Chapter 8 Integral Calculus

Section 8.1 Antiderivatives

8.1.3 Exercises

Exercise 8.1.11

Specify an antiderivative:

$\int (12 x^{2} - 4 x^{7}) d x$ $=$
.
We have

$\int (12 x^{2} - 4 x^{7}) d x = 12 \cdot \frac{1}{3} \cdot x^{3} - 4 \cdot \frac{1}{8} \cdot x^{8} + C = 4 \cdot x^{3} - \frac{1}{2} \cdot x^{8} + C .$
$\int (\sin (x) + \cos (x)) d x$ $=$
.
We have

$\int (\sin (x) + \cos (x)) d x = - \cos (x) + \sin (x) + C = \sin (x) - \cos (x) + C .$
$\int \frac{1}{6 \sqrt{x}} d x$ $=$
.
We have $\frac{1}{6 \sqrt{x}} = \frac{1}{6} \cdot \frac{1}{\sqrt{x}} = \frac{1}{6} \cdot x^{- \frac{1}{2}}$ . Hence,

$\int \frac{1}{6 \sqrt{x}} d x = \frac{1}{6} \cdot \int x^{- \frac{1}{2}} d x = \frac{1}{6} \cdot \frac{1}{- \frac{1}{2} + 1} \cdot x^{- \frac{1}{2} + 1} + C = \frac{1}{6} \cdot 2 \cdot x^{\frac{1}{2}} + C = \frac{1}{3} \cdot \sqrt{x} + C .$

Exercise 8.1.12

Find an antiderivative:

$\int e^{x + 2} d x$ $=$
.
We have $\int e^{x + 2} d x = \int e^{x} \cdot e^{2} d x = e^{2} \cdot \int e^{x} d x = e^{2} \cdot e^{x} + C = e^{x} \cdot e^{2} + C = e^{x + 2} + C$ .
$\int 3 x \cdot \sqrt[4]{x} d x$ $=$
.
We have $\int 3 x \cdot \sqrt[4]{x} d x = 3 \cdot \int x \cdot x^{\frac{1}{4}} d x = 3 \cdot \int x^{\frac{5}{4}} d x = 3 \cdot \frac{4}{9} \cdot x^{\frac{9}{4}} + C = \frac{4}{3} \cdot x^{\frac{9}{4}} + C$ .

iExpInputHint

Exercise 8.1.13

Decide whether the following statements about real-valued functions are true.

True?	Statement:
	$F$ with $F (x) = - \frac{\cos (π x) + 2}{π}$ is an antiderivative of $f$ with $f (x) = \sin (π x) + 2$ .
	$F$ with $F (x) = - \frac{\cos (π x) + 2}{π}$ is an antiderivative of $f$ with $f (x) = \sin (π x)$ .
	$F$ with $F (x) = - 7$ is an antiderivative of $f$ with $f (x) = - 7 x$ for $x \in ℝ$ .
	$F$ with $F (x) = (\sin (x {))}^{2}$ is an antiderivative of $f$ with $f (x) = 2 \sin (x) \cos (x)$ .
	If $F$ is an antiderivative of $f$ , and $G$ is an antiderivative of $g$ , then $F + G$ is an antiderivative of $f + g$ .

The derivative of $F$ with $F (x) = - \frac{\cos (π x) + 2}{π}$ is $F' (x) = \sin (π x) \neq \sin (π x) + 2 = f (x)$ . Therefore $F$ is not an antiderivative of $f$ .
The derivative of $F$ with $F (x) = - \frac{\cos (π x) + 2}{π}$ is $F' (x) = \sin (π x) = f (x)$ . $F$ is an antiderivative of $f$ .
The derivative of $F$ with $F (x) = - 7$ is $F' (x) = 0 \neq - 7 x = f (x)$ (for $x \neq 0$ ). Hence, $F$ is not an antiderivative of $f$ .
The derivative of $F$ with $F (x) = (\sin (x {))}^{2}$ is, using the chain rule, $F' (x) = 2 \cdot \sin (x) \cdot \cos (x) = f (x)$ . Thus, $F$ is an antiderivative of $f$ .
If $F$ is an antiderivative of $f$ , and $G$ is an antiderivative of $g$ , then $F$ and $G$ are differentiable, where $F' = f$ and $G' = g$ . Thus, $F + G$ is differentiable, and we have $(F + G)' (x) = F' (x) + G' (x) = f (x) + g (x) = (f + g) (x)$ , i.e. $F + G$ is an antiderivative of $f + g$ .

Exercise 8.1.14

Find an antiderivative of

$f (x) : = \frac{8 x^{3} - 6 x^{2}}{x^{4}}$
$g (x) : = \frac{18 x^{2}}{3 \sqrt{x^{5}}}$
$h (x) : = \frac{x + 2 \sqrt{x}}{4 x}$

for

x > 0

, after rewriting the terms as reduced sums of fractions:

With the simplification $f (x)$ $=$

we have the antiderivative $F (x)$ $=$
for $x > 0$ .

The function can be rewritten as $f (x) = \frac{8}{x} - \frac{6}{x^{2}} = \frac{8}{x} - 6 x^{- 2}$ which results in the antiderivative

$F (x) = 8 \cdot \ln (x) + \frac{6}{x} + C$
for $x > 0$ and $C \in ℝ$ .
With the simplification $g (x)$ $=$

we have the antiderivative $G (x)$ $=$
for $x > 0$ .

The function can be rewritten as $g (x) = \frac{6}{\sqrt{x}} = 6 \cdot x^{- \frac{1}{2}}$ which results in the antiderivative

$G (x) = 6 \cdot 2 \cdot \sqrt{x} + C = 12 \cdot \sqrt{x} + C$
for $x > 0$ and $C \in ℝ$ .
With the simplification $h (x)$ $=$

we have the antiderivative $H (x)$ $=$
for $x > 0$ .

The function can be rewritten as $h (x) = \frac{1}{4} + \frac{1}{2 \cdot \sqrt{x}}$ which results in the antiderivative

$H (x) = \frac{1}{4} \cdot x + \sqrt{x} + C$
for $x > 0$ and $C \in ℝ$ .

Enter, for example,

\sqrt{x}

as sqrt(x).

Exercise 8.1.15

Consider a function

f

with

f (x) = \frac{1}{x}

for

x > 0

. Moreover, the functions

F_{1}

and

F_{2}

with

F_{1} (x) = \ln (7 x)

F_{2} (x) = \ln (x + 7)

for

x > 0

are given. Calculate the derivatives of

F_{1}

and

F_{2}

, and state whether

F_{1}

and

F_{2}

are antiderivatives of

f

:
We have

F_{1}' (x)

=

and

F_{2}' (x)

=

.
Check the correct answer(s).

F_{1}

is an antiderivative of

f

F_{2}

is an antiderivative of

f

The derivative of

F_{1}

with

F_{1} (x) = \ln (7 x)

for

x > 0

F_{1}' (x) = \frac{1}{7 x} \cdot 7 = \frac{1}{x} = f (x)

. Thus,

F_{1}

is an antiderivative of

f

.
The function

F_{2}

with

F_{2} (x) = \ln (x + 7)

for

x > 0

has the derivative

F_{2}' (x) = \frac{1}{x + 7} \cdot 1 = \frac{1}{x + 7} \neq \frac{1}{x}

for all

x > 0

. Hence,

F_{2}

is not an antiderivative of

f

Exercise 8.1.16

Assume that

F

is an antiderivative of

f

with

f (x) = 1 + x^{2}

, and

F

has the function value

F (0) = 1

F (3)

equals
.

By assumption,

F

is an antiderivative of

f

with

f (x) = 1 + x^{2}

. Thus, we have

F' (x) = f (x) = 1 + x^{2} = x^{0} + x^{2}

, from which

F (x) = \frac{1}{0 + 1} \cdot x^{0 + 1} + \frac{1}{2 + 1} \cdot x^{2 + 1} + C = x + \frac{1}{3} \cdot x^{3} + C

follows, for a real number

C

. Moreover, we have

F (0) = 1

which tells us

1 = F (0) = 0 + \frac{1}{3} \cdot 0^{3} + C = C

. This results in

F (x) = \frac{1}{3} \cdot x^{3} + x + 1

. Substituting

x = 3

gives the required value

F (3) = \frac{1}{3} \cdot 3^{3} + 3 + 1 = 13

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

8.1.3 Exercises

Exercise 8.1.11

Exercise 8.1.12

Exercise 8.1.13

Exercise 8.1.14

Exercise 8.1.15

Exercise 8.1.16