Indefinite Integral - Area Function

Evaluation of Definite Integral

The definite integral is used to calculate the area under a curve between two points.
It gives a numerical value as the result.
Mathematically, if F'(x) = f(x), then the definite integral of f(x) from a to b is denoted as ∫ [a, b] f(x) dx. Example:
Evaluate the definite integral ∫ [1, 3] (4x^2 + 3x) dx. Solution:
∫ [1, 3] (4x^2 + 3x) dx = [ (x^3) + (3/2)x^2 ] [1, 3]
Plug in the values: (3^3 + (3/2)(3^2)) - (1^3 + (3/2)(1^2))
Simplifying, we get 54 - 4 = 50.

Properties of Indefinite Integrals

Linearity: ∫ (af(x) + bg(x)) dx = a∫ f(x) dx + b∫ g(x) dx
Constant Multiple: ∫ af(x) dx = a∫ f(x) dx (where a is a constant)
Sum/Difference Rule: ∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x) dx
Power Rule: ∫ x^n dx = (1/(n+1)) * x^(n+1) + C (where n is any real number) Example:
Evaluate ∫ (3x^2 + 5x - 2) dx. Solution:
∫ (3x^2 + 5x - 2) dx = (3/3) * x^3 + (5/2) * x^2 - 2x + C
Simplifying, we get x^3 + (5/2) * x^2 - 2x + C.

Integration by Substitution

Integration by substitution is used to simplify integrals by substituting variables.
It requires expressing a function in terms of another variable.
The integral of f(g(x))*g'(x) can be transformed by substituting u = g(x). Example:
Evaluate ∫ 2x sin(x^2) dx. Solution:
Let u = x^2.
Differentiating, we get du = 2x dx.
Substituting in the integral, we have ∫ sin(u) du = -cos(u) + C.
Simplifying, we get -cos(x^2) + C.

Integration by Parts

Integration by parts is used to evaluate the integral of a product of two functions.
It involves splitting the integral into two parts and applying the product rule of differentiation.
The formula for integration by parts is ∫ u dv = uv - ∫ v du. Example:
Evaluate ∫ x cos(x) dx. Solution:
Let u = x and dv = cos(x) dx.
Differentiating u, we get du = dx.
Integrating dv, we get v = sin(x).
Applying the formula, we have ∫ x cos(x) dx = x sin(x) - ∫ sin(x) dx.
Evaluating the integral, we get x sin(x) + cos(x) + C.

Integration of Trigonometric Functions

The integrals of trigonometric functions can be evaluated using standard identities.
Some common integrals include:
- ∫ sin(x) dx = -cos(x) + C
- ∫ cos(x) dx = sin(x) + C
- ∫ sec^2(x) dx = tan(x) + C
- ∫ csc^2(x) dx = -cot(x) + C
- ∫ sec(x) tan(x) dx = sec(x) + C
- ∫ csc(x) cot(x) dx = -csc(x) + C Example:
Evaluate ∫ sin^3(x) cos^2(x) dx. Solution:
Using the identity sin^2(x) = 1 - cos^2(x), we can rewrite the integral as ∫ sin(x) (1 - cos^2(x)) cos^2(x) dx.
Expanding and simplifying, we get ∫ sin(x) cos^2(x) dx - ∫ sin(x) cos^4(x) dx.
Applying the known integrals, we have -(1/3) cos^3(x) + (1/5) cos^5(x) + C.

Integration of Exponential and Logarithmic Functions

The integrals of exponential and logarithmic functions can be evaluated using specific rules.
Some common integrals include:
- ∫ e^x dx = e^x + C
- ∫ a^x dx = (a^x) / ln(a) + C
- ∫ log(a, x) dx = x (log(a, x) - 1) + C
- ∫ ln(x) dx = x ln(x) - x + C Example:
Evaluate ∫ e^x ln(e^x) dx. Solution:
Using the property ln(a^x) = x ln(a), we can rewrite the integral as ∫ x e^x dx.
Applying the known integral ∫ x e^x dx = x e^x - e^x + C.

Integration of Rational Functions

Rational functions are the ratios of two polynomials.
Integrating rational functions involves partial fraction decomposition.
The steps for integrating rational functions include:
1. Factorizing the denominator.
2. Setting up the partial fraction decomposition.
3. Finding the constants.
4. Integrating each term separately. Example:
Evaluate ∫ (x^2 + 2x + 1) / (x^3 + 3x^2 + 3x) dx. Solution:
Factorizing the denominator, we get (x)(x+1)^2.
Setting up the partial fraction decomposition, we have A/(x) + B/(x+1) + C/(x+1)^2.
Finding the constants, we find A = -1, B = 3, and C = -2.
Integrating each term separately, we get -ln|x| + 3ln|x+1| - 2/(x+1) + C.

Indefinite Integral - AREA FUNCTION

The indefinite integral is a reverse process of differentiation.
It finds the antiderivative or the original function from the derivative of a given function.
Mathematically, if F'(x) = f(x), then F(x) is an antiderivative of f(x}. Example:
Find the antiderivative of f(x) = 2x + 3. Solution:
F(x) = ∫ (2x + 3) dx
F(x) = ∫ 2x dx + ∫ 3 dx
F(x) = x^2 + 3x + C, where C is the constant of integration.

Evaluation of Definite Integral

The definite integral is used to calculate the area under a curve between two points.
It gives a numerical value as the result.
Mathematically, if F'(x) = f(x), then the definite integral of f(x) from a to b is denoted as ∫ [a, b] f(x) dx. Example:
Evaluate the definite integral ∫ [1, 3] (4x^2 + 3x) dx. Solution:
∫ [1, 3] (4x^2 + 3x) dx = [ (x^3) + (3/2)x^2 ] [1, 3]
Plug in the values: (3^3 + (3/2)(3^2)) - (1^3 + (3/2)(1^2))
Simplifying, we get 54 - 4 = 50.

Properties of Indefinite Integrals

Linearity: ∫ (af(x) + bg(x)) dx = a∫ f(x) dx + b∫ g(x) dx
Constant Multiple: ∫ af(x) dx = a∫ f(x) dx (where a is a constant)
Sum/Difference Rule: ∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x) dx
Power Rule: ∫ x^n dx = (1/(n+1)) * x^(n+1) + C (where n is any real number) Example:
Evaluate ∫ (3x^2 + 5x - 2) dx. Solution:
∫ (3x^2 + 5x - 2) dx = (3/3) * x^3 + (5/2) * x^2 - 2x + C
Simplifying, we get x^3 + (5/2) * x^2 - 2x + C.

Integration by Substitution

Integration by substitution is used to simplify integrals by substituting variables.
It requires expressing a function in terms of another variable.
The integral of f(g(x))*g'(x) can be transformed by substituting u = g(x). Example:
Evaluate ∫ 2x sin(x^2) dx. Solution:
Let u = x^2.
Differentiating, we get du = 2x dx.
Substituting in the integral, we have ∫ sin(u) du = -cos(u) + C.
Simplifying, we get -cos(x^2) + C.

Integration by Parts

Integration by parts is used to evaluate the integral of a product of two functions.
It involves splitting the integral into two parts and applying the product rule of differentiation.
The formula for integration by parts is ∫ u dv = uv - ∫ v du. Example:
Evaluate ∫ x cos(x) dx. Solution:
Let u = x and dv = cos(x) dx.
Differentiating u, we get du = dx.
Integrating dv, we get v = sin(x).
Applying the formula, we have ∫ x cos(x) dx = x sin(x) - ∫ sin(x) dx.
Evaluating the integral, we get x sin(x) + cos(x) + C.

Integration of Trigonometric Functions

The integrals of trigonometric functions can be evaluated using standard identities.
Some common integrals include:
- ∫ sin(x) dx = -cos(x) + C
- ∫ cos(x) dx = sin(x) + C
- ∫ sec^2(x) dx = tan(x) + C
- ∫ csc^2(x) dx = -cot(x) + C
- ∫ sec(x) tan(x) dx = sec(x) + C
- ∫ csc(x) cot(x) dx = -csc(x) + C Example:
Evaluate ∫ sin^3(x) cos^2(x) dx. Solution:
Using the identity sin^2(x) = 1 - cos^2(x), we can rewrite the integral as ∫ sin(x) (1 - cos^2(x)) cos^2(x) dx.
Expanding and simplifying, we get ∫ sin(x) cos^2(x) dx - ∫ sin(x) cos^4(x) dx.
Applying the known integrals, we have -(1/3) cos^3(x) + (1/5) cos^5(x) + C.

Integration of Exponential and Logarithmic Functions

The integrals of exponential and logarithmic functions can be evaluated using specific rules.
Some common integrals include:
- ∫ e^x dx = e^x + C
- ∫ a^x dx = (a^x) / ln(a) + C
- ∫ log(a, x) dx = x (log(a, x) - 1) + C
- ∫ ln(x) dx = x ln(x) - x + C Example:
Evaluate ∫ e^x ln(e^x) dx. Solution:
Using the property ln(a^x) = x ln(a), we can rewrite the integral as ∫ x e^x dx.
Applying the known integral ∫ x e^x dx = x e^x - e^x + C.

Integration Techniques

There are various techniques to evaluate integrals.
Some common techniques include:
- Integration by parts
- Integration by substitution
- Trigonometric substitution
- Partial fractions
- Trigonometric identities
- Logarithmic differentiation

Integration by Trigonometric Substitution

Trigonometric substitution is used to simplify integrals with radicals involving trigonometric functions.
The substitution involves using appropriate trigonometric identities.
Some common substitutions include:
- x = a sin(t)
- x = a cos(t)
- x = a tan(t) Example:
Evaluate ∫ √(4 - x^2) dx. Solution:
Using the substitution x = 2 sin(t), we have dx = 2 cos(t) dt.
Substituting in the integral, we get ∫ √(4 - 4sin^2(t)) * 2cos(t) dt.
Simplifying, we get ∫ 2 cos^2(t) dt.
Applying the trigonometric identity cos^2(t) = (1 + cos(2t))/2, we have ∫ (1 + cos(2t)) dt.
Evaluating the integral, we get t + (sin(2t))/2 + C.
Substituting back t = sin^(-1)(x/2), we have the final answer as sin^(-1)(x/2) + (sin(x))/2 + C.

Integration of Algebraic Fractions

Algebraic fractions are ratios of polynomials.
Integrating algebraic fractions involves decomposing the fractions into partial fractions.
The steps include:
1. Factorize the denominator.
2. Express the fraction as the sum of partial fractions.
3. Determine the unknown coefficients.
4. Integrate each term separately. Example:
Evaluate ∫ (7x - 1) / (x^2 - x - 2) dx. Solution:
Factoring the denominator, we have (x - 2)(x + 1).
Expressing the fraction as partial fractions, we have ∫ [A / (x - 2)] + [B / (x + 1)] dx.
Determining the coefficients, we find A = -3 and B = 4.
Integrating each term separately, we get -3 ln|x - 2| + 4 ln|x + 1| + C.

Integration of Rational Functions with Quadratic Denominators

Integrating rational functions with quadratic denominators involves partial fraction decomposition.
The steps include:
1. Factorize the denominator.
2. Express the fraction as the sum of partial fractions.
3. Determine the unknown coefficients.
4. Integrate each term separately. Example:
Evaluate ∫ (5x + 3) / (x^2 + 4x + 3) dx. Solution:
Factoring the denominator, we have (x + 1)(x + 3).
Expressing the fraction as partial fractions, we have ∫ [A / (x + 1)] + [B / (x + 3)] dx.
Determining the coefficients, we find A = 2 and B = 1.
Integrating each term separately, we get 2 ln|x + 1| + ln|x + 3| + C.

Integration of Rational Functions with Repeated Linear Factors

Integrating rational functions with repeated linear factors involves partial fraction decomposition.
The steps include:
1. Factorize the denominator, considering repeated factors.
2. Express the fraction as the sum of partial fractions.
3. Determine the unknown coefficients.
4. Integrate each term separately. Example:
Evaluate ∫ (x^2 - 3x + 2) / (x^3 - 6x^2 + 11x - 6) dx. Solution:
Factoring the denominator, we have (x - 1)^2 (x - 2).
Expressing the fraction as partial fractions, we have `