21. Antiderivatives, Areas and the FTC
We have learned how to take derivatives. We now learn how to do this backwards.
a1. Definition of Antiderivatives
An antiderivative of a function \(f(x)\) is any function \(F(x)\) whose derivative is \(f(x)\): \[ \dfrac{dF}{dx}=f(x) \]
We tend to use lower case letters for the original function and the corresponding upper case letter for the antiderivative.
Find an antiderivative of \(f(x)=x^3\).
We know that the derivative of \(x^4\) is \(4x^3\). This is \(4\) times too much. So we take \(\dfrac{1}{4}\) of it. So an antiderivative of \(f(x)=x^3\) is \(F(x)=\dfrac{1}{4}x^4\).
We check this by differentiating: \[ \dfrac{d}{dx}\left[\dfrac{1}{4}x^4\right]=\dfrac{1}{4}(4x^3)=x^3 \]
Find an antiderivative of \(\sin(3x)\).
\(\dfrac{-1}{3}\cos(3x)\)
We know that the derivative of \(\cos(x)\) is \(-\sin(x)\). This has a minus and is missing the \(3\). So we try \(-\cos(3x)\) whose derivative is \(\sin(3x)\cdot3\) which is still off be a \(3\) factor. So we take \(\dfrac{1}{3}\) of our guess. So an antiderivative of \(f(x)=\sin(3x)\) is \(F(x)=\dfrac{-1}{3}\cos(3x)\).
We check this by differentiating: \[ \dfrac{d}{dx}\left[\dfrac{-1}{3}\cos(3x)\right] =\dfrac{-1}{3}\left[-\sin(3x)\cdot3\right]=\sin(3x) \]
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