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Definite Integrals

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Core Definition

The limit of Riemann sums representing the net signed area bounded by a function f(x)f(x) on an interval [a,b][a, b]. If the limit exists:

∫abf(x) dx=lim⁑nβ†’βˆžβˆ‘i=1nf(xiβˆ—)Ξ”x\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x
Calculated via the **Fundamental Theorem of Calculus (FTOC)**:
∫abf(x) dx=F(b)βˆ’F(a)whereΒ Fβ€²(x)=f(x)\int_{a}^{b} f(x) \, dx = F(b) - F(a) \quad \text{where } F'(x) = f(x)

Concept Explanation

Definite integration accumulates continuous variable quantities over an interval. If a graph is positive, the definite integral matches the area under the curve. Negative regions subtract from the total sum, yielding a net signed area.

Visual / Geometric Intuition

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Engineering Applications

Engineers use integrals to sum variable rates over time. Mechanical engineers integrate variable force to compute work (W=∫F dxW = \int F \, dx), electrical engineers integrate power to compute total energy, and civil engineers integrate loads to design bridge supports.

Example Problem

Evaluate the definite integral:

∫13(3x2βˆ’1) dx\int_{1}^{3} (3x^2 - 1) \, dx

Solution

Step 1: Find the antiderivative F(x)F(x)

F(x)=∫(3x2βˆ’1) dx=x3βˆ’xF(x) = \int (3x^2 - 1) \, dx = x^3 - x

Step 2: Apply the Fundamental Theorem of Calculus

∫13(3x2βˆ’1) dx=F(3)βˆ’F(1)\int_{1}^{3} (3x^2 - 1) \, dx = F(3) - F(1)
F(3)=33βˆ’3=27βˆ’3=24F(3) = 3^3 - 3 = 27 - 3 = 24
F(1)=13βˆ’1=1βˆ’1=0F(1) = 1^3 - 1 = 1 - 1 = 0

Step 3: Subtract

∫13(3x2βˆ’1) dx=24βˆ’0=24\int_{1}^{3} (3x^2 - 1) \, dx = 24 - 0 = 24

Connections & References