Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 20/refcontrol
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
TheoremCreate referencenumber
Proof
Due to the product rule, the function is a primitive functionMDLD/primitive function (R) 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 Create referencenumber
We determine a primitive functionMDLD/primitive function (R) for the natural logarithmMDLD/natural logarithm (R) , with integration by parts. We write , and we integrate the constant function , and we differentiate the logarithm. Then
So a primitive function is .
== Example Example 20.3
change==
A primitive functionMDLD/primitive function (R) for the sine functionMDLD/sine function (R) 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
TheoremTheorem 20.4 change
Let denote a bijectiveMDLD/bijective differentiable function,MDLD/differentiable function (R) and let denote a primitive functionMDLD/primitive function (R) for . Then
is a primitive function for the inverse functionMDLD/inverse function .
Proof
Differentiating,MDLD/Differentiating (R) using Lemma 14.7 and Theorem 14.8 , yields
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 Create referencenumber
We compute a primitive functionMDLD/primitive function (R) for , using Theorem 20.4 . A primitive function of tangent is
Hence,
is a primitive function for .
- Substitution
TheoremTheorem 20.6 change
Suppose that denotes a real interval,MDLD/real interval and let
denote a continuous function.MDLD/continuous function Let
be a continuously differentiableMDLD/continuously differentiable (R) function. Then
holds.
Proof
Since is continuous and is continuously differentiable, both integrals exist. Let denote a primitive functionMDLD/primitive function (R) for , which exists, due to Theorem 18.17 . Because of the chain rule, the composite function
has the derivative . Therefore,
Example Create referencenumber
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.
CorollaryCreate referencenumber
Suppose that
is a continuous function,MDLD/continuous function (R) and let
denote a bijectiveMDLD/bijective continuously differentiableMDLD/continuously differentiable (R) function.MDLD/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.
- Replace by .
- Replace by .
- Replace the integration bounds and by and .
To remember the second step, think of
which in the framework "differential forms“, has a meaning.
Example Create referencenumber
The upper curve of the unit circle is the set
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 functionMDLD/primitive function (R) for . Therefore,
Example Create referencenumber
We determine a primitive functionMDLD/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 Create referencenumber
We want to find a primitive functionMDLD/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
- ↑ The inverse function of the hyperbolic cosine is called area hyperbolic cosine, and is denoted by .
<< | Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I | >> PDF-version of this lecture Exercise sheet for this lecture (PDF) |
---|