Definition, order, and degree, general and particular solutions of a differential equation An equation with a derivative of one or more dependent variables, with one or more independent variables, is a differential equation (DE). Differential Equations are classified by type, order, and linearity of the equation. There are two main types of differential equations: “ordinary” … Read more

The area between any of the two of lines, circles/parabolas/ellipses (the region should be clearly identifiable) Find the area enclosed by two sections: a line at y = x – 1 and the parabola y2 = 2x + 6. By solving the two equations, we find that the points of intersection are (-1, -2) and … Read more

Applications in finding the area under simple curves, especially lines, circles/parabolas/ellipses (in standard form only) How do we find areas under a curve, but above the x-axis? As the number of rectangles used to approximate the area of the region increases, the approximation becomes more accurate.It is possible to find the exact area by letting … Read more

Basic properties of definite integrals and evaluation of definite integrals Properties of Basic Integrals If y = f(x) is considered as a nonnegative and integrable function over a closed interval [a,b], then the area under the curve of y = f(x) from point a to b is the integral of f from point a to b, … Read more

Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof) If f(x) is a function defined for a ≤ x ≤ b, we divide the interval [a,b] into n subintervals of equal width Δx = (b-a)/n. We assume x0 =a, x1, x2,…, xn (=b) which denote the endpoints of these subintervals and … Read more

Some improper integrals are discussed below: $\int{\frac{1}{x}dx=ln\vert x\vert }+C$ $\int{\frac{1}{x^{2}}dx=-\frac{1}{x}}+C$ $\int{\frac{dx}{x^{2}-a^{2}}}=\frac{1}{2a}\log \vert \frac{x-a}{x+a}\vert +c$ Proof: we know that: $\frac{1}{x^{2}-a^{2}}=\frac{1}{(x+a)(x-a)}=\frac{1}{2a}\lbrack \frac{(x+a)-(x-a)}{(x+a)(x-a)}\rbrack =\frac{1}{2a}\lbrack \frac{1}{x-a}-\frac{1}{x+a}\rbrack $ Therefore, $\int{\frac{dx}{x^{2}-a^{2}}}=\frac{1}{2a}\lbrack \int{\frac{1}{x-a}-\int{\frac{1}{x+a}}}\rbrack $ Using the antiderivative of 1/x, we obtain: $\int{\frac{dx}{x^{2}-a^{2}}}=\frac{1}{2a}\lbrack \log \vert x-a\vert \rbrack -\log \vert x+a\vert +C$ $=\frac{1}{2a}\log \vert \frac{x-a}{x+a}\vert +c$ $\int{\frac{dx}{a^{2}-x^{2}}}=\frac{1}{2a}\log \vert \frac{a+x}{a-x}\vert +C$ Proof: As the above … Read more

Integration is the process that involves either the evaluation of an indefinite integral or a definite integral. The indefinite integral is a function g with derivative Dx [g(x)] =f(x). Notice that integration is the inverse process of differentiation. Contrary to differentiating a function, we are given the derivative of a function and asked to find … Read more