Chapter 10.1, Problem 31E

To determine

To calculate: The power series for ${\displaystyle {\int }_a^{x}f\left(t\right)dt}$ by the series and the function.

Answer to Problem 31E

The power series is ${\displaystyle {\int }_a^{x}f\left(t\right)dt}=\left[x+x^2+\frac{4}{3}{x}^3+\frac{8}{4}{x}^4+\cdots \right]$ , where $-\frac{1}{2}<x<\frac{1}{2}$ .

Explanation of Solution

Given Information:

The series is ${\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}$ .

The function is $f\left(x\right)=\frac{1}{1-2x}$ .

Formula Used:

Power Rule:

Let the function be $f\left(x\right)=x^n$ .

The derivative is $f^{\prime }\left(x\right)=n{x}^{n-1}$ .

Calculation:

The function is $f\left(x\right)=\frac{1}{1-2x}$ .

Rewrite the function as:

  $f\left(x\right)=\frac{1}{1-2x}={\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}$

Simplify the series.

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}\\ =1+2x+4x^2+8x^3+\cdots \end{array}$ , where $-\frac{1}{2}<x<\frac{1}{2}$

Now, the function becomes,

  $f\left(x\right)=1+2x+4x^2+8x^3+\cdots$

Replace $x$ with $t$ .

  $f\left(t\right)=1+2t+4t^2+8t^3+\cdots$

Find the power series.

Integrate both sides with respect to $x$ .

  $\begin{array}{c}{\displaystyle {\int }_a^{x}f\left(t\right)dt}={\displaystyle {\int }_a^{x}1+2t+4t^2+8t^3+\cdots }\\ ={\displaystyle {\int }_a^{x}1+2t+4t^2+8t^3+\cdots }\\ ={\left[t+t^2+\frac{4}{3}{t}^3+\frac{8}{4}{t}^4+\cdots \right]}_a^x\end{array}$

Substitute the limits.

  

Hence, the power series is where