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Working with Taylor Series

Write the terms of the binomial series.

The Binomial Series

Our first goal in this section is to determine the Maclaurin series for the function \(f(x)={(1+x)}^{r}\) for all real numbers \(r.\) The Maclaurin series for this function is known as the binomial series. We begin by considering the simplest case: \(r\) is a nonnegative integer. We recall that, for \(r=0,1,2,3,4,f(x)={(1+x)}^{r}\) can be written as

\[\begin{array}{l} \\ f(x)={(1+x)}^{0}=1, \\ f(x)={(1+x)}^{1}=1+x, \\ f(x)={(1+x)}^{2}=1+2x+{x}^{2}, \\ f(x)={(1+x)}^{3}=1+3x+3{x}^{2}+{x}^{3}, \\ f(x)={(1+x)}^{4}=1+4x+6{x}^{2}+4{x}^{3}+{x}^{4}.\end{array}\]

The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer \(r,\) the binomial coefficient of \({x}^{n}\) in the binomial expansion of \({(1+x)}^{r}\) is given by

\[(\begin{array}{l}r \\ n\end{array})=\frac{r\text{!}}{n\text{!}(r-n)\text{!}}\]

and

\[\begin{array}{ll}f(x) & ={(1+x)}^{r} \\ & =(\begin{array}{l}r \\ 0\end{array})1+(\begin{array}{l}r \\ 1\end{array})x+(\begin{array}{l}r \\ 2\end{array}){x}^{2}+(\begin{array}{l}r \\ 3\end{array}){x}^{3}+\text{\cdots }+(\begin{array}{l}r \\ r-1\end{array}){x}^{r-1}+(\begin{array}{l}r \\ r\end{array}){x}^{r} \\ & =\sum _{n=0}^{r}(\begin{array}{l}r \\ n\end{array}){x}^{n}.\end{array}\]

For example, using this formula for \(r=5,\) we see that

\[\begin{array}{ll}f(x) & ={(1+x)}^{5} \\ & =(\begin{array}{l}5 \\ 0\end{array})1+(\begin{array}{l}5 \\ 1\end{array})x+(\begin{array}{l}5 \\ 2\end{array}){x}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{3}+(\begin{array}{l}5 \\ 4\end{array}){x}^{4}+(\begin{array}{l}5 \\ 5\end{array}){x}^{5} \\ & =\frac{5\text{!}}{0\text{!}5\text{!}}1+\frac{5\text{!}}{1\text{!}4\text{!}}x+\frac{5\text{!}}{2\text{!}3\text{!}}{x}^{2}+\frac{5\text{!}}{3\text{!}2\text{!}}{x}^{3}+\frac{5\text{!}}{4\text{!}1\text{!}}{x}^{4}+\frac{5\text{!}}{5\text{!}0\text{!}}{x}^{5} \\ & =1+5x+10{x}^{2}+10{x}^{3}+5{x}^{4}+{x}^{5}.\end{array}\]

We now consider the case when the exponent \(r\) is any real number, not necessarily a nonnegative integer. If \(r\) is not a nonnegative integer, then \(f(x)={(1+x)}^{r}\) cannot be written as a finite polynomial. However, we can find a power series for \(f.\) Specifically, we look for the Maclaurin series for \(f.\) To do this, we find the derivatives of \(f\) and evaluate them at \(x=0.\)

We conclude that the coefficients in the binomial series are given by

\[\frac{{f}^{(n)}(0)}{n\text{!}}=\frac{r(r-1)(r-2)\text{\cdots }(r-n+1)}{n\text{!}}.\]\[(\begin{array}{l}r \\ n\end{array})=\frac{r(r-1)(r-2)\text{\cdots }(r-n+1)}{n\text{!}}.\]\[\underset{n\to \infty }{\text{lim}}\frac{|{a}_{n+1}|}{|{a}_{n}|}=|x|<1\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Common Functions Expressed as Taylor Series

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form \(f(x)={(1+x)}^{r}.\) In , we summarize the results of these series. We remark that the convergence of the Maclaurin series for \(f(x)=\text{ln}\ (1+x)\) at the endpoint \(x=1\) and the Maclaurin series for \(f(x)={\text{tan}}^{-1}x\) at the endpoints \(x=1\) and \(x=-1\) relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Function	Maclaurin Series	Interval of Convergence
\(f(x)=\frac{1}{1-x}\)	\(\sum _{n=0}^{\infty }{x}^{n}\)	\(-1
\(f(x)={e}^{x}\)	\(\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}\)	\(\text{-}\infty
\(f(x)=\text{sin}\ x\)	\(\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2n+1}}{(2n+1)\text{!}}\)	\(\text{-}\infty
\(f(x)=\text{cos}\ x\)	\(\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2n}}{(2n)\text{!}}\)	\(\text{-}\infty
\(f(x)=\text{ln}\ (1+x)\)	\(\sum _{n=1}^{\infty }{(-1)}^{n+1}\frac{{x}^{n}}{n}\)	\(-1
\(f(x)={\text{tan}}^{-1}x\)	\(\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2n+1}}{2n+1}\)	\(-1\le x\le 1\)
\(f(x)={(1+x)}^{r}\)	\(\sum _{n=0}^{\infty }(\begin{array}{l}r \\ n\end{array}){x}^{n}\)	\(-1

Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in , to create Maclaurin series for other functions.

We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In , we differentiate the binomial series for \(\sqrt{1+x}\) term by term to find the binomial series for \(\frac{1}{\sqrt{1+x}}.\) Note that we could construct the binomial series for \(\frac{1}{\sqrt{1+x}}\) directly from the definition, but differentiating the binomial series for \(\sqrt{1+x}\) is an easier calculation.

Example

Try it.

Use the binomial series for \(\sqrt{1+x}\) to find the binomial series for \(\frac{1}{\sqrt{1+x}}.\)

Solution

The two functions are related by

\[\frac{d}{dx}\sqrt{1+x}=\frac{1}{2\sqrt{1+x}},\]

so the binomial series for \(\frac{1}{\sqrt{1+x}}\) is given by

\[\begin{array}{ll}\frac{1}{\sqrt{1+x}} & =2\frac{d}{dx}\sqrt{1+x} \\ & =1+\sum _{n=1}^{\infty }\frac{{(-1)}^{n}}{n\text{!}}\ \frac{1\cdot 3\cdot 5\text{\cdots }(2n-1)}{{2}^{n}}{x}^{n}.\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Solving Differential Equations with Power Series

Consider the differential equation

\[{y}^{'}(x)=y.\]

Recall that this is a first-order separable equation and its solution is \(y=C{e}^{x}.\) This equation is easily solved using techniques discussed earlier in the text. For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form \(y=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}\) and determine what the coefficients would need to be. In the next example, we consider an initial-value problem involving \({y}^{'}=y\) to illustrate the technique.

Example

Try it.

Use power series to solve the initial-value problem

\[{y}^{'}=y,\ y(0)=3.\]

Solution

Suppose that there exists a power series solution

\[y(x)=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\text{\cdots }.\]

Differentiating this series term by term, we obtain

\[{y}^{'}={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\text{\cdots }.\]

If y satisfies the differential equation, then

\[{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{\cdots }={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{3}{x}^{3}+\text{\cdots }.\]

Using on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,

\[\begin{array}{l}{c}_{0}={c}_{1}, \\ {c}_{1}=2{c}_{2}, \\ {c}_{2}=3{c}_{3}, \\ {c}_{3}=4{c}_{4}, \\ \text{⋮}.\end{array}\]

Using the initial condition \(y(0)=3\) combined with the power series representation

\[y(x)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{\cdots },\]

we find that \({c}_{0}=3.\) We are now ready to solve for the rest of the coefficients. Using the fact that \({c}_{0}=3,\) we have

\[\begin{array}{l} \\ \\ {c}_{1}={c}_{0}=3=\frac{3}{1\text{!}}, \\ {c}_{2}=\frac{{c}_{1}}{2}=\frac{3}{2}=\frac{3}{2\text{!}}, \\ {c}_{3}=\frac{{c}_{2}}{3}=\frac{3}{3\cdot 2}=\frac{3}{3\text{!}}, \\ {c}_{4}=\frac{{c}_{3}}{4}=\frac{3}{4\cdot 3\cdot 2}=\frac{3}{4\text{!}}.\end{array}\]

Therefore,

\[\begin{array}{ll}y & =3[1+\frac{1}{1\text{!}}x+\frac{1}{2\text{!}}{x}^{2}+\frac{1}{3\text{!}}{x}^{3}+\frac{1}{4\text{!}}{x}^{4}+\text{\cdots }] \\ & =3\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}.\end{array}\]

You might recognize

\[\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}\]

as the Taylor series for \({e}^{x}.\) Therefore, the solution is \(y=3{e}^{x}.\)

We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation

Condensed — the full section is in OpenStax Calculus Volume 2.

Evaluating Nonelementary Integrals

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

One integral that arises often in applications in probability theory is \(\int {e}^{\text{-}{x}^{2}}dx.\) Unfortunately, the antiderivative of the integrand \({e}^{\text{-}{x}^{2}}\) is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with noncomplicated function. For example, the function \(f(x)=\sqrt{{x}^{2}-3x}+{e}^{{x}^{3}}-\text{sin}(5x+4)\) is an elementary function, although not a particularly simple-looking function. Any integral of the form \(\int f(x)\ dx\) where the antiderivative of \(f\) cannot be written as an elementary function is considered a nonelementary integral.

Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering \(\int {e}^{\text{-}{x}^{2}}dx.\)

Example

Try it.

Express \(\int {e}^{\text{-}{x}^{2}}dx\) as an infinite series.
Evaluate \({\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx\) to within an error of \(0.01.\)

Solution

The Maclaurin series for \({e}^{\text{-}{x}^{2}}\) is given by
\[\begin{array}{ll}{e}^{\text{-}{x}^{2}} & =\sum _{n=0}^{\infty }\frac{{(\text{-}{x}^{2})}^{n}}{n\text{!}} \\ & =1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{\cdots } \\ & =\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2n}}{n\text{!}}.\end{array}\]
Therefore,
\[\begin{array}{ll}\int {e}^{\text{-}{x}^{2}}dx & =\int (1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{\cdots })\ dx \\ & =C+x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5\cdot 2\text{!}}-\frac{{x}^{7}}{7\cdot 3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{x}^{2n+1}}{(2n+1)n\text{!}}+\text{\cdots }.\end{array}\]
Using the result from part a. we have
\[{\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-\text{\cdots }.\]
The sum of the first four terms is approximately \(0.74.\) By the alternating series test, this estimate is accurate to within an error of less than \(\frac{1}{216}\approx 0.0046296<0.01.\)

(See .)

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

The binomial series is the Maclaurin series for \(f(x)={(1+x)}^{r}.\) It converges for \(|x|<1.\)
Taylor series for functions can often be derived by algebraic operations with a known Taylor series or by differentiating or integrating a known Taylor series.
Power series can be used to solve differential equations.
Taylor series can be used to help approximate integrals that cannot be evaluated by other means.

Working with Taylor Series

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

In the following exercises, use the substitution \({(b+x)}^{r}={(b+a)}^{r}{(1+\frac{x-a}{b+a})}^{r}\) in the binomial expansion to find the Taylor series of each function with the given center.

In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most \(1\text{/}1000.\)

In the following exercises, use the binomial approximation \(\sqrt{1-x}\approx 1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{128}-\frac{7{x}^{5}}{256}\) for \(|x|<1\) to approximate each number. Compare this value to the value given by a scientific calculator.

In the following exercises, use the expansion \({(1+x)}^{1\text{/}3}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4}+\text{\cdots }\) to write the first five terms (not necessarily a quartic polynomial) of each expression.

In the following exercises, find the Maclaurin series of each function.

In the following exercises, find the Maclaurin series of \(F(x)={\int }_{0}^{x}f(t)\ dt\) by integrating the Maclaurin series of \(f\) term by term. If \(f\) is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than \(\frac{1}{100}.\)
Compare the accuracy of the polynomial integral estimate with the remainder estimate.

Condensed — the full section is in OpenStax Calculus Volume 2.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Find the binomial series for \(f(x)=\sqrt{1+x}.\)
2. Use the third-order Maclaurin polynomial \({p}_{3}(x)\) to estimate \(\sqrt{1.5}.\) Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of \(f\) and \({p}_{3}.\)
Առաջարկել պատասխանը
1. Here \(r=\frac{1}{2}.\) Using the definition for the binomial series, we obtain
  \[\begin{array}{ll}\sqrt{1+x} & =1+\frac{1}{2}x+\frac{(1\text{/}2)(\text{-}1\text{/}2)}{2\text{!}}{x}^{2}+\frac{(1\text{/}2)(\text{-}1\text{/}2)(\text{-}3\text{/}2)}{3\text{!}}{x}^{3}+\text{\cdots } \\ & =1+\frac{1}{2}x-\frac{1}{2\text{!}}\ \frac{1}{{2}^{2}}{x}^{2}+\frac{1}{3\text{!}}\ \frac{1\cdot 3}{{2}^{3}}{x}^{3}-\text{\cdots }+\frac{{(-1)}^{n+1}}{n\text{!}}\ \frac{1\cdot 3\cdot 5\text{\cdots }(2n-3)}{{2}^{n}}{x}^{n}+\text{\cdots } \\ & =1+\sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}}{n\text{!}}\ \frac{1\cdot 3\cdot 5\text{\cdots }(2n-3)}{{2}^{n}}{x}^{n}.\end{array}\]
2. From the result in part a. the third-order Maclaurin polynomial is
  \[{p}_{3}(x)=1+\frac{1}{2}x-\frac{1}{8}{x}^{2}+\frac{1}{16}{x}^{3}.\]
  Therefore,
  \[\begin{array}{ll}\sqrt{1.5} & =\sqrt{1+0.5} \\ & \approx 1+\frac{1}{2}(0.5)-\frac{1}{8}{(0.5)}^{2}+\frac{1}{16}{(0.5)}^{3} \\ & \approx 1.2266.\end{array}\]
  From Taylor’s theorem, the error satisfies
  \[{R}_{3}(0.5)=\frac{{f}^{(4)}(c)}{4\text{!}}{(0.5)}^{4}\]
  for some \(c\) between \(0\) and \(0.5.\) Since \({f}^{(4)}(x)=-\frac{15}{{2}^{4}{(1+x)}^{7\text{/}2}},\) and the maximum value of \(|{f}^{(4)}(x)|\) on the interval \((0,0.5)\) occurs at \(x=0,\) we have
  \[|{R}_{3}(0.5)|\le \frac{15}{4\text{!}{2}^{4}}{(0.5)}^{4}\approx 0.00244.\]
  The function and the Maclaurin polynomial \({p}_{3}\) are graphed in .
Find the binomial series for \(f(x)=\frac{1}{{(1+x)}^{2}}.\)

Առաջարկել պատասխանը

\(\sum _{n=0}^{\infty }{(-1)}^{n}(n+1){x}^{n}\)
Find the Maclaurin series of each of the following functions by using one of the series listed in .
1. \(f(x)=\text{cos}\sqrt{x}\)
2. \(f(x)=\text{sinh}\ x\)
Առաջարկել պատասխանը
1. Using the Maclaurin series for \(\text{cos}\ x\) we find that the Maclaurin series for \(\text{cos}\sqrt{x}\) is given by
  \[\begin{array}{ll}\sum _{n=0}^{\infty }\frac{{(-1)}^{n}{(\sqrt{x})}^{2n}}{(2n)\text{!}} & =\sum _{n=0}^{\infty }\frac{{(-1)}^{n}{x}^{n}}{(2n)\text{!}} \\ & =1-\frac{x}{2\text{!}}+\frac{{x}^{2}}{4\text{!}}-\frac{{x}^{3}}{6\text{!}}+\frac{{x}^{4}}{8\text{!}}-\text{\cdots }.\end{array}\]
  This series converges to \(\text{cos}\sqrt{x}\) for all \(x\) in the domain of \(\text{cos}\sqrt{x};\) that is, for all \(x\ge 0.\)
2. To find the Maclaurin series for \(\text{sinh}\ x,\) we use the fact that
  \[\text{sinh}\ x=\frac{{e}^{x}-{e}^{\text{-}x}}{2}.\]
  Using the Maclaurin series for \({e}^{x},\) we see that the \(n\text{th}\) term in the Maclaurin series for \(\text{sinh}\ x\) is given by
  \[\frac{{x}^{n}}{n\text{!}}-\frac{{(\text{-}x)}^{n}}{n\text{!}}.\]
  For \(n\) even, this term is zero. For \(n\) odd, this term is \(\frac{2{x}^{n}}{n\text{!}}.\) Therefore, the Maclaurin series for \(\text{sinh}\ x\) has only odd-order terms and is given by
  \[\sum _{n=0}^{\infty }\frac{{x}^{2n+1}}{(2n+1)\text{!}}=x+\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}}+\text{\cdots }.\]
Find the Maclaurin series for \(\text{sin}({x}^{2}).\)

Առաջարկել պատասխանը

\(\sum _{n=0}^{\infty }\frac{{(-1)}^{n}{x}^{4n+2}}{(2n+1)\text{!}}\)
Use the binomial series for \(\sqrt{1+x}\) to find the binomial series for \(\frac{1}{\sqrt{1+x}}.\)

Առաջարկել պատասխանը

The two functions are related by
\[\frac{d}{dx}\sqrt{1+x}=\frac{1}{2\sqrt{1+x}},\]
so the binomial series for \(\frac{1}{\sqrt{1+x}}\) is given by
\[\begin{array}{ll}\frac{1}{\sqrt{1+x}} & =2\frac{d}{dx}\sqrt{1+x} \\ & =1+\sum _{n=1}^{\infty }\frac{{(-1)}^{n}}{n\text{!}}\ \frac{1\cdot 3\cdot 5\text{\cdots }(2n-1)}{{2}^{n}}{x}^{n}.\end{array}\]
Find the binomial series for \(f(x)=\frac{1}{{(1+x)}^{3\text{/}2}}\)

Առաջարկել պատասխանը

\(\sum _{n=1}^{\infty }\frac{{(-1)}^{n}}{n\text{!}}\ \frac{1\cdot 3\cdot 5\text{\cdots }(2n-1)}{{2}^{n}}{x}^{n}\)
Use power series to solve the initial-value problem
\[{y}^{'}=y,\ y(0)=3.\]

Առաջարկել պատասխանը

Suppose that there exists a power series solution
\[y(x)=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\text{\cdots }.\]
Differentiating this series term by term, we obtain
\[{y}^{'}={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\text{\cdots }.\]
If y satisfies the differential equation, then
\[{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{\cdots }={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{3}{x}^{3}+\text{\cdots }.\]
Using on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,
\[\begin{array}{l}{c}_{0}={c}_{1}, \\ {c}_{1}=2{c}_{2}, \\ {c}_{2}=3{c}_{3}, \\ {c}_{3}=4{c}_{4}, \\ \text{⋮}.\end{array}\]
Using the initial condition \(y(0)=3\) combined with the power series representation
\[y(x)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{\cdots },\]
we find that \({c}_{0}=3.\) We are now ready to solve for the rest of the coefficients. Using the fact that \({c}_{0}=3,\) we have
\[\begin{array}{l} \\ \\ {c}_{1}={c}_{0}=3=\frac{3}{1\text{!}}, \\ {c}_{2}=\frac{{c}_{1}}{2}=\frac{3}{2}=\frac{3}{2\text{!}}, \\ {c}_{3}=\frac{{c}_{2}}{3}=\frac{3}{3\cdot 2}=\frac{3}{3\text{!}}, \\ {c}_{4}=\frac{{c}_{3}}{4}=\frac{3}{4\cdot 3\cdot 2}=\frac{3}{4\text{!}}.\end{array}\]
Therefore,
\[\begin{array}{ll}y & =3[1+\frac{1}{1\text{!}}x+\frac{1}{2\text{!}}{x}^{2}+\frac{1}{3\text{!}}{x}^{3}+\frac{1}{4\text{!}}{x}^{4}+\text{\cdots }] \\ & =3\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}.\end{array}\]
You might recognize
\[\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}\]
as the Taylor series for \({e}^{x}.\) Therefore, the solution is \(y=3{e}^{x}.\)
Use power series to solve \({y}^{'}=2y,\ y(0)=5.\)

Առաջարկել պատասխանը

\(y=5{e}^{2x}\)
Use power series to solve
\[{y}^{″}-xy=0\]
with the initial conditions \(y(0)=a\) and \(y'(0)=b.\)

Առաջարկել պատասխանը

We look for a solution of the form
\[y=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\text{\cdots }.\]
Differentiating this function term by term, we obtain
\[\begin{array}{lll}{y}^{'} & = & {c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\text{\cdots }, \\ {y}^{″} & = & 2\cdot 1{c}_{2}+3\cdot 2{c}_{3}x+4\cdot 3{c}_{4}{x}^{2}+\text{\cdots }.\end{array}\]
If y satisfies the equation \({y}^{″}=xy,\) then
\[2\cdot 1{c}_{2}+3\cdot 2{c}_{3}x+4\cdot 3{c}_{4}{x}^{2}+\text{\cdots }=x({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{\cdots }).\]
Using on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,
\[\begin{array}{l}2\cdot 1{c}_{2}=0, \\ 3\cdot 2{c}_{3}={c}_{0}, \\ 4\cdot 3{c}_{4}={c}_{1}, \\ 5\cdot 4{c}_{5}={c}_{2}, \\ \text{⋮}.\end{array}\]
More generally, for \(n\ge 3,\) we have \(n\cdot (n-1){c}_{n}={c}_{n-3}.\) In fact, all coefficients can be written in terms of \({c}_{0}\) and \({c}_{1}.\) To see this, first note that \({c}_{2}=0.\) Then
\[\begin{array}{l} \\ \\ {c}_{3}=\frac{{c}_{0}}{3\cdot 2}, \\ {c}_{4}=\frac{{c}_{1}}{4\cdot 3}.\end{array}\]
For \({c}_{5},{c}_{6},{c}_{7},\) we see that
\[\begin{array}{l} \\ \\ {c}_{5}=\frac{{c}_{2}}{5\cdot 4}=0, \\ {c}_{6}=\frac{{c}_{3}}{6\cdot 5}=\frac{{c}_{0}}{6\cdot 5\cdot 3\cdot 2}, \\ {c}_{7}=\frac{{c}_{4}}{7\cdot 6}=\frac{{c}_{1}}{7\cdot 6\cdot 4\cdot 3}.\end{array}\]
Therefore, the series solution of the differential equation is given by
\[y={c}_{0}+{c}_{1}x+0\cdot {x}^{2}+\frac{{c}_{0}}{3\cdot 2}{x}^{3}+\frac{{c}_{1}}{4\cdot 3}{x}^{4}+0\cdot {x}^{5}+\frac{{c}_{0}}{6\cdot 5\cdot 3\cdot 2}{x}^{6}+\frac{{c}_{1}}{7\cdot 6\cdot 4\cdot 3}{x}^{7}+\text{\cdots }.\]
The initial condition \(y(0)=a\) implies \({c}_{0}=a.\) Differentiating this series term by term and using the fact that \({y}^{'}(0)=b,\) we conclude that \({c}_{1}=b.\) Therefore, the solution of this initial-value problem is
\[y=a(1+\frac{{x}^{3}}{3\cdot 2}+\frac{{x}^{6}}{6\cdot 5\cdot 3\cdot 2}+\text{\cdots })+b(x+\frac{{x}^{4}}{4\cdot 3}+\frac{{x}^{7}}{7\cdot 6\cdot 4\cdot 3}+\text{\cdots }).\]
Use power series to solve \({y}^{″}+{x}^{2}y=0\) with the initial condition \(y(0)=a\) and \({y}^{'}(0)=b.\)

Առաջարկել պատասխանը

\(y=a(1-\frac{{x}^{4}}{3\cdot 4}+\frac{{x}^{8}}{3\cdot 4\cdot 7\cdot 8}-\text{\cdots })+b(x-\frac{{x}^{5}}{4\cdot 5}+\frac{{x}^{9}}{4\cdot 5\cdot 8\cdot 9}-\text{\cdots })\)
1. Express \(\int {e}^{\text{-}{x}^{2}}dx\) as an infinite series.
2. Evaluate \({\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx\) to within an error of \(0.01.\)
Առաջարկել պատասխանը
1. The Maclaurin series for \({e}^{\text{-}{x}^{2}}\) is given by
  \[\begin{array}{ll}{e}^{\text{-}{x}^{2}} & =\sum _{n=0}^{\infty }\frac{{(\text{-}{x}^{2})}^{n}}{n\text{!}} \\ & =1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{\cdots } \\ & =\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2n}}{n\text{!}}.\end{array}\]
  Therefore,
  \[\begin{array}{ll}\int {e}^{\text{-}{x}^{2}}dx & =\int (1-{x}^{2}+\frac{{x}^{4}}{2\text{!}}-\frac{{x}^{6}}{3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{x}^{2n}}{n\text{!}}+\text{\cdots })\ dx \\ & =C+x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5\cdot 2\text{!}}-\frac{{x}^{7}}{7\cdot 3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{x}^{2n+1}}{(2n+1)n\text{!}}+\text{\cdots }.\end{array}\]
2. Using the result from part a. we have
  \[{\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}+\frac{1}{216}-\text{\cdots }.\]
  The sum of the first four terms is approximately \(0.74.\) By the alternating series test, this estimate is accurate to within an error of less than \(\frac{1}{216}\approx 0.0046296<0.01.\)
Express \(\int \text{cos}\sqrt{x}dx\) as an infinite series. Evaluate \({\int }_{0}^{1}\text{cos}\sqrt{x}dx\) to within an error of \(0.01.\)

Առաջարկել պատասխանը

\(C+\sum _{n=1}^{\infty }{(-1)}^{n+1}\frac{{x}^{n}}{n(2n-2)\text{!}}\) The definite integral is approximately \(0.514\) to within an error of \(0.01.\)
Suppose a set of standardized test scores are normally distributed with mean \(\mu =100\) and standard deviation \(\sigma =50.\) Use and the first six terms in the Maclaurin series for \({e}^{\text{-}{x}^{2}\text{/}2}\) to approximate the probability that a randomly selected test score is between \(x=100\) and \(x=200.\) Use the alternating series test to determine how accurate your approximation is.

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Since \(\mu =100,\sigma =50,\) and we are trying to determine the area under the curve from \(a=100\) to \(b=200,\) integral becomes
\[\frac{1}{\sqrt{2\pi }}{\int }_{0}^{2}{e}^{\text{-}{z}^{2}\text{/}2}dz.\]
The Maclaurin series for \({e}^{\text{-}{x}^{2}\text{/}2}\) is given by
\[\begin{array}{ll}{e}^{\text{-}{x}^{2}\text{/}2} & =\sum _{n=0}^{\infty }\frac{{(-\frac{{x}^{2}}{2})}^{n}}{n\text{!}} \\ & =1-\frac{{x}^{2}}{{2}^{1}\cdot 1\text{!}}+\frac{{x}^{4}}{{2}^{2}\cdot 2\text{!}}-\frac{{x}^{6}}{{2}^{3}\cdot 3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{x}^{2n}}{{2}^{n}\cdot n\text{!}}+\text{\cdots } \\ & =\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2}{}^{n}}{{2}^{n}\cdot n\text{!}}.\end{array}\]
Therefore,
\[\begin{array}{lll}\frac{1}{\sqrt{2\pi }}\int {e}^{\text{-}{z}^{2}\text{/}2}dz & = & \frac{1}{\sqrt{2\pi }}\int (1-\frac{{z}^{2}}{{2}^{1}\cdot 1\text{!}}+\frac{{z}^{4}}{{2}^{2}\cdot 2\text{!}}-\frac{{z}^{6}}{{2}^{3}\cdot 3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{z}^{2n}}{{2}^{n}\cdot n\text{!}}+\text{\cdots })\ dz \\ & = & \frac{1}{\sqrt{2\pi }}(C+z-\frac{{z}^{3}}{3\cdot {2}^{1}\cdot 1\text{!}}+\frac{{z}^{5}}{5\cdot {2}^{2}\cdot 2\text{!}}-\frac{{z}^{7}}{7\cdot {2}^{3}\cdot 3\text{!}}+\text{\cdots }+{(-1)}^{n}\frac{{z}^{2n+1}}{(2n+1){2}^{n}\cdot n\text{!}}+\text{\cdots }) \\ \frac{1}{\sqrt{2\pi }}{\int }_{0}^{2}{e}^{\text{-}{z}^{2}\text{/}2}dz & = & \frac{1}{\sqrt{2\pi }}(2-\frac{8}{6}+\frac{32}{40}-\frac{128}{336}+\frac{512}{3456}-\frac{{2}^{11}}{11\cdot {2}^{5}\cdot 5\text{!}}+\text{\cdots }).\end{array}\]
Using the first five terms, we estimate that the probability is approximately \(0.4922.\) By the alternating series test, we see that this estimate is accurate to within
\[\frac{1}{\sqrt{2\pi }}\ \frac{{2}^{13}}{13\cdot {2}^{6}\cdot 6\text{!}}\approx 0.00546.\]
Use the first five terms of the Maclaurin series for \({e}^{\text{-}{x}^{2}\text{/}2}\) to estimate the probability that a randomly selected test score is between \(100\) and \(150.\) Use the alternating series test to determine the accuracy of this estimate.

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The estimate is approximately \(0.3414.\) This estimate is accurate to within \(0.0000094.\)
The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length \(L\) that makes a maximum angle \({\theta }_{\text{max}}\) with the vertical, its period \(T\) is given by
\[T=4\sqrt{\frac{L}{g}}{\int }_{0}^{\pi \text{/}2}\frac{d\theta }{\sqrt{1-{k}^{2}{\text{sin}}^{2}\theta }}\]
where \(g\) is the acceleration due to gravity and \(k=\text{sin}(\frac{{\theta }_{\text{max}}}{2})\) (see ). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and \(\text{sin}\ \theta\) is approximated by \(\theta .)\) Use the binomial series
\[\frac{1}{\sqrt{1+x}}=1+\sum _{n=1}^{\infty }\frac{{(-1)}^{n}}{n\text{!}}\ \frac{1\cdot 3\cdot 5\text{\cdots }(2n-1)}{{2}^{n}}{x}^{n}\]
to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if
1. you use only the first term in the binomial series, and
2. you use the first two terms in the binomial series.
Առաջարկել պատասխանը
We use the binomial series, replacing \(x\) with \(\text{-}{k}^{2}{\text{sin}}^{2}\theta .\) Then we can write the period as
\[T=4\sqrt{\frac{L}{g}}{\int }_{0}^{\pi \text{/}2}(1+\frac{1}{2}{k}^{2}{\text{sin}}^{2}\theta +\frac{1\cdot 3}{2\text{!}{2}^{2}}{k}^{4}{\text{sin}}^{4}\theta +\text{\cdots })\ d\theta .\]
1. Using just the first term in the integrand, the first-order estimate is
  \[T\approx 4\sqrt{\frac{L}{g}}{\int }_{0}^{\pi \text{/}2}d\theta =2\pi \sqrt{\frac{L}{g}}.\]
  If \({\theta }_{\text{max}}\) is small, then \(k=\text{sin}(\frac{{\theta }_{\text{max}}}{2})\) is small. We claim that when \(k\) is small, this is a good estimate. To justify this claim, consider
  \[{\int }_{0}^{\pi \text{/}2}(1+\frac{1}{2}{k}^{2}{\text{sin}}^{2}\theta +\frac{1\cdot 3}{2\text{!}{2}^{2}}{k}^{4}{\text{sin}}^{4}\theta +\text{\cdots })\ d\theta .\]
  Since \(|\text{sin}\ x|\le 1,\) this integral is bounded by
  \[{\int }_{0}^{\pi \text{/}2}(\frac{1}{2}{k}^{2}+\frac{1.3}{2\text{!}{2}^{2}}{k}^{4}+\text{\cdots })\ d\theta <\frac{\pi }{2}(\frac{1}{2}{k}^{2}+\frac{1\cdot 3}{2\text{!}{2}^{2}}{k}^{4}+\text{\cdots }).\]
  Furthermore, it can be shown that each coefficient on the right-hand side is less than \(1\) and, therefore, that this expression is bounded by
  \[\frac{\pi {k}^{2}}{2}(1+{k}^{2}+{k}^{4}+\text{\cdots })=\frac{\pi {k}^{2}}{2}\cdot \frac{1}{1-{k}^{2}},\]
  which is small for \(k\) small.
2. For larger values of \({\theta }_{\text{max}},\) we can approximate \(T\) by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate
  \[\begin{array}{ll}T & \approx 4\sqrt{\frac{L}{g}}{\int }_{0}^{\pi \text{/}2}(1+\frac{1}{2}{k}^{2}{\text{sin}}^{2}\theta )d\theta \\ & =2\pi \sqrt{\frac{L}{g}}(1+\frac{{k}^{2}}{4}).\end{array}\]
\({(1-x)}^{1\text{/}3}\)
\({(1+{x}^{2})}^{-1\text{/}3}\)

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\({(1+{x}^{2})}^{-1\text{/}3}=\sum _{n=0}^{\infty }(\begin{array}{l}-\frac{1}{3} \\ n\end{array}){x}^{2n}\)
\({(1-x)}^{1.01}\)
\({(1-2x)}^{2\text{/}3}\)

Առաջարկել պատասխանը

\({(1-2x)}^{2\text{/}3}=\sum _{n=0}^{\infty }{(-1)}^{n}{2}^{n}(\begin{array}{l}\frac{2}{3} \\ n\end{array}){x}^{n}\)
\(\sqrt{x+2}\) at \(a=0\)
\(\sqrt{{x}^{2}+2}\) at \(a=0\)

Առաջարկել պատասխանը

\(\sqrt{2+{x}^{2}}=\sum _{n=0}^{\infty }{2}^{(1\text{/}2)-n}(\begin{array}{l}\frac{1}{2} \\ n\end{array}){x}^{2n};(|{x}^{2}|<2)\)
\(\sqrt{x+2}\) at \(a=1\)
\(\sqrt{2x-{x}^{2}}\) at \(a=1\) (Hint: \(2x-{x}^{2}=1-{(x-1)}^{2})\)

Առաջարկել պատասխանը

\(\sqrt{2x-{x}^{2}}=\sqrt{1-{(x-1)}^{2}}\) so \(\sqrt{2x-{x}^{2}}=\sum _{n=0}^{\infty }{(-1)}^{n}(\begin{array}{l}\frac{1}{2} \\ n\end{array}){(x-1)}^{2n}\)
\({(x-8)}^{1\text{/}3}\) at \(a=9\)
\(\sqrt{x}\) at \(a=4\)

Առաջարկել պատասխանը

\(\sqrt{x}=2\sqrt{1+\frac{x-4}{4}}\) so \(\sqrt{x}=\sum _{n=0}^{\infty }{2}^{1-2n}(\begin{array}{l}\frac{1}{2} \\ n\end{array}){(x-4)}^{n}\)
\({x}^{1\text{/}3}\) at \(a=27\)
\(\sqrt{x}\) at \(\alpha =9\)

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\(\sqrt{x}=\sum _{n=0}^{\infty }{3}^{1-2n}(\frac{1}{2}n){(x-9)}^{n}\)
[T] \({(15)}^{1\text{/}4}\) using \({(16-x)}^{1\text{/}4}\)
[T] \({(1001)}^{1\text{/}3}\) using \({(1000+x)}^{1\text{/}3}\)

Առաջարկել պատասխանը

\(10{(1+\frac{x}{1000})}^{1\text{/}3}=\sum _{n=0}^{\infty }{10}^{1-3n}(\begin{array}{l}\frac{1}{3} \\ n\end{array}){x}^{n}.\) Using, for example, a fourth-degree estimate at \(x=1\) gives \(\begin{array}{ll}{(1001)}^{1\text{/}3} & \approx 10(1+(\begin{array}{l}\frac{1}{3} \\ 1\end{array}){10}^{-3}+(\begin{array}{l}\frac{1}{3} \\ 2\end{array}){10}^{-6}+(\begin{array}{l}\frac{1}{3} \\ 3\end{array}){10}^{-9}+(\begin{array}{l}\frac{1}{3} \\ 4\end{array}){10}^{-12}) \\ & =10(1+\frac{1}{{3.10}^{3}}-\frac{1}{{9.10}^{6}}+\frac{5}{{81.10}^{9}}-\frac{10}{{243.10}^{12}})=10.00333222...\end{array}\) whereas \({(1001)}^{1\text{/}3}=10.00332222839093....\) Two terms would suffice for three-digit accuracy.
[T] \(\frac{1}{\sqrt{2}}\) using \(x=\frac{1}{2}\) in \({(1-x)}^{1\text{/}2}\)
[T] \(\sqrt{5}=5\ \times \ \frac{1}{\sqrt{5}}\) using \(x=\frac{4}{5}\) in \({(1-x)}^{1\text{/}2}\)

Առաջարկել պատասխանը

The approximation is \(2.3152;\) the CAS value is \(2.23\text{\ldots }.\)
[T] \(\sqrt{3}=\frac{3}{\sqrt{3}}\) using \(x=\frac{2}{3}\) in \({(1-x)}^{1\text{/}2}\)
[T] \(\sqrt{6}\) using \(x=\frac{5}{6}\) in \({(1-x)}^{1\text{/}2}\)

Առաջարկել պատասխանը

The approximation is \(2.583\text{\ldots };\) the CAS value is \(2.449\text{\ldots }.\)
Integrate the binomial approximation of \(\sqrt{1-x}\) to find an approximation of \({\int }_{0}^{x}\sqrt{1-t}dt.\)
[T] Recall that the graph of \(\sqrt{1-{x}^{2}}\) is an upper semicircle of radius \(1.\) Integrate the binomial approximation of \(\sqrt{1-{x}^{2}}\) up to order \(8\) from \(x=-1\) to \(x=1\) to estimate \(\frac{\pi }{2}.\)

Առաջարկել պատասխանը

\(\sqrt{1-{x}^{2}}=1-\frac{{x}^{2}}{2}-\frac{{x}^{4}}{8}-\frac{{x}^{6}}{16}-\frac{5{x}^{8}}{128}+\text{\cdots }.\) Thus
\({\int }_{-1}^{1}\sqrt{1-{x}^{2}}dx=x-\frac{{x}^{3}}{6}-\frac{{x}^{5}}{40}-\frac{{x}^{7}}{7\cdot 16}-\frac{5{x}^{9}}{9\cdot 128}+\text{\cdots }{|}_{-1}^{1}\approx 2-\frac{1}{3}-\frac{1}{20}-\frac{1}{56}-\frac{10}{9\cdot 128}+\text{error}=1.590...\) whereas \(\frac{\pi }{2}=1.570...\)
\({(1+4x)}^{1\text{/}3};a=0\)
\({(1+4x)}^{4\text{/}3};a=0\)

Առաջարկել պատասխանը

\(\begin{array}{l}{(1+4x)}^{4/3}=(1+4x){(1+4x)}^{1/3} \\ =(1+4x)(1+\frac{4x}{3}-\frac{16{x}^{3}}{9}+\frac{320{x}^{3}}{81}-\frac{2560{x}^{4}}{243}) \\ =1+\frac{16}{3}x+\frac{32}{9}{x}^{2}-\frac{256}{81}{x}^{3}+\frac{1280}{243}{x}^{4}-\frac{10240}{243}{x}^{5}\end{array}\)
\({(3+2x)}^{1\text{/}3};a=-1\)
\({({x}^{2}+6x+10)}^{1\text{/}3};a=-3\)

Առաջարկել պատասխանը

\({(1+{(x+3)}^{2})}^{1\text{/}3}=1+\frac{1}{3}{(x+3)}^{2}-\frac{1}{9}{(x+3)}^{4}+\frac{5}{81}{(x+3)}^{6}-\frac{10}{243}{(x+3)}^{8}+\text{\cdots }\)
Use \({(1+x)}^{1\text{/}3}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4}+\text{\cdots }\) with \(x=1\) to approximate \({2}^{1\text{/}3}.\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

How to: Working with Taylor Series

Write the terms of the binomial series.
Recognize the Taylor series expansions of common functions.
Recognize and apply techniques to find the Taylor series for a function.
Use Taylor series to solve differential equations.
Use Taylor series to evaluate nonelementary integrals.
Find the binomial series for
Use the third-order Maclaurin polynomial
Here

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Ցուցադրել Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests