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Integration

Integration is the reverse process of differentiation and is used to find areas, volumes, and accumulations of quantities. It plays a fundamental role in physics, engineering, and many other fields.

1. The Indefinite Integral

The indefinite integral of a function \( f(x) \) is a function \( F(x) \) whose derivative is \( f(x) \). It is written as:

\[ \int f(x) \, dx = F(x) + C \]

where \( C \) is the constant of integration, representing an infinite family of solutions.

Basic Integration Rules:

  • Power Rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 \)
  • Constant Rule: \( \int c \, dx = cx + C \)
  • Sum Rule: \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \)
  • Constant Multiple Rule: \( \int c f(x) \, dx = c \int f(x) \, dx \)

Example 1: Computing an Indefinite Integral

Evaluate:

\[ \int (3x^2 - 4x + 5) \, dx \]

Using the power rule:

\[ \int 3x^2 \, dx - \int 4x \, dx + \int 5 \, dx \] \[ = \frac{3x^3}{3} - \frac{4x^2}{2} + 5x + C \] \[ = x^3 - 2x^2 + 5x + C \]

2. The Definite Integral

The definite integral computes the net area under a curve from \( x = a \) to \( x = b \):

\[ \int_a^b f(x) \, dx = F(b) - F(a) \]

This is known as the Fundamental Theorem of Calculus, which states that integration and differentiation are inverse processes.

Example 2: Computing a Definite Integral

Find:

\[ \int_1^3 (2x^2 - 3x + 4) \, dx \]

Step 1: Compute the antiderivative:

\[ F(x) = \frac{2x^3}{3} - \frac{3x^2}{2} + 4x \]

Step 2: Evaluate at \( x = 3 \) and \( x = 1 \):

\[ F(3) = \frac{2(3)^3}{3} - \frac{3(3)^2}{2} + 4(3) \] \[ = \frac{2(27)}{3} - \frac{3(9)}{2} + 12 \] \[ = 18 - \frac{27}{2} + 12 = \frac{36}{2} - \frac{27}{2} + \frac{24}{2} = \frac{33}{2} \] \[ F(1) = \frac{2(1)^3}{3} - \frac{3(1)^2}{2} + 4(1) \] \[ = \frac{2}{3} - \frac{3}{2} + 4 \] \[ = \frac{4}{6} - \frac{9}{6} + \frac{24}{6} = \frac{19}{6} \]

Step 3: Compute the result:

\[ \int_1^3 (2x^2 - 3x + 4) \, dx = \frac{33}{2} - \frac{19}{6} = \frac{99}{6} - \frac{19}{6} = \frac{80}{6} = \frac{40}{3} \]

3. The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(x) \) is an antiderivative of \( f(x) \), then:

\[ \frac{d}{dx} \left( \int_a^x f(t) \, dt \right) = f(x) \]

It connects differentiation and integration.

4. Integration by Substitution

Substitution is used when an integral involves a composition of functions. The general approach is to let \( u = g(x) \), so that \( du = g'(x) dx \).

Example 3: Using Substitution

Evaluate:

\[ \int (3x^2 + 2x) e^{x^3 + x^2} \, dx \]

Step 1: Let \( u = x^3 + x^2 \), then \( du = (3x^2 + 2x) dx \).

Step 2: Rewrite the integral:

\[ \int e^u \, du \]

Step 3: Integrate:

\[ e^u + C \]

Step 4: Substitute back \( u = x^3 + x^2 \):

\[ e^{x^3 + x^2} + C \]

Practice Problems

Problem 1: Compute:

\[ \int (4x^3 - 6x + 2) \, dx \]

Solution:

\[ \frac{4x^4}{4} - \frac{6x^2}{2} + 2x + C \] \[ = x^4 - 3x^2 + 2x + C \]

Problem 2: Compute:

\[ \int_0^2 (x^2 - 2x + 3) \, dx \]

Solution:

Find the antiderivative:

\[ F(x) = \frac{x^3}{3} - x^2 + 3x \]

Evaluate at \( x = 2 \) and \( x = 0 \):

\[ F(2) = \frac{2^3}{3} - (2)^2 + 3(2) = \frac{8}{3} - 4 + 6 = \frac{14}{3} \] \[ F(0) = 0 \] \[ \int_0^2 (x^2 - 2x + 3) \, dx = \frac{14}{3} - 0 = \frac{14}{3} \]

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