Integral Calculator & Area Under Curve (Fundamental Theorem & Solids of Revolution)

Compute definite integrals, geometric areas with automatic sign-change root splitting, enclosed areas between two functions, and volume of solids of revolution with interactive 2D graph and a 4-level trainer.

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Musterbeispiele:
Funktionsgleichung f(x) = a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀
x⁴ +x³ +x² +x +
Integrationsintervall [a, b]
Untere Grenze a:
Obere Grenze b:

2D-Integrationsfläche & Kurvenverlauf

Visualizer Optionen:
xya=-3b=3
f(x) Funktion
Positive Teilfläche (+A)
Negative Teilfläche (-A)
Geometrischer Flächeninhalt A
15.3333 FE

Summe aller Beträge (3 Teilflächen)

Orientiertes Integral
-6.0001

Netto-Flächenbilanz mit Vorzeichen

Nullstellen im Intervall
-2, 2

Vorzeichenwechsel-Trennpunkte

Vollständiger mathematischer Lösungsweg & Formeln

GegebeneFunktion:f(x)=x2−4imIntervall[−3,3]Gegebene Funktion: f(x) = x^2 - 4 im Intervall [-3, 3]
1.NullstellenimIntegrationsinteralsuchen(Vorzeichenwechselpru¨fen):1. Nullstellen im Integrationsinteral suchen (Vorzeichenwechsel prüfen):
GefundeneNullstellen:xN=−2,xN=2Gefundene Nullstellen: x_N = -2, x_N = 2
DasIntervall[−3,3]wirdandenNullstellenin3Teilintervallezerlegt.Das Intervall [-3, 3] wird an den Nullstellen in 3 Teilintervalle zerlegt.
2.Teilfla¨chenberechnenundBetra¨gesummieren:A=∑i∣Ai∣=∑i∣∫xixi+1f(x) dx∣2. Teilflächen berechnen und Beträge summieren: A = \sum_{i} |A_i| = \sum_{i} \left| \int_{x_i}^{x_{i+1}} f(x)\,dx \right|
Teilintervall[−3,−2]:∫−3−2f(x) dx=F(−2)−F(−3)=5.3333−(3)=2.3333  ⟹  A1=∣2.3333∣=2.3333  FETeilintervall [-3, -2]: \int_{-3}^{-2} f(x)\,dx = F(-2) - F(-3) = 5.3333 - (3) = 2.3333 \implies A_{1} = |2.3333| = 2.3333\;\text{FE}
Teilintervall[−2,2]:∫−22f(x) dx=F(2)−F(−2)=−5.3333−(5.3333)=−10.6667  ⟹  A2=∣−10.6667∣=10.6667  FETeilintervall [-2, 2]: \int_{-2}^{2} f(x)\,dx = F(2) - F(-2) = -5.3333 - (5.3333) = -10.6667 \implies A_{2} = |-10.6667| = 10.6667\;\text{FE}
Teilintervall[2,3]:∫23f(x) dx=F(3)−F(2)=−3−(−5.3333)=2.3333  ⟹  A3=∣2.3333∣=2.3333  FETeilintervall [2, 3]: \int_{2}^{3} f(x)\,dx = F(3) - F(2) = -3 - (-5.3333) = 2.3333 \implies A_{3} = |2.3333| = 2.3333\;\text{FE}
3.GesamtergeometrischerFla¨cheninhalt:Ages=2.3333+10.6667+2.3333=15.3333  FE3. Gesamter geometrischer Flächeninhalt: A_{ges} = 2.3333 + 10.6667 + 2.3333 = 15.3333\;\text{FE}
(ZumVergleich:DasorientierteIntegralohneBetragbetra¨gt∫−33f(x) dx=−6.0001)(Zum Vergleich: Das orientierte Integral ohne Betrag beträgt \int_{-3}^{3} f(x)\,dx = -6.0001)

Historical Origin & The Challenge Solved

Why was this mathematics developed, where does it come from, and what real-world problem did it solve?

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Epoche & ZeitalterAntiquity to Renaissance (c. 250 BCE – 1675 CE)
Pioniere: Archimedes of Syracuse, Johannes Kepler, Isaac Newton, Gottfried Wilhelm Leibniz

1. Curved Areas & Barrel Volumes

In 1612, Johannes Kepler bought wine barrels for his wedding and was frustrated by merchants' inaccurate estimates for curved barrels. Archimedes had previously sought exact methods to determine areas bounded by parabolas and circles.

2. Fundamental Theorem of Calculus

Newton and Leibniz discovered the fundamental connection: integration (summing infinite thin slices) is the exact inverse of differentiation (finding slopes). Kepler pioneered slicing solids into infinitesimal discs.

3. Structural Engineering, Quantum Mechanics & Energy

From calculating battery electrical work to airplane wing lift distribution and quantum wave function probabilities, modern technology rests on integrals.

„It is useful to divide the whole into its smallest parts and assemble them again.“ — Gottfried Wilhelm Leibniz

Real-World Applications of Integral Calculus

Integration is fundamental for accumulating quantities, physics kinematics, structural volumes, and financial economics.

Physics & Kinematics

Total distance from variable velocity

The accumulated distance s(t) of any moving body is the definite integral of its velocity function v(t) over time.

Praxis-Beispiel:

Integrating acceleration a(t) yields velocity v(t); a second integral computes total displacement s = ∫ v(t) dt.

Economics & Finance

Consumer & Producer Surplus

Integrals between market demand and supply curves measure total economic welfare at market equilibrium.

Praxis-Beispiel:

Consumer surplus equals the area between demand curve p(x) and equilibrium price p* on [0, x*].

Engineering & Architecture

Volume of Revolution Tanks

The storage capacity of cylindrical, conical, or parabolic storage vessels is computed via V = π ∫ [f(x)]² dx.

Praxis-Beispiel:

A cooling tower with contour f(x) has concrete volume calculated exactly through definite integration.

Mathematical Guide to Integral Calculus

Integral calculus reverses differentiation and enables exact calculations of curved areas, averages, and physical quantities.

Fundamental Theorem of Calculus (FTC)

If f is continuous on [a, b] and F is an antiderivative with F'(x) = f(x), then ∫_a^b f(x) dx = F(b) - F(a).

Geometric Area with Root Splitting

When roots x_i lie inside [a, b], the integral splits at each root to sum absolute values: A = ∫_a^(x_1) |f(x)| dx + ∫_(x_1)^b |f(x)| dx.

Area Between Two Curves

The area enclosed by two curves between intersection bounds is given by A = ∫_(x_1)^(x_2) |f(x) - g(x)| dx.

Solid of Revolution Volume (x-Axis)

Rotating f(x) about the x-axis on [a, b] produces a solid of volume V = π · ∫_a^b [f(x)]² dx.

Frequently Asked Questions (FAQ) on Integral Calculus

What is the difference between a definite integral and geometric area?▼

A definite integral yields net signed area (areas below the x-axis cancel areas above). Geometric area takes absolute magnitudes of all sub-areas.

How do you find the antiderivative of a polynomial term?▼

Apply the power rule of integration: x^n becomes (1 / (n + 1)) x^(n + 1). For example, 3x² becomes x³ + C.

Why do we need to find roots for area calculation?▼

Functions cross the x-axis at roots, changing sign. Splitting at roots prevents positive and negative areas from canceling each other out.

How do you find the enclosed area between two parabolas?▼

1. Set f(x) = g(x) to find intersection points. 2. Define d(x) = f(x) - g(x). 3. Integrate d(x) between intersections and take the absolute value.