notes/education/math/MATH1210 (calc 1)/Integrals.md

Antiderivatives

An antiderivative is useful when you know the rate of change, and you want to find a point from that rate of change

A function F is said to be an antiderivative of f if F'(x) = f(x)

Notation

The collection of all antiderivatives of a function f is referred to as the indefinite integral of f with respect to $x$, and is denoted by:

\int f(x) dx

Examples

Find the antiderivative of the function y = x^2

1. We know that to find the derivative of the above function, you'd multiply by the exponent (2), and subtract 1 from the exponent.
2. To perform this operation in reverse:
   1. Add 1 to the exponent
   2. Multiply by \dfrac{1}{n + 1}
3. This gives us an antiderivative of \dfrac{1}{3}x^3
4. To check our work, work backwards.
5. The derivative of \dfrac{1}{3}x^3 is \dfrac{1}{3} (3x^2)
6. = \dfrac{3}{3} x^2

Formulas

Differentiation Formula	Integration Formula
\dfrac{d}{dx} x^n = nx^{x-1}	\int x^n dx = \dfrac{1}{n+1}x^{n+1}+ C for n \ne -1
\dfrac{d}{dx} kx = k	\int k \space dx = kx + C
\dfrac{d}{dx} \ln \|x\| = \dfrac{1}{x}	\int \dfrac{1}{x}dx = \ln \|x\| + C
\dfrac{d}{dx} e^x = e^x	\int e^x dx = e^x + C
\dfrac{d}{dx} a^x = (\ln{a}) a^x	\int a^xdx = \ln \|x\| + C
\dfrac{d}{dx} \sin x = \cos x	\int \cos(x) dx = \sin (x) + C
\dfrac{d}{dx} \cos x = -\sin x	\int \sin(x)dx = \sin x + C
\dfrac{d}{dx} \tan{x} = \sec^2 x	\int \sec^2(x)dx = \tan(x) + C
\dfrac{d}{dx} \sec x = \sec x \tan x	\int sec^2(x) dx = \sec(x) + C
\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}	\int \sec(x) \tan(x) dx = \sec x + C
\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2}	\int \dfrac{1}{\sqrt{1+x^2}}dx = \tan^{-1}x + C
\dfrac{d}{dx} k f(x) = k f'(x)	\int k*f(x)dx = k\int f(x)dx
\dfrac{d}{dx} f(x) \pm g(x) = f'(x) \pm g'(x)	\int (f(x) \pm g(x))dx = \int f(x) dx \pm \int g(x) dx

Area Under a Curve

The area under the curve y = f(x) can be approximated by the equation \sum_{i = 1}^n f(\hat{x_i})\Delta x where \hat{x_i} is any point on the interval [x_{i - 1}, x_i], and the curve is divided into n equal parts of width \Delta x

Any sum of this form is referred to as a Reimann Sum.

To summarize:

• The area under a curve is equal to the sum of the area of n rectangular subdivisions where each rectangle has a width of \Delta x and a height of f(x).

Definite Integrals

Let f be a continuous function on the interval [a, b]. Divide [a, b] into n equal parts of width \Delta x = \dfrac{b - a}{n} . Let x_0, x_1, x_2, \cdots, x_3 be the endpoints of the subdivision.

The definite integral of f(x) with respect to x from x = a to x = b can be denoted:

\int_{a}^b f(x) dx

And can be defined as:

\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i = 1}^n f(x_i)\Delta x

f(x_i) is the height of each sub-interval, and \Delta x is the change in the x interval, so f(x_i) \Delta x is solving for the area of each sub-interval.

Examples

Find the exact value of the integral \int_0^1 5x \space dx

Relevant formulas:

\sum_{i = 1}^n = \dfrac{(n)(n + 1)}{2}

$ \Delta x = \dfrac{1 - 0$

1. \int_0^1 5x \space dx = \lim_{n \to \infty} \sum_{i=1}^n 5(x_i) * \Delta x
2. = \lim_{n \to \infty} \sum_{i=1}^n 5(\frac{1}{n} \cdot i) \cdot \frac{1}{n}
3. = \lim_{n \to \infty} \sum_{i = 1}^n \dfrac{5}{n^2}\cdot i
4. = \lim_{n \to \infty} \dfrac{5}{n^2} \sum_{i = 1}^n i
5. = \lim_{x \to \infty} \dfrac{5}{n^2} \cdot \dfrac{n(n + 1)}{2}
6. = \lim_{n \to \infty} \dfrac{5n^2 + 5n}{2n^2}
7. = \dfrac{5}{2}