Indefinite Integral

  • In calculus, the indefinite integral represents a family of functions called antiderivatives.
  • It is denoted by the symbol ∫ (integral sign).
  • The indefinite integral calculates the reverse process of differentiation, finding a function whose derivative is equal to the given function.

Notation

  • The indefinite integral of a function f(x) with respect to x is represented as ∫f(x) dx.
  • Here, f(x) is the integrand, and dx denotes the integration variable.
  • The result of the integral is expressed as F(x) + C, where F(x) is the antiderivative of f(x) and C is the constant of integration.

Basic Rules

  1. Linearity: ∫(af(x) + bg(x)) dx = a∫f(x) dx + b∫g(x) dx, where a and b are constants.
  1. Power Rule: ∫x^n dx = (x^(n+1))/(n+1) + C, where n ≠ -1.
  1. Constant Rule: ∫k dx = kx + C, where k is a constant.
  1. Sum/Difference Rule: ∫(f(x) ± g(x)) dx = ∫f(x) dx ± ∫g(x) dx.

Example

Find the indefinite integral of the function f(x) = 3x^2 + 5x - 4.

  • Solution:
    • Using the power rule, we have:
      • ∫(3x^2 + 5x - 4) dx = (3x^3/3) + (5x^2/2) - 4x + C
      • Simplifying further:
        • ∫(3x^2 + 5x - 4) dx = x^3 + (5/2)x^2 - 4x + C

Integration by Substitution

  • Integration by substitution is a technique used to simplify the integrand by substituting a new variable.
  • The aim is to transform the integral into a new form that is easier to evaluate.
  • Common substitutions include u-substitution, trigonometric substitution, and exponential substitution.

u-Substitution

  • Steps for u-substitution:
    1. Let u = g(x), where g(x) is a function within the integrand.
    2. Calculate du/dx, the derivative of u with respect to x.
    3. Rewrite the integral in terms of u and du.
    4. Evaluate the new integral with respect to u.
    5. Replace u with g(x) in the final answer.

Example

Evaluate the integral ∫(2x + 1)^3 dx using u-substitution.

  • Solution:
    1. Let u = 2x + 1.
    2. Calculate du/dx: du/dx = 2.
    3. Rewrite the integral in terms of u and du: ∫u^3 (1/2) du.
    4. Evaluate the integral with respect to u: (1/2) ∫u^3 du = (1/2) * u^4/4 + C.
    5. Replace u with 2x + 1 in the final answer: (1/2) * (2x + 1)^4/4 + C.

Integration by Parts

  • Integration by parts is a technique that allows us to integrate the product of two functions.
  • It is based on the product rule of differentiation.
  • The formula for integration by parts is given by: ∫u dv = uv - ∫v du, where u and v are functions of x.

Formula

The formula for integration by parts can be written as: ∫u dv = uv - ∫v du

  • Where:
    • u is the first function chosen for integration,
    • dv is the differential of the second function,
    • du is the differential of u,
    • v is the antiderivative of dv.

Example

Evaluate the integral ∫x sin(x) dx using integration by parts.

  • Solution:
    • Choose u = x and dv = sin(x) dx.
    • Calculate du = dx (differential of u) and v = -cos(x) (antiderivative of dv).
    • Apply the integration by parts formula:
      • ∫x sin(x) dx = -x cos(x) - ∫-cos(x) dx.
    • Simplify the integral:
      • ∫x sin(x) dx = -x cos(x) + sin(x) + C.

Indefinite Integral - Integral of some particular functions

  • When finding the indefinite integral of certain functions, we use specific rules and formulas.
  • These functions include:
    • Constant function: ∫c dx = cx + C, where c is a constant.
    • Exponential function: ∫e^x dx = e^x + C.
    • Natural logarithm function: ∫(1/x) dx = ln|x| + C.
    • Trigonometric functions: ∫sin(x) dx = -cos(x) + C, ∫cos(x) dx = sin(x) + C, ∫sec^2(x) dx = tan(x) + C, etc.

Indefinite Integral - Trigonometric Functions

  • The integrals of trigonometric functions can be found using trigonometric identities.
  • Some common integrals include:
    • ∫sin^2(x) dx = (1/2)(x - sin(x)cos(x)) + C,
    • ∫cos^2(x) dx = (1/2)(x + sin(x)cos(x)) + C,
    • ∫tan(x) dx = -ln|cos(x)| + C.

Indefinite Integral - Rational Functions

  • Rational functions involve a fraction of polynomials.
  • To find the indefinite integral of rational functions, we can use techniques such as partial fractions or long division.
  • Examples:
    • ∫(2x^3 + 3x^2 - 4)/(x^2 - 1) dx = ∫(2x + 1) dx + ∫(5x + 3)/(x^2 - 1) dx
    • ∫(x^2 + 3)/(x^3 - 2x) dx = ∫(1/x - 2/(x^2) + 3/(x - 2)) dx

Indefinite Integral - Trigonometric Substitution

  • Trigonometric substitution is useful for simplifying integrals involving certain algebraic expressions.
  • Common trigonometric substitutions include:
    • √(a^2 - x^2) -> x = a sinθ
    • √(x^2 + a^2) -> x = a tanθ
    • √(x^2 - a^2) -> x = a secθ

Indefinite Integral - Exponential/Substitution

  • Exponential and substitution methods are used to solve more complex integrals.
  • Examples:
    • ∫e^(3x) sin(2x) dx: We can use the substitution method to simplify the integral.
    • ∫e^x / (1 + e^x)^2 dx: We can use the substitution method along with manipulation of fractions to solve this integral.

Definite Integral

  • The definite integral calculates the area under a curve between two specified limits.
  • It is denoted by ∫[a, b] f(x) dx, where a and b are the lower and upper limits, f(x) is the integrand, and dx is the integration variable.
  • The result of the definite integral is a number that represents the area enclosed between the curve and the x-axis over the specified interval.

Properties of Definite Integrals

  • Linearity: ∫[a, b] (af(x) + bg(x)) dx = a∫[a, b] f(x) dx + b∫[a, b] g(x) dx
  • Additivity: ∫[a, b] f(x) dx + ∫[b, c] f(x) dx = ∫[a, c] f(x) dx

Fundamental Theorem of Calculus

  • The Fundamental Theorem of Calculus connects differentiation and integration.
  • It states that if f(x) is continuous on the interval [a, b] and F(x) is an antiderivative of f(x), then:
    • ∫[a, b] f(x) dx = F(b) - F(a)
  • This theorem allows us to evaluate definite integrals without finding the antiderivative explicitly.

Example

Evaluate the definite integral ∫[1, 2] (x^2 + 3x - 2) dx.

  • Solution:
    • First, find the antiderivative of the integrand:
      • F(x) = (1/3)x^3 + (3/2)x^2 - 2x
    • Using the Fundamental Theorem of Calculus:
      • ∫[1, 2] (x^2 + 3x - 2) dx = F(2) - F(1)
      • Plug in the values:
      • = [(1/3)(2)^3 + (3/2)(2)^2 - 2(2)] - [(1/3)(1)^3 + (3/2)(1)^2 - 2(1)]
      • Simplify to find the result.

Conclusion

  • Indefinite and definite integrals are essential concepts in calculus.
  • The indefinite integral finds the antiderivative of a function, while the definite integral calculates the area under a curve.
  • Various techniques such as substitution, integration by parts, and the Fundamental Theorem of Calculus are used to evaluate integrals.
  • Practice and understanding these methods will help in solving a wide range of integral problems.

Indefinite Integral - Integral of Some Particular Functions

  • The indefinite integral of particular functions can be found using specific rules and formulas.
  • These functions include:
    • Exponential functions
    • Natural logarithm functions
    • Trigonometric functions
    • Hyperbolic functions
  • Memorizing these integral forms can greatly simplify the process of evaluating integrals.

Indefinite Integral - Exponential Functions

  • The integral forms of exponential functions are as follows:
    • ∫e^ax dx = (1/a)e^ax + C, where a ≠ 0
    • ∫e^x dx = e^x + C
    • ∫a^x dx = (1/ln a) a^x + C, where a > 0 and a ≠ 1 Note: The constant ‘a’ can be any real number.

Indefinite Integral - Natural Logarithm Function

  • The integral form of the natural logarithm function is as follows:
    • ∫(1/x) dx = ln |x| + C, where x ≠ 0 Note: The absolute value |x| is used to ensure the function is defined for both positive and negative values of x.

Indefinite Integral - Trigonometric Functions

  • The integral forms of trigonometric functions are as follows:
    • ∫sin(x) dx = -cos(x) + C
    • ∫cos(x) dx = sin(x) + C
    • ∫sec^2(x) dx = tan(x) + C
    • ∫cosec^2(x) dx = -cot(x) + C
    • ∫tan(x) dx = -ln|cos(x)| + C
    • ∫cot(x) dx = ln|sin(x)| + C Note: These integral forms assume the angles are measured in radians.

Indefinite Integral - Hyperbolic Functions

  • The integral forms of hyperbolic functions are as follows:
    • ∫sinh(x) dx = cosh(x) + C
    • ∫cosh(x) dx = sinh(x) + C Note: The hyperbolic sine (sinh) and hyperbolic cosine (cosh) are the hyperbolic analogs of the trigonometric functions.

Example

Evaluate the integral ∫e^3x dx.

  • Solution:
    • Using the integral form of the exponential function:
      • ∫e^3x dx = (1/3)e^3x + C
    • Therefore, the value of the integral is (1/3)e^3x + C.

Example

Evaluate the integral ∫ln(x) dx.

  • Solution:
    • Using the integral form of the natural logarithm function:
      • ∫ln(x) dx = x ln(x) - x + C
    • Therefore, the value of the integral is x ln(x) - x + C.

Example

Evaluate the integral ∫sin(2x) dx.

  • Solution:
    • Using the integral form of the sine function:
      • ∫sin(2x) dx = -cos(2x)/2 + C
    • Therefore, the value of the integral is -cos(2x)/2 + C.

Example

Evaluate the integral ∫tan(x) dx.

  • Solution:
    • Using the integral form of the tangent function:
      • ∫tan(x) dx = -ln|cos(x)| + C
    • Therefore, the value of the integral is -ln|cos(x)| + C.

Conclusion

  • The indefinite integral of particular functions has specific integral forms.
  • Recognizing these forms can significantly simplify the process of evaluating integrals.
  • Examples of such functions include exponential functions, natural logarithm functions, trigonometric functions, and hyperbolic functions.
  • Understanding these integral forms will aid in solving a wide range of integral problems.