\(F\) is an antiderivative of \(f\) if \(F'(x)=f(x).\) Note that \(F+2019\) and \(F\) have the same derivative, so \(F+2019\) is also an antiderivative of \(f\). Hence antiderivatives are not unique.
Example 1. If \(f(x)=x^2\), an antiderivative of \(f\) is \(F(x)=\frac{1}{3}x^3\). Another antiderivative is \(G(x)=\frac{1}{3}x^3+666.\) In fact, for any constant \(C\), \(F(x)+C\) is an antiderivative of \(f\).
Question: other than \(F(x)+C\), do there exist other classes of functions, say, \(F\cdot g\) or \(F+g\) for some function \(g\), as antiderivatives of \(f\)?
Theorem 1. If \(F\) is an antiderivative of \(f\), then the most general antiderivative of \(f\) is \[F(x)+C\] where \(C\) is an arbitrary constant.
Theorem 2. Assume \(F'=f\) and \(G'=g\). Then (1). \(F+G\) is an antiderivative of \(f+g.\;\) (2). for any constant \(c\), \(cF\) is an antiderivative of \(cf\).
Ex 2. Find the most general antiderivative of each of the following functions.
(1). \(f(x)=\dfrac{3721}{x}+1\quad\)
(2). \(f(x)=\sin x + 521x\quad\)
(3). \(f(x)=\dfrac{x^6+2\sqrt{x}}{x}\)
Sol:
(1).If \(F_1=3721\ln|x|\), then \(F_1'=\dfrac{3721}{x}\). If \(F_2=x\), then \(F_2'=1\).
Hence \((F_1+F_2)'=f.\) The general antiderivative is \(F=C+F_1+F_2=C+3721\ln|x|+x\).
(2). An antiderivative of \(\sin x\) is ____ ? (Good mistake: \(\cos x\); no mistake: \(-\cos x\))
\[F=C+(-\cos x) + 521\cdot \dfrac{x^2}{2}=C-\cos x + \dfrac{521x^2}{2}.\]
(3). First we rewrite \(f\) as \(f(x)=x^5+2x^{-1/2}.\)
Then we see that \[F=C+\dfrac{x^6}{5+1}+2\cdot \dfrac{x^{1/2}}{-1/2+1} = C+\dfrac{x^6}{6}+4x^{1/2}\]
You can verify that indeed \(F'=x^5+4\cdot\frac{1}{2}x^{x^{-1/2}}=f.\)
Tips. Make sure to check your result by differentiating the antiderivative!
In some cases, when the function value of the antiderivative at certain point is given, then the constant \(C\) can be uniquely determined.
Example 3.
Find \(f\) if \(f'(x)=e^x+3(1+x^2)^{-1}\) and \(f(0)=1111.\)
Sol:
The general antiderivative of \(f'\) is
\[f(x)=C+e^x+3\tan^{-1}x.\]
We need to determine the value of \(C\) using the fact \(f(0)=1111\).
\[f(0)=C+e^0+3\tan^{-1}0=C+1+0=1111\]
Thus \(C=1110\),
so the particular solution is
\[f(x)=1000+e^x+3\tan^{-1}x.\]
Ex. Find \(f\) if \(f'(x)= x^3-2x+1\) and \(f(1)=0.\)
(Answer: \(\; f(x)=-\frac{1}{4}+\frac{1}{4}x^4-x^2+x\))
Sketch an antiderivative \(F\) from \(f\):
\(f>0\) ==> \(F\) increasing;
\(f<0\) ==> \(F\) decreasing;
\(f(a)=0\) ==> \(F\) has a horizontal tangent line at \(a\)