4.1: Introduction to Antiderivatives (2024)

Definition 4.1.1.

A function \(F(x)\) that satisfies

\begin{align*} \dfrac{d}{dx} F(x) &= f(x) \end{align*}

is called an antiderivative of \(f(x)\text{.}\)

Lemma 4.1.2.

Let \(F(x)\) be an antiderivative of \(f(x)\text{,}\) then for any constant \(c\text{,}\) the function \(F(x)+c\) is also an antiderivative of \(f(x)\text{.}\)

Example 4.1.3 Antiderivative of \(3x^5-7x^2+2x+3 + x^{-1} - x^{-2}\).

Let \(f(x) = 3x^5-7x^2+2x+3 + x^{-1} - x^{-2}\text{.}\) Then the antiderivative of \(f(x)\) is

\begin{align*} F(x) &= \frac{3}{6}x^6 - \frac{7}{3}x^3 + \frac{2}{2}x^2 + 3x +\log|x| - \frac{1}{-1} x^{-1} + c & \text{clean it up}\\ &= \frac{1}{2}x^6 - \frac{7}{3}x^3 + x^2 + 3x +\log|x| + x^{-1} + c \end{align*}

Now to check we should differentiate and hopefully we get back to where we started

\begin{align*} F'(x) &= \frac{6}{2}x^5 - \frac{7}{3} \cdot 3 x^2 + 2 x + 3 + \frac{1}{x} - x^{-2}\\ &= 3 x^5 - 7 x^2 + 2 x + 3 + \frac{1}{x} - x^{-2} & \checkmark \end{align*}

Example 4.1.4 Antiderivative of \(\sin x, \cos 2x\) and \(\frac{1}{1+4x^2}\).

Consider the functions

\begin{align*} f(x) &= \sin x + \cos 2x & g(x) &= \frac{1}{1+4x^2} \end{align*}

Find their antiderivatives.

Solution The first one we can almost just look up our table. Let \(F\) be the antiderivative of \(f\text{,}\) then

\begin{align*} F(x) &= -\cos x + \sin 2x + c & \text{is not quite right}. \end{align*}

When we differentiate to check things, we get a factor of two coming from the chain rule. Hence to compensate for that we multiply \(\sin2x\) by \(\frac{1}{2}\text{:}\)

\begin{align*} F(x) &= -\cos x + \frac{1}{2} \sin 2x + c \end{align*}

Differentiating this shows that we have the right answer.

Similarly, if we use \(G\) to denote the antiderivative of \(g\text{,}\) then it appears that \(G\) is nearly \(\arctan x\text{.}\) To get this extra factor of \(4\) we need to substitute \(x \mapsto 2x\text{.}\) So we try

\begin{align*} G(x) &= \arctan(2x)+c & \text{which is nearly correct}. \end{align*}

Differentiating this gives us

\begin{align*} G'(x) &= \frac{2}{1+(2x)^2} = 2g(x) \end{align*}

Hence we should multiply by \(\frac{1}{2}\text{.}\) This gives us

\begin{align*} G(x) &= \frac{1}{2} \arctan(2x) + c. \end{align*}

We can then check that this is, in fact, correct just by differentiating.

Example 4.1.15 Position as antiderivativeof velocity.

Suppose that we are driving to class. We start at \(x=0\) at time \(t=0\text{.}\) Our velocity is \(v(t) = 50\sin(t)\text{.}\) The class is at \(x=25\text{.}\) When do we get there?

Solution Let's denote by \(x(t)\) our position at time \(t\text{.}\) We are told that

• \(x(0) = 0\)
• \(x'(t) = 50\sin t\)

We have to determine \(x(t)\) and then find the time \(T\) that obeys \(x(T)=25\text{.}\) Now armed with our table above we know that the antiderivative of \(\sin t\) is just \(-\cos t\text{.}\) We can check this:

\begin{align*} \dfrac{d}{dt}\left(-\cos t\right) &= \sin t \end{align*}

We can then get the factor of \(50\) by multiplying both sides of the above equation by 50:

\begin{align*} \dfrac{d}{dt}\left(-50\cos t\right) &= 50\sin t \end{align*}

And of course, this is just an antiderivative of \(50\sin t\text{;}\) to write down the general antiderivative we just add a constant \(c\text{:}\)

\begin{align*} \dfrac{d}{dt}\left(-50\cos t + c\right) &= 50\sin t \end{align*}

Since \(v(t) = \dfrac{d}{dt}x(t)\text{,}\) this antiderivative is \(x(t)\text{:}\)

\begin{align*} x(t) &= -50\cos t + c \end{align*}

To determine \(c\) we make use of the other piece of information we are given, namely

\begin{align*} x(0) &= 0.\\ \end{align*}

Substituting this in gives us

\begin{align*} x(0) &= -50\cos 0 + c = -50+c \end{align*}

Hence we must have \(c=50\) and so

\begin{align*} x(t) &= -50\cos t + 50 = 50(1-\cos t). \end{align*}

Now that we have our position as a function of time, we can determine how long it takes us to arrive there. That is, we can find the time \(T\) so that \(x(T)=25\text{.}\)

\begin{align*} 25 = x(T) &= 50(1-\cos T) & \text{so}\\ \frac{1}{2} &= 1-\cos T\\ -\frac{1}{2} &= -\cos T\\ \frac{1}{2} &= \cos T. \end{align*}

Recalling our special triangles, we see that \(T=\frac{\pi}{3}\text{.}\)

Example 4.1.6 3.3.2 revisited.

Back in Section 3.3 we encountered a simple differential equation, namely equation 3.3.1. We were able to solve this equation by guessing the answer and then checking it carefully. We can derive the solution more systematically by using antiderivatives.

Recall equation 3.3.1:

\begin{align*} \dfrac{dQ}{dt} &= -k Q \end{align*}

where \(Q(t)\) is the amount of radioactive material at time \(t\) and we assume \(Q(t) \gt 0\text{.}\) Take this equation and divide both sides by \(Q(t)\) to get

\begin{align*} \frac{1}{Q(t)} \dfrac{dQ}{dt} &= -k \end{align*}

At this point we should ¹think that the left-hand side is familiar. Now is a good moment to look back at logarithmic differentiation in Section 2.10.

The left-hand side is just the derivative of \(\log Q(t)\text{:}\)

\begin{align*} \dfrac{d}{dt}\left( \log Q(t) \right) &= \frac{1}{Q(t)} \dfrac{dQ}{dt}\\ &= -k \end{align*}

So to solve this equation, we are really being asked to find all functions \(\log Q(t)\) having derivative \(-k\text{.}\) That is, we need to find all antiderivatives of \(-k\text{.}\) Of course that is just \(-kt + c\text{.}\) Hence we must have

\begin{align*} \log Q(t) &= -kt +c \end{align*}

and then taking the exponential of both sides gives

\begin{align*} Q(t) &= e^{-kt+c} = e^c \cdot e^{-kt} = C e^{-kt} \end{align*}

where \(C = e^c\text{.}\) This is precisely Theorem 3.3.2.

Example 4.1.7 Volume of a cone.

We know (especially if one has revised the material in the appendix and Appendix B.5.2 in particular) that the volume of a right-circular cone is

\begin{align*} V &= \frac{\pi}{3} r^2h \end{align*}

where \(h\) is the height of the cone and \(r\) is the radius of its base. Now, the derivation of this formula given in Appendix B.5.2 is not too simple. We present an alternate proof here that uses antiderivatives.

Consider cutting off a portion of the cone so that its new height is \(x\) (rather than \(h\)). Call the volume of the resulting smaller cone \(V(x)\text{.}\) We are going to determine \(V(x)\) for all \(x\ge 0\text{,}\) including \(x=h\text{,}\) by first evaluating \(V'(x)\) and \(V(0)\) (which is obviously \(0\)).

Call the radius of the base of the new smaller cone \(y\) (rather than \(r\)). By similar triangles we know that

\begin{align*} \frac{r}{h} &= \frac{y}{x}. \end{align*}

Now keep \(x\) and \(y\) fixed and consider cutting off a little more of the cone so its height is \(X\text{.}\) When we do so, the radius of the base changes from \(y\) to \(Y\) and again by similar triangles we know that

\begin{align*} \frac{Y}{X} &= \frac{y}{x} = \frac{r}{h} \end{align*}

The change in volume is then

\begin{gather*} V(x) - V(X) \end{gather*}

Of course if we knew the formula for the volume of a cone, then we could compute the above exactly. However, even without knowing the volume of a cone, it is easy to derive upper and lower bounds on this quantity. The piece removed has bottom radius \(y\) and top radius \(Y\text{.}\) Hence its volume is bounded above and below by the cylinders of height \(x-X\) and with radius \(y\) and \(Y\) respectively. Hence

\begin{align*} \pi Y^2 (x-X) & \leq V(x)-V(X) \leq \pi y^2 (x-X) \end{align*}

since the volume of a cylinder is just the area of its base times its height. Now massage this expression a little

\begin{align*} \pi Y^2 & \leq \frac{V(x)-V(X)}{x-X} \leq \pi y^2 \end{align*}

The middle term now looks like a derivative; all we need to do is take the limit as \(X \to x\text{:}\)

\begin{align*} \lim_{X\to x} \pi Y^2 & \leq \lim_{X\to x} \frac{V(x)-V(X)}{x-X} \leq \lim_{X\to x} \pi y^2 \end{align*}

The rightmost term is independent of \(X\) and so is just \(\pi y^2\text{.}\) In the leftmost term, as \(X \to x\text{,}\) we must have that \(Y \to y\text{.}\) Hence the leftmost term is just \(\pi y^2\text{.}\) Then by the squeeze theorem (Theorem 1.4.18) we know that

\begin{align*} \dfrac{dV}{dx} = \lim_{X\to x} \frac{V(x)-V(X)}{x-X} &= \pi y^2. \end{align*}

But we know that

\begin{align*} y &= \frac{r}{h} \cdot x \end{align*}

so

\begin{align*} \dfrac{dV}{dx} &= \pi \left( \frac{r}{h} \right)^2 x^2 \end{align*}

Now we can antidifferentiate to get back to \(V\text{:}\)

\begin{align*} V(x) &= \frac{\pi}{3} \left( \frac{r}{h} \right)^2 x^3 + c \end{align*}

To determine \(c\) notice that when \(x=0\) the volume of the cone is just zero and so \(c=0\text{.}\) Thus

\begin{align*} V(x) &= \frac{\pi}{3} \left( \frac{r}{h} \right)^2 x^3 \end{align*}

and so when \(x=h\) we are left with

\begin{align*} V(h) &= \frac{\pi}{3} \left( \frac{r}{h} \right)^2 h^3 = \frac{\pi}{3} r^2 h \end{align*}

as required.