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Guided Notes 5 Integrals (Integralrechnung)
by Ugorotti
Learning goals (Lernziele). After this booklet you should be able to:
- explain what an integral (Integral) means as an area-content object (Fl¨acheninhalt),
- use the fundamental theorems (Haupts¨atze) to compute definite integrals quickly,
- work confidently with an antiderivative (Stammfunktion),
- apply the two main techniques: substitution (Substitution/Transformationssatz) and in- tegration by parts (partielle Integration),
- know what Riemann integrable (Riemann-integrierbar) means at idea level (Ober- /Untersummen).
1. The integral as an area-content problem (Fl¨achenproblem)
Bounded function (beschr¨ankte Funktion). A function f : I → R is bounded on an interval I if there exists M > 0 such that
|f (t)| ≤ M for all t ∈ I.
We write f ∈ B(I) for the set of bounded functions on I.
Idea. For a bounded function f on an interval [α, β] ⊂ I, we want a number that represents the (signed) area-content (Fl¨acheninhalt) under the graph. In the script this is formalized via an area-content function μα,β (f ) and then written as an integral (Integral).
Integral notation (Integral-Schreibweise). For f ∈ B(I) and α, β ∈ I one writes Z (^) β
α
f :=
Z (^) β
α
f (t) dt := μα,β (f ).
The variable is a dummy variable: Z (^) β
α
f (t) dt =
Z (^) β
α
f (x) dx.
Also, reversing limits changes the sign: Z (^) β
α
f = −
Z (^) α
β
f.
Additivity over intervals (Zerlegungseigenschaft). For all f ∈ B(I) and α, β, γ ∈ I: Z (^) β
α
f +
Z (^) γ
β
f =
Z (^) γ
α
f.
Mini-check (write in your notes). Explain in one sentence why Z (^5)
2
f =
Z 3
2
f +
Z 5
3
f
is exactly the same type of rule as splitting a sum (Summe) into two parts.
- Antiderivatives (Stammfunktionen)
Antiderivative (Stammfunktion). Let f : I → R. A function F is an antiderivative (Stammfunktion, SF) of f if
- F is continuous (stetig) on I,
- F is differentiable (differenzierbar) in the interior of I,
- and F ′(t) = f (t).
Uniqueness up to constants. If F and Fe are both Stammfunktionen of f , then there exists c ∈ R such that Fe = F + c.
Example. For r ̸= −1:
d dt
r + 1 tr+
= tr, so
Z
tr^ dt =
r + 1 tr+1^ + C.
Also (^) dtd ln(t) = (^1) t (for t > 0), so
R 1
t dt^ = ln^ |t|^ +^ C.
- The Fundamental Theorems (Haupts¨atze)
First Fundamental Theorem (Erster Hauptsatz der Differential- und Integralrech- nung). Let f be continuous (stetig) on I and fix a ∈ I. Define
F (t) :=
Z (^) t
a
f.
Then F is a Stammfunktion of f , i.e.
F ′(t) = f (t).
Moreover, F (a) = 0.
Second Fundamental Theorem (Zweiter Hauptsatz der Differential- und Integral- rechnung). Let f be continuous on I and let F be any Stammfunktion of f. Then for all α, β ∈ I: Z (^) β
α
f (t) dt = F (β) − F (α).
- Integration techniques (Integrationstechniken)
5.1 Integration by parts (partielle Integration)
Integration by parts (Satz: Partielle Integration). If f and g are continuously differentiable (stetig differenzierbar) on an interval, then Z (^) b
a
f ′(t) g(t) dt = −
Z (^) b
a
f (t) g′(t) dt +
f (t)g(t)
(^) b a
Equivalently: (^) Z b a
u v′^ = uv
b a
Z (^) b
a
u′^ v.
Practical recipe (Merkregel mit Pfeilen). Choose
u (differentiate / ableiten) and dv (integrate / aufleiten).
Then compute du and v, and apply Z u dv = uv
b a
Z
v du.
Worked example 2. Compute
Z (^) π
0
t cos(t) dt.
Choice. Let u = t (ableiten) and dv = cos(t) dt (aufleiten). Then du = dt and v = sin(t). Apply partielle Integration. Z (^) π
0
t cos(t) dt = t sin(t)
π 0
Z (^) π
0
1 · sin(t) dt.
Compute the boundary term:
t sin(t)
π 0 = π sin(π) − 0 · sin(0) = 0.
Compute the remaining integral: Z (^) π
0
sin(t) dt = − cos(t)
π 0 = (− cos π) − (− cos 0) = (1) − (−1) = 2.
Therefore (^) Z (^) π
0
t cos(t) dt = 0 − 2 = − 2.
5.2 Substitution (Substitution / Transformationssatz)
Substitution rule (Substitutionsregel / Transformationssatz). Let φ : [a, b] → J be continuously differentiable (stetig differenzierbar) and let f : J → R be continuous. Then (^) Z φ(b)
φ(a)
f (s) ds =
Z (^) b
a
f (φ(t)) φ′(t) dt.
Practical recipe. To compute
R
f (φ(t))φ′(t) dt:
- Set s = φ(t) (new variable / neue Variable).
- Then ds = φ′(t) dt.
- Change limits: t = a 7 → s = φ(a) and t = b 7 → s = φ(b).
Worked example 3. Compute
Z 1
0
2 t (1 + t^2 )^5 dt.
Substitution. Let s = 1 + t^2. Then ds = 2t dt. Change limits. If t = 0, then s = 1. If t = 1, then s = 2. Compute. (^) Z 1
0
2 t(1 + t^2 )^5 dt =
Z 2
1
s^5 ds =
s^6
2 1
(2^6 − 16 ) =
5.3 Choosing the method quickly (Methodenwahl)
Decision guide (schnelle Methodenwahl).
- If you see f (φ(x))φ′(x): try substitution (Substitution).
- If you see a product like x · ex, x · sin x, ln x · xn: try integration by parts (partielle Integration).
- If it is a rational function P Q^ ((xx)) : try algebra first (Polynomdivision), then possibly partial fractions (Partialbruchzerlegung) if covered/allowed in your exam set.
- Riemann integrability (Riemann-Integrierbarkeit) — idea level
Why this matters. Up to now, we used continuity (Stetigkeit) a lot. The script later formalizes integrals for broader classes of functions using the Riemann approach.
Partition (Zerlegung) and upper/lower sums (Ober-/Untersummen). For a partition Z = {x 0 < · · · < xn} of [α, β] and a bounded function f :
U (f, Z) =
X^ n
k=
sup [xk− 1 ,xk ]
f
(xk − xk− 1 ), L(f, Z) =
X^ n
k=
inf [xk− 1 ,xk ]
f
(xk − xk− 1 ).
(Names: Obersumme U , Untersumme L.)
Riemann integrable (Riemann-integrierbar) — criterion idea. A bounded function is Riemann-integrierbar if upper and lower sums can be made arbitrarily close:
∀ε > 0 ∃Z : U (f, Z) − L(f, Z) < ε.
Then the common value is the integral.
Important take-away. Every continuous function on a compact interval (kompaktes Inter- vall) is Riemann-integrierbar. So in most exam tasks, showing continuity is enough to justify integrability.
Practice H (Partielle Integration twice). Compute: Z (^) π
0
t^2 sin(t) dt.
Requirement: Apply integration by parts (partielle Integration) twice and simplify completely.
Practice I (Substitution + exponential). Compute:
Z (^1)
0
e^2 x^ dx,
Z 1
0
x ex 2 dx.
Requirement: For the second integral, you must explicitly use substitution (Substitution) u = x^2.
Practice J (Mixed: algebra first, then integrate). Compute:
Z (^2)
1
x^2 + 1 x
dx.
Requirement: First simplify the integrand by algebraic division (Polynomdivision / Umformen), then integrate termwise.
Practice K (Rational function: partial fraction template / Partialbruchzerlegung). Compute: (^) Z 1 x^2 − 1 dx.
Requirement: Decompose into partial fractions (Partialbruchzerlegung) and state the domain restrictions.
Practice L (Improper integral / uneigentliches Integral). Investigate Konvergenz and compute if it converges: (^) Z (^) ∞
1
xp^
dx (parameter p ∈ R).
Requirement: Give the condition on p and justify using a limit (Grenzwert) of antiderivatives.
Practice M (Improper at 0 : comparison / Vergleich). Investigate Konvergenz:
Z (^1)
0
x
dx,
Z 1
0
ln(x) √ x
dx.
Requirement: For the second integral, you must use substitution (Substitution) x = u^2 and discuss the limit at u → 0.
Practice N (Mean value theorem for integrals / Mittelwertsatz der Integration). Let f : [0, 2] → R be continuous and assume 0 ≤ f (x) ≤ 3 for all x ∈ [0, 2]. Show that
Z 2
0
f (x) dx ≤ 6 ,
and state precisely what the Mittelwertsatz der Integration guarantees about the existence of some ξ ∈ [0, 2] with
R 2
0 f^ = 2f^ (ξ).
Practice O (Fundamental theorem + parameter). Define
F (t) =
Z (^) t
1
1 + x^2
dx.
- Compute F ′(t) using the first fundamental theorem (1. Hauptsatz).
- Show that F is increasing (monoton steigend).
- Evaluate F (1) and compute F (t) explicitly (hint: arctan).
Practice P (Estimate an integral using bounds / Absch¨atzung). Show that for x ∈ [0, 1] we have 1 − x ≤ (^) 1+^1 x ≤ 1 and use this to bound
Z (^1)
0
1 + x
dx
between two explicit rational numbers. Requirement: State which inequality (Ungleichung) you use and integrate the bounds (absch¨atzen und integrieren).
End of Guided Notes: Integrals (Integralrechnung)
