Motivating Questions
If we have a function given by a formula, is there a way to write down the antiderivative?
How can we find the area between two curves?
If we have a function given by a formula, is there a way to write down the antiderivative?
How can we find the area between two curves?
The general problem of finding an antiderivative is difficult. In part, this is due to the fact that we are trying to undo the process of differentiating, and the undoing is much more difficult than the doing. For example, while it is evident that an antiderivative of \(f(x) = \sin(x)\) is \(F(x) = -\cos(x)\) and that an antiderivative of \(g(x) = x^2\) is \(G(x) = \frac{1}{3} x^3\text{,}\) combinations of \(f\) and \(g\) can be far more complicated. Consider the functions
In the above cases, the antiderivative might not be obvious and in some cases, for example, \(\sin(x^2)\) the antiderivative cannot be written down using elementary functions. This warrants the question, how do we find an antiderivative from a function given by a formula in general? We will begin to answer that question in this chapter.
What is involved in trying to find an antiderivative for each? From our experience with derivative rules, we know that derivatives of sums and constant multiples of basic functions are simple to execute, but derivatives involving products, quotients, and composites of familiar functions are more complicated. Therefore, it stands to reason that antidifferentiating products, quotients, and composites of basic functions may be even more challenging. We defer our study of all but the most elementary antiderivatives to later in the text.
Before proceeding any further it will be useful to state two properties.
Let \(f\) and \(g\) be functions that have an antiderivative and \(c\) a constant then
In other words, the antiderivative of a sum is the sum of the antiderivatives.
We do note that whenever we know the derivative of a function, we have a function-derivative pair, so we also know the antiderivative of a function. For instance, since we know that
we also know that \(F(x) = -\cos(x)\) is an antiderivative of \(f(x) = \sin(x)\text{.}\) \(F\) and \(f\) together form a function-derivative pair. Clearly, every basic derivative rule leads us to such a pair, and thus to a known antiderivative.
In Example5.14, we will construct a list of the basic antiderivatives we know at this time. Those rules will help us antidifferentiate sums and constant multiples of basic functions. For example, since \(-\cos(x)\) is an antiderivative of \(\sin(x)\) and \(\frac{1}{3}x^3\) is an antiderivative of \(x^2\text{,}\) it follows that
is an antiderivative of \(f(x) = 5\sin(x) - 4x^2\text{.}\)
Finally, before proceeding to build a list of common functions whose antiderivatives we know, we recall that each function has more than one antiderivative. Because the derivative of any constant is zero, we may add a constant of our choice to any antiderivative. For instance, we know that \(G(x) = \frac{1}{3}x^3\) is an antiderivative of \(g(x) = x^2\text{.}\) But we could also have chosen \(G(x) = \frac{1}{3}x^3 + 7\text{,}\) since in this case as well, \(G'(x) = x^2\text{.}\) If \(g(x) = x^2\text{,}\) we say that the general antiderivative of \(g\) is
where \(C\) represents an arbitrary real number constant. Regardless of the formula for \(g\text{,}\) including \(+C\) in the formula for its antiderivative \(G\) results in the most general possible antiderivative.
In the following example, we work to build a list of basic functions whose antiderivatives we already know.
Use your knowledge of derivatives of basic functions to complete the above table of antiderivatives. For each entry, your task is to find a function \(F\) whose derivative is the given function \(f\text{.}\)
given function, \(f(x)\) | antiderivative, \(F(x)\) |
\(k\text{,}\) (\(k\) is constant) | |
\(x^n\text{,}\) \(n \ne -1\) | |
\(\frac{1}{x}\) | |
\(\sin(x)\) | |
\(\cos(x)\) | |
\(\sec(x) \tan(x)\) | |
\(\csc(x) \cot(x)\) | |
\(\sec^2 (x)\) | |
\(\csc^2 (x)\) | |
\(e^x\) | |
\(a^x\) \((a \gt 1)\) | |
\(\frac{1}{1+x^2}\) | |
\(\frac{1}{\sqrt{1-x^2}}\) |
The antiderivative of \(\frac{1}{x}\) is \(\ln|x|+C\text{.}\) The absolute value sign comes about by considering two cases separately. For \(x>0\text{,}\) we have that
On the other hand, for \(x< 0\text{,}\) \(\ln|x|=\ln(-x)\text{,}\) so
You might start by constructing a list of all the basic functions whose derivative you know.
given function, \(f(x)\) | antiderivative, \(F(x)\) |
\(k\text{,}\) (\(k \ne 0\)) | \(kx+C\) |
\(x^n\text{,}\) \(n \ne -1\) | \(\frac{1}{n+1}x^{n+1}+C\) |
\(\frac{1}{x}\) | \(\ln|x|+C\) |
\(\sin(x)\) | \(-\cos(x)+C\) |
\(\cos(x)\) | \(\sin(x)+C\) |
\(\sec(x) \tan(x)\) | \(\sec(x)+C\) |
\(\csc(x) \cot(x)\) | \(-\csc(x)+C\) |
\(\sec^2 (x)\) | \(\tan(x)+C\) |
\(\csc^2 (x)\) | \(\cot(x)+C\) |
\(e^x\) | \(e^x+C\) |
\(a^x\) \((a \gt 1)\) | \(\frac{1}{\ln(a)} a^x+C\) |
\(\frac{1}{1+x^2}\) | \(\arctan(x)+C\) |
\(\frac{1}{\sqrt{1-x^2}}\) | \(\arcsin(x)+C\) |
given function, \(f(x)\) | antiderivative, \(F(x)\) |
\(k\text{,}\) (\(k \ne 0\)) | \(kx+C\) |
\(x^n\text{,}\) \(n \ne -1\) | \(\frac{1}{n+1}x^{n+1}+C\) |
\(\frac{1}{x}\) | \(\ln|x|+C\) |
\(\sin(x)\) | \(-\cos(x)+C\) |
\(\cos(x)\) | \(\sin(x)+C\) |
\(\sec(x) \tan(x)\) | \(\sec(x)+C\) |
\(\csc(x) \cot(x)\) | \(-\csc(x)+C\) |
\(\sec^2 (x)\) | \(\tan(x)+C\) |
\(\csc^2 (x)\) | \(\cot(x)+C\) |
\(e^x\) | \(e^x+C\) |
\(a^x\) \((a \gt 1)\) | \(\frac{1}{\ln(a)} a^x+C\) |
\(\frac{1}{1+x^2}\) | \(\arctan(x)+C\) |
\(\frac{1}{\sqrt{1-x^2}}\) | \(\arcsin(x)+C\) |
Now that we have some basic antiderivatives in hand we can proceed to solve problems involving the definite integral when the integrand is an expression. Specifically, we seek to use antiderivatives and the Fundamental Theorem of Calculus to evaluate expressions like \(\displaystyle \int_0^1 \left(x^3 - x - e^x + 2\right) \,dx\text{.}\)
Since our current interest in antiderivatives is so that we can evaluate definite integrals by the Fundamental Theorem of Calculus the constant \(C\) is actually irrelevant, and mathematicians genrally omit it in these types of problems. To see why, consider the definite integral
For the integrand \(g(x) = x^2\text{,}\) suppose we find and use the general antiderivative \(G(x) = \frac{1}{3} x^3 + C\text{.}\) Then, by the Fundamental Theorem of Calculus,
Observe that the \(C\)-values appear as opposites in the evaluation of the integral and thus do not affect the definite integral's value.
Use the FTC and the results in Table5.16 to evaluate the three given definite integrals.
\(\displaystyle \int_0^1 \left(x^3 - x - e^x + 2\right) \,dx\)
\(\displaystyle \int_0^{\frac{\pi}{3}} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt\)
\(\displaystyle \int_0^1 (\sqrt{x}-x^2) \, dx\)
Be sure to recall the sum and constant multiple rules, which work not only for differentiating, but also for antidifferentiating.
\(\int_0^1 \left(x^3 - x - e^x + 2\right) \,dx = \frac{11}{4} - 3\text{.}\)
\(\int_0^{\frac{\pi}{3}} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt = 1 - \sqrt{3} - \frac{\pi^2}{3}\text{.}\)
\(\int_0^1 (\sqrt{x} - x^2) \, dx = \frac{1}{3}\text{.}\)
By standard antiderivative rules and the FTC,
Calculating the needed antiderivative and applying the FTC,
Noting that \(\frac{d}{dx}[\frac{2}{3} x^{\frac{3}{2}}] = x^{\frac{1}{2}}\text{,}\) we find that
Integration allows us to find the area between a curve and the \(x\)-axis. In fact, by visually inspecting the graphs of two different curves we can develop a technique for finding the area between two curves. Consider the graphs of the curves \(y=x^2+1\) and \(y=x+3\) displayed in Figure5.18. We might ask if there is a way to find the area enclosed by these two curves as is colored in the figure.
In fact, using the knowledge we have about integration, we can figure out this quantity exactly.
What is the exact area enclosed by the curves given by the functions \(f(x)=x^2+1\) and \(g(x)=x+3\text{?}\)
It might help to answer the following questions:
\(\frac{9}{2}\)
The first step in solving this problem is to determine where the curves intersect. We can find the point of intersection by setting the two curves equal to each other.
Therefore, the curves intersect at two \(x\)-values, \(x=-1,2\text{.}\) This means were are concerned with the interval \([-2,2]\text{.}\)
We next note that on the interval in question, \([-2,2]\text{,}\) the curve \(g(x)=x+3\) is above the curve \(f(x)=x^2+1\text{.}\) We also note that \(\int_{-1}^{2}(x+3)dx\) would give the area shown in Figure5.20 and \(\int_{-1}^{2}(x^2+1)dx\) Figure5.21.
By overlaying Figure5.20 and Figure5.21 as in Figure5.22 you can see that Figure5.18 should be given by subtracting \(\int_{-1}^{2}(x^2+1)dx\) from \(\int_{-1}^{2}(x+3)dx\text{.}\)
Therefore, the area between the curves \(y=x^2+1\) and \(y=x+3\) is given by the following quantity:
The instantaneous velocity (in meters per minute) of a moving object is given by the function \(v\) as pictured in Figure5.23. Assume that on the interval \(0 \le t \le 4\text{,}\) \(v(t)\) is given by \(v(t) = -\frac{1}{4}t^3 + \frac{3}{2}t^2 + 1\text{,}\) and that on every other interval \(v\) is piecewise linear, as shown.
Determine the exact distance traveled by the object on the time interval \(0 \le t \le 4\text{.}\)
What is the object's average velocity on \([12,24]\text{?}\)
At what time is the object's acceleration greatest?
Suppose that the velocity of the object is increased by a constant value \(c\) for all values of \(t\text{.}\) What value of \(c\) will make the object's total distance traveled on \([12,24]\) be 210 meters?
A function \(f\) is given piecewise by the formula
Determine the exact value of the net signed area enclosed by \(f\) and the \(x\)-axis on the interval \([2,5]\text{.}\)
Compute the exact average value of \(f\) on \([0,5]\text{.}\)
Find a formula for a function \(g\) on \(5 \le x \le 7\) so that if we extend the above definition of \(f\) so that \(f(x) = g(x)\) if \(5 \le x \le 7\text{,}\) it follows that \(\int_0^7 f(x) \, dx = 0\text{.}\)