Answer the question.
-Estimate the error if $\mathrm { e } ^ { - 3 \mathrm { x } ^ { 2 } }$ is approximated by $1 - 3 \mathrm { x } ^ { 2 } + \frac { 3 ^ { 2 } \mathrm { x } ^ { 4 } } { 2 ! }$ in the integral of $\int _ { 0 } ^ { 1 } \mathrm { e } ^ { - 3 \mathrm { x } ^ { 2 } } \mathrm { dx }$.
A) $\frac { 3 ^ { 4 } } { 5 ( 2 ! ) ^ { 2 } }$
B) $\frac { 3 ^ { 5 } } { 6 ( 3 ! ) }$
C) $\frac { 3 ^ { 6 } } { 7 ( 3 ! ) }$
D) $\frac { 3 ^ { 6 } } { 7 ( 3 ! ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

The error formula for the Taylor series approximation of a function is given by:
$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$
where $f$ is the function being approximated, $n$ is the degree of the Taylor polynomial, $c$ is some point between $x$ and $a$, and $a$ is the point of expansion.
In this case, we are using the second degree Taylor polynomial to approximate $f(x) = e^{-3x^2}$ around $a=0$, so we have:
$$R_2(x) = \frac{f^{(3)}(c)}{3!}x^3$$
where $f^{(3)}(x) = 54xe^{-3x^2} - 36x^3e^{-3x^2}$.
To estimate the error in the given integral, we need to bound $R_2(x)$ for $x$ between 0 and 1:
$$|R_2(x)| \leq \max_{0 \leq c \leq 1}\left|\frac{54c}{6}x^3 - \frac{36c}{6}x^3\right|$$
$$\leq \frac{3}{2}x^3$$
The maximum value of $|R_2(x)|$ over the interval $[0,1]$ occurs at $x=1$, so:
$$|R_2(1)| \leq \frac{3}{2} \cdot 1^3 = \frac{3}{2}$$
Therefore, the error in the approximation is bounded by $\frac{3}{2}$, and the best choice is D, which gives this value with a common denominator:
$$\frac{3^6}{7(3!)^2} = \frac{3}{2} \cdot \frac{3^4}{7 \cdot 6} = \frac{81}{140}$$
which is less than 1, so the approximation is good.

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Answer the Question $\mathrm { e } ^ { - 3 \mathrm { x } ^ { 2 } }$