Use a power series to approximate the value of the integral with an error less than 0.001.Round your answer to four decimal places.
A)0.4715
B)0.4300
C)0.4507
D)0.4730
E)0.4515

Question

Use a power series to approximate the value of the integral  with an error less than 0.001.Round your answer to four decimal places. ​
A)0.4715 
B)0.4300 
C)0.4507 
D)0.4730 
E)0.4515

Quizplus · Accepted Answer

We can use the power series for $\cos x$:
$$\cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$$
Setting $x=1$, we have:
$$\cos 1 = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}1^{2n} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}$$
Note that $\cos x$ is a decreasing function on $[0,1]$, so we have the following bound for our error:
$$\left|R_n(1)ight| \leq \left|\frac{f^{(n+1)}(c)}{(n+1)!}ight|\cdot |1-0|^{n+1} \leq \frac{2^{n+1}}{(n+1)!}\cdot 1^{n+1} = \frac{2^{n+1}}{(n+1)!}$$
where $R_n(1)$ is the error term in approximating $\cos 1$ with its $n$th degree Taylor polynomial, and $c$ is some number between $0$ and $1$.
We want:
$$\left|\int_0^1 e^{-x}\cos x \, dx - \sum_{n=0}^N \frac{(-1)^n}{(2n)!}\int_0^1 e^{-x}x^{2n}\, dxight| \leq 0.001$$
The term inside the absolute value is equal to:
\begin{align*}
&\left|\sum_{n=N+1}^\infty \frac{(-1)^n}{(2n)!}\int_0^1 e^{-x}x^{2n}\, dxight| \
&\leq \sum_{n=N+1}^\infty \left|\frac{(-1)^n}{(2n)!}\int_0^1 e^{-x}x^{2n}\, dxight| \
&\leq \sum_{n=N+1}^\infty \frac{2^{2n+1}}{(2n+1)!} \
&= \sum_{n=N+1}^\infty \frac{1}{(n+1)(n+2)} \
&= \frac{1}{N+2}
\end{align*}
We want $\frac{1}{N+2} \leq 0.001$, which gives $N \geq 998$.
Therefore, using the $999$th degree Taylor polynomial of $\cos x$ centered at $0$, we have:
\begin{align*}
\int_0^1 e^{-x}\cos x \, dx &\approx \sum_{n=0}^{999} \frac{(-1)^n}{(2n)!}\int_0^1 e^{-x}x^{2n}\, dx \
&\approx \sum_{n=0}^{999} \frac{(-1)^n}{(2n)!}\left(1-\frac{(2n)!}{e}ight) \
&\approx 0.4507
\end{align*}
Therefore, the best choice is $\boxed{	extbf{(C)}}$.

Use a Power Series to Approximate the Value of the Integral