Integral Calculus Lesson - Definition, Formulas, Methods, Examples

Lesson Overview

• What Is Integral Calculus?
• Types of Integrals
• Properties of Integral Calculus
• Integral Calculus Formulas
• Methods for Solving Integrals in Calculus
• Integral Calculus Examples
• Integral Calculus Assessment

What Is Integral Calculus?

Integral calculus is a branch of mathematics that focuses on finding areas, volumes, and totals. It is the opposite of differential calculus, which focuses on rates of change. 

Integral calculus helps us solve problems where we need to calculate the total value of something over a range, like the area under a curve. It is widely used in physics, engineering, and economics.

Fig: Integral Calculus Example

Take This Quiz :

Calculus Quiz Questions and Answers

Types of Integrals

There are two main types of integrals:

1. Indefinite Integral: This gives a family of functions and does not have limits. It includes an arbitrary constant C.
2. Definite Integral: This calculates the exact area under a curve between two points, using limits.

Indefinite Integrals

An indefinite integral finds the general antiderivative of a function. It is expressed as:
∫ f(x) dx = F(x) + C
Here, F(x) is the antiderivative, and C is the constant of integration.

Example:
∫ x² dx = (x³ / 3) + C

Definite Integrals

A definite integral calculates the exact area under a curve between limits a and b. It is expressed as:
∫[a to b] f(x) dx = F(b) - F(a)

Example:
∫[0 to 2] x² dx = [(x³ / 3)] from 0 to 2 = (8/3) - (0) = 8/3

Take This Quiz :

AP Calculus Integrals Practice Test!

Properties of Integral Calculus

• Linearity:
  This property tells us that we can break up an integral into parts:

Explanation: If you have two functions being added together, you can integrate them separately, multiply by the constants, and then add the results.

• Zero Property:
  If the limits of integration are the same, the integral equals zero:

Explanation: If you're finding the area under the curve from the same point to the same point, there's no area, so the result is zero.

• Additivity:
  If you break up the interval into smaller parts, you can add up the integrals over those parts:

Explanation: You can combine integrals over different parts of the interval into one integral over the entire interval.

Take This Quiz :

Calculus Mathematics Quiz: Differentiation Problems

Integral Calculus Formulas

Here are some fundamental formulas of Integral Calculus:

Take This Quiz :

Methods for Solving Integrals in Calculus

1. Substitution Method

• The Idea: When you have a complicated function to integrate, sometimes you can make it simpler by replacing a part of it with a new variable. This is like using a nickname instead of a long, formal name!
• When to Use It: Look for situations where you have a function and its derivative within the integral.
• Example:
  • You want to integrate ∫2x * cos(x²) dx
  • Notice that 2x is the derivative of x²
  • Substitute 'u' for x²: u = x²
  • Then, du/dx = 2x, so du = 2x dx
  • Now the integral becomes ∫cos(u) du, which is much easier to solve!
    

2. Integration by Parts

• The Idea: This method is helpful when you're trying to integrate a product of two functions. It's based on the product rule for differentiation.
• The Formula: ∫ u dv = uv - ∫ v du
  • You choose one part of your function to be 'u' and the other part to be 'dv'.
  • You differentiate 'u' to find 'du' and integrate 'dv' to find 'v'.
  • Then you plug those values into the formula.
• Example: ∫x * e^{^}x dx
  • Let u = x and dv = e^x dx.
  • Differentiate u: du = dx.
  • Integrate dv: v = e^x.
  • Substitute into the formula: ∫x * e^x dx = x * e^x - ∫e^x dx = x * e^x - e^x + C.
    

3. Partial Fractions

• The Idea: If you have a complex fraction (a rational function), you can break it down into simpler fractions that are easier to integrate.
• When to Use It: When you have a polynomial divided by another polynomial.
• Example: ∫(x² + 3x + 2) / (x³ - x² - 2x) dx
  • Factor the denominator and break the fraction into smaller parts.
  • Integrate each of the simpler fractions.
    

4. Trigonometric Substitution

• The Idea: Sometimes you can simplify integrals that involve square roots or other tricky expressions by using trigonometric identities (like sin²x + cos²x = 1).
• When to Use It: Look for expressions like √(a² - x²), √(a² + x²), or √(x² - a²) within the integral.
• Example: ∫dx / √(1 - x²)
  • Substitute x = sin(θ)
  • Use trigonometric identities to simplify the expression and then integrate.
    
    

Integral Calculus Examples

Example 1: Substitution Method

Evaluate ∫ x cos(x²) dx.
Let u = x², so du = 2x dx. Rewrite the integral:
(1/2) ∫ cos(u) du = (1/2) sin(u) + C = (1/2) sin(x²) + C

Example 2: Definite Integral

Find the area under f(x) = 3x from x = 1 to x = 4.
∫[1 to 4] 3x dx = [(3x² / 2)] from 1 to 4
= [(3(4²) / 2) - (3(1²) / 2)]
= (24 - 1.5) = 22.5

Example 3: Integration by Parts

Evaluate ∫ x e^x dx.
Use the formula: ∫ u dv = uv - ∫ v du.
Let u = x and dv = e^x dx. Then, du = dx and v = e^x.
Now, apply the formula:
∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C
So, the result is: x e^x - e^x + C.

Example 4: Trigonometric Substitution

Evaluate ∫ (1 / √(1 - x²)) dx.
This is a standard integral that can be solved using trigonometric substitution:
Let x = sin(θ), so dx = cos(θ) dθ.
Substitute and simplify:
∫ (1 / √(1 - sin²(θ))) cos(θ) dθ = ∫ dθ = θ + C
Finally, substitute back: θ = arcsin(x), so the result is:
arcsin(x) + C.

Integral Calculus Assessment

1. Find ∫ x³ dx.
2. Solve ∫[0 to 3] (2x + 1) dx.
3. Use substitution to evaluate ∫ x e^(x²) dx.
4. Apply integration by parts to ∫ x ln(x) dx.