Calculus BC ยท Unit 6
Build the general antiderivative, then use a starting value when one is given.
Power rule, then add C
For \(n\neq -1\),
\(\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C\)
An indefinite integral is a family of functions.
The four basic transcendental rules
\(\int \dfrac{1}{x}\,dx=\ln|x|+C\), \(\int e^x\,dx=e^x+C\)
\(\int \sin x\,dx=-\cos x+C\), \(\int \cos x\,dx=\sin x+C\)
Two inverse-trig antiderivatives
\(\int \dfrac{1}{1+x^2}\,dx=\arctan x+C\)
\(\int \dfrac{1}{\sqrt{1-x^2}}\,dx=\arcsin x+C\)
A starting value picks one function
Find the general antiderivative, then use \(f(a)=b\) to solve for \(C\).
One rule, plus the constant.
\(\int 6x^2\,dx = 2x^3\)
\(\int -4x\,dx = -2x^2\)
\(\int 1\,dx = x\)
\(2x^3-2x^2+x+C\)
Answer: \(2x^3-2x^2+x+C\)
\(\dfrac{4x^{4}}{4}+C\)
\(x^4+C\)
Answer: \(x^4+C\)
\(5x+C\)
Answer: \(5x+C\)
\(3e^x+C\)
Answer: \(3e^x+C\)
\(4\ln|x|+C\)
Answer: \(4\ln|x|+C\)
\(2\sin x+C\)
Answer: \(2\sin x+C\)
A negative power, a second basic rule, or one initial condition.
\(\int \sin x\,dx = -\cos x\)
\(\int \sec^2 x\,dx = \tan x\)
\(-4\cos x+\tan x+C\)
Answer: \(-4\cos x+\tan x+C\)
\(\int x^{-3}\,dx = \dfrac{x^{-2}}{-2}+C\)
\(-\dfrac{1}{2x^2}+C\)
Answer: \(-\dfrac{1}{2x^2}+C\)
\(f(x)=2x^3+C\)
\(f(1)=2+C\)
\(2+C=5\)
\(C=3\)
\(f(x)=2x^3+3\)
Answer: \(2x^3+3\)
\(3\arctan x+C\)
Answer: \(3\arctan x+C\)
\(\dfrac{x^3+2}{x} = x^2+\dfrac{2}{x}\)
\(\int x^2\,dx = \dfrac{x^3}{3}\)
\(\int \dfrac{2}{x}\,dx = 2\ln|x|\)
\(\dfrac{x^3}{3}+2\ln|x|+C\)
Answer: \(\dfrac{x^3}{3}+2\ln|x|+C\)
\(f(x)=e^x+2x+C\)
\(f(0)=1+C\)
\(1+C=4\)
\(C=3\)
\(f(x)=e^x+2x+3\)
Answer: \(e^x+2x+3\)
Rewrite first, or match a scaled inverse-trig form.
\(-\dfrac{6}{x^2} = -6x^{-2}\)
\(\int -6x^{-2}\,dx = -6\cdot \dfrac{x^{-1}}{-1}\)
\(\dfrac{6}{x}\)
\(x^3+\dfrac{6}{x}+2x+C\)
Answer: \(x^3+\dfrac{6}{x}+2x+C\)
\(\int \dfrac{1}{a^2+x^2}\,dx = \dfrac{1}{a}\arctan\dfrac{x}{a}+C\)
\(a=3\)
\(\dfrac{1}{3}\arctan\dfrac{x}{3}+C\)
Answer: \(\dfrac{1}{3}\arctan\dfrac{x}{3}+C\)
The derivative of \(\arcsin(x/2)\) is \(\dfrac{1}{\sqrt{4-x^2}}\).
\(\arcsin\dfrac{x}{2}+C\)
Answer: \(\arcsin\dfrac{x}{2}+C\)
\(f'(x)=6x^2+C\)
\(f'(0)=C=5\)
\(f'(x)=6x^2+5\)
\(f(x)=2x^3+5x+D\)
\(f(0)=D=-2\)
\(f(x)=2x^3+5x-2\)
Answer: \(2x^3+5x-2\)
The domain is negative, so a bare \(\ln x\) is not defined.
The absolute value keeps the logarithm defined.
The factor 2 stays in front. It is not the same as \(\ln|2x|=\ln 2+\ln|x|\).
Answer: A. \(2\ln|x|\)
\(\int \csc^2 x\,dx = -\cot x\)
\(\int \sin x\,dx = -\cos x\)
\(-\cot x - 3(-\cos x) + C\)
\(-\cot x+3\cos x+C\)
Answer: \(-\cot x+3\cos x+C\)
Need help with antiderivatives?
Book a Session with Amy