Question

Use a power series to approximate the value of the integral with an error of less than 0.0001. (Round your answer to four decimal places.)
integral^1_0 sin(x)/x dx
integral^1_0 sin(x)/x dx approximately equal to summation^2_n=0 (-1)^n x^2n+1 (-1)^n (x-1)^n/(2n + 1)!

Answer 1

Answer:

The answer is "0.9461"

Step-by-step explanation:

Given:

$\int^1_0 \frac{sin(x)}{x} dx$

$\int^1_0 \frac{sin(x)}{x} dx$ ≈$\sum^2_{n=0}\frac{(-1)^n x^{2n+1} (-1)^n (x-1)^n}{(2n + 1)!}$

$\therefore$

$\to \sin x = \sum^{\infty}_{n=0} (-1)^n\frac{x^{(2n+1)}}{(2n + 1)!}$

$\because \\\\ \to \frac{\sin x}{x} =\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1-1}}{(2n+1)!}$

           $=\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n}}{(2n+1)!}$

The value is in between 0 and 1 then:

 $\to \int^1_0 \frac{sin(x)}{x} = =\sum^{2}_{n=0} \frac{(-1)^n}{ (2n+1) (2n+1)!}$

The above-given series is an alternative series, and it will give an error, when the nth term is bounded by its absolute value, that can be described as follows:

$\to \frac{1}{(2n+3) (2n+3)!}< 0.0001\\\\\to (2n+3) (2n+3)!> 0.0001\\\\\to n\geq 2$

So,

$\int^1_0 \frac{sin(x)}{x} dx$ ≈$1 - \frac{1}{3 \cdot 3!}+\frac{1}{5 \cdot 5!}$

                  $\approx 1 - \frac{1}{3 \cdot 6}+\frac{1}{5 \cdot 120}\\\\\approx 1 - \frac{1}{18}+\frac{1}{600}\\\\\approx 1 - \frac{1}{18}+\frac{1}{600}\\\\\approx \frac{18-1}{18}+\frac{1}{600}\\\\\approx \frac{17}{18}+\frac{1}{600}\\\\\approx \frac{1700+3}{1800}\\\\ \approx \frac{1703}{1800}\\\\\approx 0.9461$