Integration Methods and Examples PDF

## Part 4: Integrations ### علم التفاضل (Calculus) A graph of a function is shown. The area under the curve between x1 and x2 is: $A = \int_{x_1}^{x_2} sin(x) dx$ The area under the curve between x1 and x3 is: $A = \sum_{i = 1}^{3} x_i$ ### علم التكامل (Integral Calculus) #### Definite I...

## Part 4: Integrations ### علم التفاضل (Calculus) A graph of a function is shown. The area under the curve between x1 and x2 is: $A = \int_{x_1}^{x_2} sin(x) dx$ The area under the curve between x1 and x3 is: $A = \sum_{i = 1}^{3} x_i$ ### علم التكامل (Integral Calculus) #### Definite Integral The definite integral of f(x) from a to b is represented as: $\int_{a}^{b} f(x) dx$ **Example:** $I = \int_{a}^{b} sin^2(x) dx = g(b)-g(a)$ Where g(x) is the antiderivative of $sin^2(x)$. #### Indefinite Integral The indefinite integral of f(x) is represented by: $\int f(x) dx$ **Example:** $ I = \int sin^2(x) dx = g(x) + C$ ### Types of Integrals - **Direct Integrals** - **Methods of Integrals** - **Fractions** - **Parts** - **Substitution** #### Standend Integrals **Examples:** - $I = \int x^3 + 6x + 1 dx = \frac{1}{4}x^4 + 6*\frac{1}{2}x^2 + x + C$ - $I = \int sin(x) dx = - cos(x) + C$ - $I = \int sec^2(x) dx = tan(x) + C$ - $I = \int sec( x) tan(x) dx = sec(x) + C$ - $I = \int sinh(3x) dx = \frac{1}{3}cosh(3x) + C$ - $I = \int \frac{3dx}{\sqrt{1-9x^2}} = \frac{1}{3}cos^{-1}(3x) + C$ - $I = \int \frac{1}{x} dx = ln(|x|) + C$ - $I = \int \frac{f'(x)}{f(x)} dx = ln(f(x)) + C$ #### Other Methods of Integration **Example:** $I = \int \frac{1}{x^2 + 1} dx = tan^{-1}(x) + C$ **Example:** $I = \int e^{3x} dx = \frac{1}{3}e^{3x} + C$ **Example:** $I = \int cos(3x + 1) dx = \frac{1}{3}sin(3x + 1) + C$ **Example:** $I = \int sin^3(x)cos(x) dx = \frac{1}{4}sin^4(x) + C$ ### Integration By Partial Fractions **Example:** $I = \int \frac{3x}{(x+1)^2(x+1)} dx$ Let's solve $\frac{3x}{(x+1)^2(x+1)} = \frac{A}{(x+1)^2} + \frac{B}{x+1}$ Multiply both sides by $(x+1)^2(x+1):$ $3x = A(x+1) + B(x+1)^2$ $x = - 1: -3 = A(0) + B(0) \implies B = -3$ $x = 0: 0 = A(1) + B(1) \implies A = 3$ Now with our values of A and B we can integrate: $I = \int \frac{3}{(x+1)^2} - \frac{3}{(x+1)} dx$ $I = 3 \int \frac{1}{(x+1)^2} - \int \frac{3}{(x+1)} dx$ $I = 3\int (x+1)^{-2} dx - 3\int \frac{1}{(x+1)} dx$ $I = 3 \frac{(x + 1)^3}{3} - 3ln(x+1) + C$ $I = -3ln(x+1) - (x+1)^{-1} + C$ Let me know if I can help with anything else!