hence, the area of the upper half of the unit circle is the area beneath the graph of the function , above the interval , that is

Applying substitution with

(where is bijective, due to Corollary 16.14 ), we obtain, using Example 20.3 , the identities

In particular, we get that

is a primitive function for . Therefore,


Example

We determine a primitive function for , using the hyperbolic functions and , for which the relation holds. The substitution

yields[1]

A primitive function of the hyperbolic sine squared follows from

Therefore,

and hence

Due to the addition theorem for hyperbolic sine, we have , and therefore this primitive function can also be written as


Example

We want to find a primitive function for the function

We first determine the derivative of

This is

Therefore, we write as a product . We apply integration by parts where we integrate the first factor and differentiate the second factor. The derivative of the second factor is

Hence, we have



Footnotes
  1. The inverse function of the hyperbolic cosine is called area hyperbolic cosine, and is denoted by .


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