21. Antiderivatives, Areas and the FTC

a2. Non-Uniqueness of Antiderivatives

In principle, there could be many antiderivatives of a given function. In fact, there is only one, up to the addition of a constant.

Any two antiderivatives of a given function must differ by a constant. In formulas, if \(F_1(x)\) and \(F_2(x)\) are both antiderivatives of \(f(x)\), then \[ F_1(x)-F_2(x)=C \] where \(C\) is a constant.

Since \(F_1(x)\) and \(F_2(x)\) are both antiderivatives of \(f(x)\), we compute: \[ \dfrac{d}{dx}\left[F_1(x)-F_2(x)\right] =\dfrac{dF_1}{dx}-\dfrac{dF_2}{dx} =f(x)-f(x)=0 \] Since \(F_1(x)-F_2(x)\) has \(0\) derivative, its slope if \(0\) and it must be a constant function: \[ F_1(x)-F_2(x)=C \]

The general antiderivative of a function is any particular antiderivative plus a constant, usually \(C\). Thus, if \(F(x)\) is some antiderivative of \(f(x)\), then the general antiderivative is: \[ F(x)+C \]

Find the general antiderivative of \(f(x)=x^3\).

We previously found that an antiderivative of \(f(x)=x^3\) is \(\dfrac{1}{4}x^4\). So the general antiderivative is \[ F(x)=\dfrac{1}{4}x^4+C \]

Find the general antiderivative of \(g(x)=\sin x+e^{2x}\).

Since the derivative of a sum is the sum of the derivatives, the same is true for antiderivatives.
What function has \(\sin x\) as its derivative?
What function has \(e^{2x}\) as its derivative?

\(G(x)=-\cos x+\dfrac{1}{2}e^{2x}+C\)

Since the derivative of \(\cos x\) is \(-\sin x\), the antiderivative of \(\sin x\) is \(-\cos x\). Since the derivative of \(e^{2x}\) is \(2e^{2x}\), which is twice what we need, the antiderivative of \(e^{2x}\) is \(\dfrac{1}{2}e^{2x}\). Since the derivative of a sum is the sum of the derivatives, the same is true for antiderivatives. So the general antiderivative of \(g(x)=\sin x+e^{2x}\) is: \[ G(x)=-\cos x+\dfrac{1}{2}e^{2x}+C \]

Ruby says an antiderivative of \(f(x)=2\sin x\cos x\) is \(\sin^2 x\). Jose says an antiderivative is \(-\cos^2 x\). Who is right and who is wrong?

They are both right.

\(\dfrac{d\sin^2 x}{dx}=2\sin x\cos x\) and \(\dfrac{d(-\cos^2 x)}{dx}=-2\cos x(-\sin x)=2\sin x\cos x\). So they are both right! This is possible because \(\sin^2 x-(-\cos^2 x)=1\). So the two answers differ by a constant and both answers are correct.

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