Indefinite Integral - What Are Anti-Derivatives

Indefinite Integral - WHAT ARE ANTI-DERIVATIVES

In calculus, the concept of integration allows us to find the area under a curve.
Indefinite integral is a type of integration where we find the antiderivative or the primitive function of a given function.
The antiderivative of a function f(x) is a new function F(x) such that F’(x) = f(x).
The antiderivative is also known as the integral and is denoted by ∫f(x) dx.

Properties of Indefinite Integral

The indefinite integral of a constant is the constant multiplied by x: ∫k dx = kx + C, where k is a constant and C is the constant of integration.
The power rule of integration states that ∫x^n dx = (x^(n+1))/(n+1) + C, where n is any real number (except -1) and C is the constant of integration.
The linear combination rule of integration states that ∫(af(x) + bg(x)) dx = a∫f(x) dx + b∫g(x) dx, where a and b are constants and f(x), g(x) are functions.
The constant factor rule of integration states that ∫kf(x) dx = k∫f(x) dx, where k is a constant.

Methods of Evaluating Indefinite Integrals

Direct integration: In this method, we use the basic rules of integration to find the antiderivative. For example, ∫x^2 dx = (x^3)/3 + C.
Substitution method: In this method, we substitute a different variable or expression in place of x to simplify the integral. For example, if we have ∫cos(x) sin(x) dx, we can substitute u = sin(x) to simplify the integral.
Integration by parts: In this method, we use a specific formula, which is derived from the product rule of differentiation, to find the integral. For example, ∫u dv = uv - ∫v du.
Partial fractions: In this method, we decompose a rational function into simpler fractions and then integrate each term separately. For example, if we have ∫(2x+3)/(x^2+x+1) dx, we can decompose the denominator using partial fractions and then integrate each term.

Integrals of Common Functions

∫dx = x + C
∫x^n dx = (x^(n+1))/(n+1) + C, for n ≠ -1
∫e^x dx = e^x + C
∫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
∫1/(x^2 + a^2) dx = (1/a) arctan(x/a) + C, where a is a constant
∫1/(x^2 - a^2) dx = (1/(2a)) ln|(x-a)/(x+a)| + C, where a is a constant

Properties of Definite Integral

The definite integral represents the area under a curve between two limits.
The definite integral is denoted as ∫[a, b] f(x) dx, where a and b are the limits of integration.
The definite integral can be positive, negative, or zero depending on the function and the limits.
The definite integral can be evaluated using the antiderivative or the fundamental theorem of calculus.
The definite integral is additive: ∫[a, b] f(x) dx + ∫[b, c] f(x) dx = ∫[a, c] f(x) dx.

Methods of Evaluating Definite Integrals

Evaluation using antiderivative: If F(x) is the antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
Evaluation using substitution: Sometimes, substitution can be used to simplify a definite integral before evaluating it.
Evaluation using geometric interpretation: If the function represents an area or volume, we can use geometric methods to evaluate the definite integral.

Indefinite Integral - WHAT ARE ANTI-DERIVATIVES (contd.)

Indefinite integrals are often used to find the antiderivative of a given function.
The antiderivative is a function whose derivative is equal to the original function.
The antiderivative of a function can have a constant of integration, denoted by + C.

Example 1: Evaluating Indefinite Integral

Given function: f(x) = 3x^2 + 2x + 5
To find the antiderivative, we use the power rule of integration for each term.
∫(3x^2) dx = (3/3)x^3 + C1 = x^3 + C1
∫(2x) dx = (2/2)x^2 + C2 = x^2 + C2
∫(5) dx = 5x + C3
Combining the terms, the antiderivative is F(x) = x^3 + x^2 + 5x + C, where C = C1 + C2 + C3.

Example 2: Evaluating Indefinite Integral

Given function: f(x) = 2sin(x) + 3cos(x)
To find the antiderivative, we use the rules of integration for each term.
∫(2sin(x)) dx = -2cos(x) + C1
∫(3cos(x)) dx = 3sin(x) + C2
Combining the terms, the antiderivative is F(x) = -2cos(x) + 3sin(x) + C, where C = C1 + C2.

Substitution Method

The substitution method is a powerful technique used to simplify integrals.
It involves substituting a different variable or expression in place of x.
The goal is to transform the integral into a simpler form that can be easily solved.
The substitution is often chosen to cancel out complicated terms or simplify the integral.

Example: Substitution Method

Given integral: ∫(2x + 1)dx
Let u = 2x + 1 (substitution)
To find du/dx, differentiate u with respect to x: du/dx = 2
Rearranging, du = 2dx (multiply both sides by dx)
Substitute the values of u and du in the original integral:
∫(2x + 1)dx = ∫u du/2

Example (continued)

Simplifying the integral with the substitution:
∫(2x + 1)dx = ∫u du/2
∫u du/2 = (1/2)∫u du
∫u du = (1/2)(u^2) + C
Substitute back the value of u:
(1/2)(u^2) + C = (1/2)((2x + 1)^2) + C

Integration By Parts Method - Formula

The integration by parts method is used to find the antiderivative of a product of two functions.
The formula for integration by parts is:
∫u dv = uv - ∫v du
Here, u and v are the two functions, and du and dv are the differentials of u and v respectively.
The goal is to choose u and dv such that the integral on the right-hand side becomes simpler to solve.

Integration By Parts Method - Example

Given integral: ∫x sin(x)dx
Choosing u = x and dv = sin(x), we have du = dx and v = -cos(x).
Applying the integration by parts formula, we get:
∫x sin(x)dx = -x cos(x) - ∫(-cos(x))dx
Simplifying the integral on the right-hand side, we have:
∫x sin(x)dx = -x cos(x) - ∫cos(x)dx

Integration By Parts Method - Example (continued)

To solve the integral ∫cos(x)dx, we use the antiderivative of cos(x), which is sin(x).
∫cos(x)dx = sin(x) + C1
Substituting this into the previous equation, we get:
∫x sin(x)dx = -x cos(x) - (sin(x) + C1)
Simplifying further, the antiderivative is:
∫x sin(x)dx = -x cos(x) - sin(x) + C1

Partial Fractions Method

The partial fractions method is used to decompose a rational function into simpler fractions.
This technique allows us to simplify the integral of a rational function and solve it more easily.
The decomposition involves expressing the rational function as a sum of partial fractions.
Each partial fraction can then be integrated separately.

Example: Partial Fractions Method

Given integral: ∫(x^2 + 3)/(x(x+1)) dx
To solve this integral, we need to decompose the rational function into partial fractions.
Step 1: Factorize the denominator: x(x+1)
Step 2: Write the partial fraction decomposition as:
- (x^2 + 3)/(x(x+1)) = A/x + B/(x+1)
Step 3: Find the values of A and B by equating the numerators:
- x^2 + 3 = A(x+1) + Bx
Step 4: Solve for A and B by comparing coefficients:
- 1A + 1B = 0 (coefficient of x)
- 0A + 1B = 3 (constant term)
Step 5: Solve the system of equations to find A and B:
- A = -3
- B = 3
Step 6: Substitute the values of A and B back into the partial fraction decomposition.
∫(x^2 + 3)/(x(x+1)) dx = ∫(-3/x) dx + ∫(3/(x+1)) dx
Simplify and integrate each term separately.

Integration by Trig Substitution - Introduction

Trigonometric substitution is a technique used to simplify integrals involving radicals and trigonometric functions.
It involves substituting a trigonometric expression for the variable in the integral.
The goal is to transform the integral into a form that can be easily solved using trigonometric identities and properties.

Trigonometric Substitution - Example

Given integral: ∫√(x^2 + 9) dx (square root of (x^2 + 9))
We substitute x = 3tanθ, where θ is the new variable.
Since x = 3tanθ, dx = 3sec^2θ dθ (using the derivative of tangent)
Substituting the values of x and dx, the integral becomes:
∫√(9tan^2θ + 9) 3sec^2θ dθ
Simplifying the expression inside the square root, we get:
∫√(9sec^2θ) 3sec^2θ dθ
Further simplifying, we have:
∫3secθ sec^2θ dθ

Trigonometric Substitution - Example (continued)

Using the trigonometric identity: sec^2θ = tan^2θ + 1
The integral becomes:
∫3secθ (tan^2θ + 1) dθ
Simplifying, we have:
∫3tan^2θ secθ dθ + ∫3secθ dθ

Trigonometric Substitution - Example (continued)

The integral of 3tan^2θ secθ can be solved using the substitution method or by recognizing it as a known integral.
The integral of 3secθ can be solved by recognizing it as the derivative of the natural logarithm of the absolute value of secθ + tanθ.
Solve each integral separately, and then combine the results to obtain the final solution.

Integration by Tables - Introduction

Integration by tables, also known as tabular integration, is a method used for integrating certain types of functions.
It involves creating a table to perform repeated integration by parts until a pattern emerges.
The method is particularly useful for functions that result in a repeating pattern after applying integration by parts.

Integration by Tables - Example

Given integral: ∫x^3 e^x dx
To solve this integral using integration by tables, we create a table as follows: u v dv du ∫vdu x^3 e^x 3x^2 dx 3x dx 3x^2 e^x - 6x e^x - 6 ∫x^2 e^x dx 3x^2 e^x 6x dx 6x dx 6x e^x - 6 ∫x e^x dx 6x e^x 6 dx 6 dx 6 e^x - 6 ∫e^x dx 6 e^x 0 0 6 e^x
The pattern in the table shows that the integral will converge to zero after multiple iterations.
Hence, the final result is 6x^3 e^x - 6x^2 e^x - 6x e^x - 6 e^x + C.

Definite Integral - Introduction

The definite integral represents the area under a curve between two given limits.
It is denoted as ∫[a, b] f(x) dx, where a and b are the limits of integration.
The definite integral evaluates the signed area between the curve and the x-axis within the given limits.

Properties of Definite Integral

The definite integral is additive: ∫[a, b] f(x) dx + ∫[b, c] f(x) dx = ∫[a, c] f(x) dx.
The definite integral can be negative if the curve lies below the x-axis within the given limits.
The definite integral of a constant function is given by the product of the constant and the difference of the limits.
The definite integral of a function over the entire real line is denoted as ∫(-∞, ∞) f(x) dx.

Methods of Evaluating Definite Integrals

Evaluation using antiderivative: If F(x) is the antiderivative of f(x), then ∫[a, b] f(x) dx = F(b) - F(a).
Evaluation using substitution: Sometimes, substitution can be used to simplify a definite integral before evaluating it.
Evaluation using geometric interpretation: If the function represents an area or volume, we can use geometric methods to evaluate the definite integral.