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)

Examples

Find the antiderivative of the function y = x^2

1. We know that f'(x) = 2x^1

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}}
\dfrac{d}{dx} k f(x) = k f'(x)
\dfrac{d}{dx} f(x) \pm g(x) = f'(x) \pm g'(x)