Chapter 8 Integral Calculus

Section 8.2 Definite Integral

8.2.3 Calculation Rules

Partition of the Interval of Integration 8.2.5

Let

f : [a; b] \to ℝ

be an integrable function. Then for every number

z

between

a

and

b

, we have

\int_{a}^{b} f (x) d x = \int_{a}^{z} f (x) d x + \int_{z}^{b} f (x) d x .

With the definition

\int_{d}^{c} f (x) d x : = - \int_{c}^{d} f (x) d x

the rule above applies to all real numbers

z

for which the two integrals on the right-hand side of the equation exist, even if

z

does not lie between

a

and

b

. Before we demonstrate this calculation with an example, we will examine the definition above in more detail.

Exchanging the Limits of Integration 8.2.6

Let

f : [a; b] \to ℝ

be an integrable function. The integral of the function

f

from

a

b

is calculated according to the rule

\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x .

The calculation rule described above is convenient when integrating functions that involve absolute values, or piecewise-defined functions.

Example 8.2.7

The integral of the function

f : [- 4; 6] \to ℝ, x \mapsto | x |

\begin{matrix} \int_{- 4}^{6} | x | d x & = & \int_{- 4}^{0} (- x) d x + \int_{0}^{6} x d x \\ = & {[- \frac{1}{2} x^{2}]}_{- 4}^{0} + {[\frac{1}{2} x^{2}]}_{0}^{6} \\ = & (0 - (- 8)) + (18 - 0) \\ = & 26 . \end{matrix}

The integration over a sum of two functions can also be split up into two integrals:

Sum and Constant Multiple Rule 8.2.8

Let

f

and

g

be integrable functions on

[a; b]

, and let

r

be a real number. Then

\int_{a}^{b} (f (x) + g (x)) d x = \int_{a}^{b} f (x) d x + \int_{a}^{b} g (x) d x .

(8.2.2)

For constant multiples of a function, we have

\int_{a}^{b} r \cdot f (x) d x = r \cdot \int_{a}^{b} f (x) d x .

(8.2.3)

This material is above course level and is not required for the exercises and tests.

There is also a calculation rule for the integration of a product of two functions, which results from the product rule for the derivative.

Integration by Parts 8.2.9

Let

u

and

v

be differentiable functions on

[a; b]

with the continuous derivatives

u'

and

v'

. For the integral of the function

u \cdot v'

, we have

\int_{a}^{b} u (x) \cdot v' (x) d x = {[u (x) \cdot v (x)]}_{a}^{b} - \int_{a}^{b} u' (x) \cdot v (x) d x,

where

u'

is the derivative of

u

and

v

is an antiderivative of

v'

. This calculation rule is called integration by parts or partial integration.

This rule is also illustrated by an example.

Example 8.2.10

Calculate the integral

\int_{0}^{π} x \sin (x) d x

by means of integration by parts. For this purpose, we choose the functions

u

and

v'

such that

\begin{matrix} u (x) = x and v' (x) = \sin (x) . \end{matrix}

Thus, we have

\begin{matrix} u' (x) = 1 and v (x) = - \cos x . \end{matrix}

The required integral can now be calculated using integration by parts:

\begin{matrix} \int_{0}^{π} x \sin x d x & = & {[x \cdot (- \cos (x))]}_{0}^{π} - \int_{0}^{π} 1 \cdot (- \cos x) d x \\ = & π \cdot (- \cos (π)) - 0 + \int_{0}^{π} \cos (x) d x \\ = & π \cdot (- (- 1)) + {[\sin (x)]}_{0}^{π} = π . \end{matrix}

The assignments of the functions

u

and

v'

have to be appropriate. This becomes obvious if in this example the assignments of

u

and

v'

are exchanged. Readers are invited to calculate this integral with exchanged assignments of

u

and

v'

In the following two exercises we practice using the rule of integration by parts.

Exercise 8.2.11

Calculate the integral

I = \int_{1}^{4} x \cdot e^{x} d x

I =

The integrand

f

with

f (x) = x \cdot e^{x}

is a product of a polynomial

u

with

u (x) = x

and an exponential function. The derivative of

u

u' (x) = 1

, i.e. a constant function. Moreover, the antiderivative of

v'

with

v' (x) = e^{x}

v

with

v (x) = e^{x}

. Thus, integration by parts results in

\begin{matrix} \int_{1}^{4} x \cdot e^{x} d x & = & {[x \cdot e^{x}]}_{1}^{4} - \int_{1}^{4} 1 \cdot e^{x} d x \\ = & {[x \cdot e^{x}]}_{1}^{4} - {[e^{x}]}_{1}^{4} \\ = & {[x \cdot e^{x} - e^{x}]}_{1}^{4} \\ = & {[(x - 1) \cdot e^{x}]}_{1}^{4} \\ = & 3 e^{4} . \end{matrix}

Exercise 8.2.12

Calculate the integral

I = \int_{1}^{8} x \cdot \ln (x) d x

I =

The integrand

f

with

f (x) = x \cdot \ln (x) = \ln (x) \cdot x

for

1 \leq x \leq 8

is a product of a polynomial and a logarithmic function. The derivative of the logarithmic function

u

with

u (x) = \ln (x)

u' (x) = \frac{1}{x}

. Thus, the function

u'

is a "simple" rational function. Moreover, the antiderivative of the polynomial

v'

with

v' (x) = x

is known, namely

v

with

v (x) = \frac{1}{2} \cdot x^{2}

.
The product

u' \cdot v

with

u' (x) \cdot v (x) = \frac{1}{x} \cdot \frac{1}{2} x^{2} = \frac{1}{2} x

for

1 \leq x \leq 8

is a continuous function for which an antiderivative is known. Thus, the required integral can be calculated by means of integration by parts:

\begin{matrix} \int_{1}^{8} x \cdot \ln (x) d x = \int_{1}^{8} \ln (x) \cdot x d x & = & {[\ln (x) \cdot \frac{1}{2} x^{2}]}_{1}^{8} - \int_{1}^{8} \frac{1}{x} \cdot \frac{1}{2} x^{2} d x \\ = & {[\ln (x) \cdot \frac{1}{2} x^{2}]}_{1}^{8} - \int_{1}^{8} \frac{1}{2} x d x \\ = & {[\frac{1}{2} x^{2} \cdot \ln (x) - \frac{1}{4} x^{2}]}_{1}^{8} \\ = & (32 \ln (8) - 16) - (\frac{1}{2} \cdot \ln (1) - \frac{1}{4}) \\ = & 96 \ln (2) - 16 + \frac{1}{4}, \end{matrix}

where

\ln (8) = \ln (2^{3}) = 3 \cdot \ln (2)

and

\ln (1) = 0

was used.

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics

Chapter 8 Integral Calculus

8.2.3 Calculation Rules

Partition of the Interval of Integration 8.2.5

Exchanging the Limits of Integration 8.2.6

Example 8.2.7

Sum and Constant Multiple Rule 8.2.8

Integration by Parts 8.2.9

Example 8.2.10

Exercise 8.2.11

Exercise 8.2.12