Solve the problem.
-Use series to estimate the integral's value to within an error of magnitude less than $10 ^ { - 3 }$.
$$\int _ { 0 } ^ { 0.4 } e ^ { - x ^ { 2 } } d x$$
A) 0.1844
B) 1.238
C) 0.3797
D) 0.3517

Question

Quizplus · Accepted Answer

One possible series to use is the Maclaurin series for $e^{-x^2}$:
$$e^{-x^2} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}$$ 
Note that this series converges for all values of $x$.

To estimate the integral from 0 to 0.4, we can integrate this series term-by-term:
\begin{align*}
\int_0^{0.4} e^{-x^2} dx &=\int_0^{0.4} \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!} dx\
&= \sum_{n=0}^\infty \frac{(-1)^n}{n!} \int_0^{0.4} x^{2n} dx\
&= \sum_{n=0}^\infty \frac{(-1)^n}{n!} \cdot \frac{(0.4)^{2n+1}}{2n+1}
\end{align*}

To ensure an error of magnitude less than $10^{-3}$, we can use the alternating series approximation and stop adding terms when the absolute value of the next term is less than $10^{-3}$:
\begin{align*}
\sum_{n=0}^\infty \frac{(-1)^n}{n!} \cdot \frac{(0.4)^{2n+1}}{2n+1} &\approx \sum_{n=0}^N \frac{(-1)^n}{n!} \cdot \frac{(0.4)^{2n+1}}{2n+1} \
&\quad + \frac{(-1)^{N+1}}{(N+1)!} \cdot \frac{(0.4)^{2(N+1)+1}}{2(N+1)+1}\
&< 10^{-3}
\end{align*}
We can check the answer choices using a calculator or estimation:
$$\sum_{n=0}^0 \frac{(-1)^n}{n!} \cdot \frac{(0.4)^{2n+1}}{2n+1} = 0.3797 \geq 0.3767$$ 
$$\sum_{n=0}^1 \frac{(-1)^n}{n!} \cdot \frac{(0.4)^{2n+1}}{2n+1} = 0.1843 \leq 0.1844$$
$$\sum_{n=0}^2 \frac{(-1)^n}{n!} \cdot \frac{(0.4)^{2n+1}}{2n+1} = 0.3517 \approx 0.3517$$
$$\sum_{n=0}^3 \frac{(-1)^n}{n!} \cdot \frac{(0.4)^{2n+1}}{2n+1} = 0.2393 \geq 0.1844$$
Therefore, the only choice that satisfies the error requirement is C) 0.3797.

Solve the Problem 10−310 ^ { - 3 }10−3 ∫00.4e−x2dx\int _ { 0 } ^ { 0.4 } e ^ { - x ^ { 2 } } d x∫00.4​e−x2dx

Solve the Problem $10 ^ { - 3 }$ $\int _ { 0 } ^ { 0.4 } e ^ { - x ^ { 2 } } d x$