Integrate a power series term by term
The area under a series
The lower graph shows the curve \(f(t)=\dfrac1{1-t}\), which equals the series \(1+t+t^2+t^3+\cdots\) for \(|t|<1\). The teal region is the area under the curve from 0 to \(x\).
In the upper graph, the chalk curve shows this exact area as a function of \(x\). The orange curve is the area under the orange polynomial of the lower graph, which keeps the first four terms of the series.
To find the area under the polynomial, integrate each term with the power rule. The exponent rises by 1, and the coefficient is divided by the new exponent.
At \(x=0.5\), the exact area is about 0.693, and the polynomial gives about 0.682. Press the first button to keep more terms, and the orange curve moves toward the chalk curve.
Any power series
The curve in the lower graph is now any sum of terms \(c_nt^n\). The area under a sum equals the sum of the areas under its terms, because areas add.
The term \(c_nt^n\) has the area \(\dfrac{c_nx^{n+1}}{n+1}\) from 0 to \(x\). The orange curve in the upper graph is the sum of these areas.
Press the first button to start the area at a height of 2. Both upper curves move up by 2, but their slope at each \(x\) is still the height of the lower curve.
Every antiderivative of the series differs from the area by a constant, so the new series needs a constant \(C\).
Notation
Start with the series of \(f\), whose curve and polynomial are in the lower graph.
\[f(t)=\sum_{n=0}^{\infty}c_nt^n\]Integrate one term with the power rule. The exponent rises by 1, and the coefficient is divided by the new exponent.
\[\int c_nt^n\,dt=\cb{\dfrac{c_nt^{n+1}}{n+1}}\]Add one constant \(C\) to the whole sum, because any two antiderivatives of \(f\) differ by a constant.
\[F(x)=C+\sum_{n=0}^{\infty}\dfrac{c_nx^{n+1}}{n+1}\]Find \(C\) from a known value of \(F\). At \(x=0\), every term of the sum is 0, so \(F(0)=C\).
\[C=F(0)\]For a series centered at \(a\), use powers of \(x-a\). Then \(C=F(a)\). The new series has the same radius of convergence.
When it fails
Start the area at 2, but leave out the constant. The series without \(C\) is 0 at \(x=0\), so its orange curve lies 2 below the chalk curve.
The red region between the two curves shows the error. It is 2 high at every \(x\), because the series is missing the constant. Only a known value of \(F\) tells you \(C\).
Example 1: an antiderivative with a given value
Find the series of the antiderivative \(F\) of \(f(x)=\dfrac1{1+x^3}\) with \(F(0)=2\).
Write \(f\) as a geometric series. The ratio is \(-x^3\), because \(\frac1{1-r}\) with \(r=-x^3\) equals \(\frac1{1+x^3}\). The series converges for \(|x|<1\).
\[1-x^3+x^6-x^9+\cdots\]Integrate each term with the power rule, and add one constant \(C\).
\[\begin{aligned} F(x)&=C+x-\dfrac{x^4}{4} \\ &\quad+\dfrac{x^7}{7}-\dfrac{x^{10}}{10}+\cdots \end{aligned}\]At \(x=0\), every term except \(C\) is 0, so \(C=F(0)=2\). The chalk curve passes through the point \((0,2)\), and the orange series moves up to meet it.
\[F(x)=2+x-\dfrac{x^4}{4}+\dfrac{x^7}{7}-\cdots\]
Example 2: the area function of the cosine
The series of \(\cos x\) is \(1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots\). Find the series of the antiderivative \(F\) with \(F(0)=0\).
Because \(F(0)=0\), \(F(x)\) is the teal area from 0 to \(x\). At \(x=1\), the upper graph shows an area of about 0.84.
A tempting wrong move: Raise each exponent by 1. Keep each coefficient.
\[x-\dfrac{x^3}{2}+\dfrac{x^5}{24}-\cdots\]At \(x=1\), this series is about 0.54, but the teal area is about 0.84, so the red curve lies below the chalk curve.
Raise each exponent by 1, and divide each coefficient by the new exponent. Because \(F(0)=0\), the constant is 0.
\[\begin{aligned} &x-\dfrac{x^3}{2\cdot3}+\dfrac{x^5}{24\cdot5}-\cdots \\ &\quad=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots \end{aligned}\]This is the series of \(\sin x\), an antiderivative of \(\cos x\). At \(x=1\), it gives about 0.841, the teal area.
\[\cb{\sin 1\approx0.841}\]
Practice
The series of \(f(x)=xe^x\) is \(x+x^2+\dfrac{x^3}{2}+\dfrac{x^4}{6}+\cdots\). Find the series of the antiderivative \(F\) with \(F(0)=1\).
In the upper graph, the chalk curve of \(F\) has the height 1 at \(x=0\), where the orange dot is. This height is the constant \(C\).
Solution
Integrate each term with the power rule. For example, \(\frac{x^4}{6}\) gives \(\frac{x^5}{5\cdot6}=\frac{x^5}{30}\).
\[C+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{8}+\dfrac{x^5}{30}+\cdots\]Use the known value. Because \(F(0)=1\), the constant is 1, and the orange curve meets the chalk curve.
\[\begin{aligned} F(x)&=1+\dfrac{x^2}{2}+\dfrac{x^3}{3} \\ &\quad+\dfrac{x^4}{8}+\dfrac{x^5}{30}+\cdots \end{aligned}\]Integration keeps the radius of convergence, so this series converges for every \(x\), like the series of \(xe^x\).
Questions
The upper graph shows two antiderivatives of \(f(t)=\dfrac1{(1-t)^2}\), whose series is \(1+2t+3t^2+\cdots\). The solid curve is \(F\), and the dashed curve is \(G\). The dashed vertical lines mark \(x=0\) and \(x=0.5\).
What is the constant term in the series of \(F\)?
- 3
- 1
- 4
Answer
3
How far above \(G\) is \(F\) at \(x=0.5\)?
- 2
- 1
- 4
Answer
2
The series of \(f(t)=\dfrac{e^t-1}{t}\) is \(1+\dfrac t2+\dfrac{t^2}6+\dfrac{t^3}{24}+\cdots\). The antiderivative \(F\) has \(F(0)=0\). What is the coefficient of \(x^4\) in its series?
- \(\dfrac1{96}\)
- \(\dfrac1{24}\)
- \(\dfrac1{72}\)
- \(\dfrac1{120}\)
Answer
\(\dfrac1{96}\)
Summary
You integrate a power series term by term and add a constant \(C\), which you find from a known value of the antiderivative.
This works because the area under a sum is the sum of the areas, and the power rule gives the area under each power.
You use it to write a series for an antiderivative, including one that has no simple formula.
This is one of five skills in topic 10.15 of Which & Why: AP Calculus.