Attempt to approximate the integral $\int_{1}^{2} \mathrm{e}^{\frac{1} {x}} \mathrm{d} x$ using both the trapezoidal rule and Simpson's rule, and estimate the truncation error as ____.

Options:
A. $$
Trapezoidal: \int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1730
$$
$$
| R_{!} | \leq \frac{(2-1)^{3}}{12} \operatorname*{max}_{1 \leq x \leq 2} | f^{\prime\prime}(x) | = 0.6785
$$
$$
Simpson: \int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0160
$$
$$
\mid R_{2} \mid \leq \frac{\left( 2-1 \right)^{5}} {2880} \operatorname*{max}_{1 \leq x \leq 2} \mid f^{(4)}(x) \mid = 0.06750
$$
B. $$
Trapezoidal: \int_{1}^{2} \mathrm{e}^{\frac{1} {x}} \mathrm{d} x^{2} = 2.1835
$$
$$
| \, R_{1} \, | \leq \frac{( 2-1 )^{3}} {12} \operatorname*{max}_{1 \leq x \leq 2} | \, f^{\prime\prime} ( \, x ) \, | = 0.6796
$$
$$
Simpson: \int_{1}^{2} \mathrm{e}^{\frac{1} {x}} \mathrm{d} x^{2} = 2.0263
$$
$$
\mid R_{2} \mid \leq \frac{\left( 2-1 \right)^{5}} {2880} \operatorname*{max}_{1 \leq x \leq 2} \mid f^{\left( 4 \right)} \left( x \right) \mid = 0.06890
$$
C. $$
Trapezoidal: \int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1635
$$
$$
| R_{!} | \leq \frac{(2-1)^{3}}{12} \operatorname*{max}_{1 \leq x \leq 2} | f''(x) | = 0.6776
$$
$$
Simpson: \int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0463
$$
$$
| R_{2} | \leq \frac{(2-1)^{5}}{2880} \operatorname*{max}_{1 \leq x \leq 2} | f^{(4)}(x) | = 0.07380
$$
D. $$
Trapezoidal: \int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1630
$$
$$
| R_{!} | \leq \frac{(2-1)^3}{12} \max_{1 \leq x \leq 2} | f^{\prime\prime}(x) | = 0.6735
$$
$$
Simpson: \int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0460
$$
$$
\mid R_{2} \mid \leq \frac{\left( 2-1 \right)^{5}} {2880} \max_{1 \leq x \leq 2} \mid f^{(4)}(x) \mid = 0.06595
$$
E. $$
Trapezoidal: \int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1930
$$
$$
| R_{!} | \leq \frac{(2-1)^3}{12} \operatorname*{max}_{1 \leq x \leq 2} | f^{\prime\prime}(x) | = 0.6805
$$
$$
Simpson: \int_{1}^{2} \mathrm{e}^{\frac{1} {x}} \mathrm{d} x^{2} = 2.0360
$$
$$
\mid R_{2} \mid \leq \frac{\left( 2-1 \right)^{5}} {2880} \max_{1 \leq x \leq 2} \mid f^{(4)}(x) \mid = 0.07030
$$
F. Trapezoidal Rule:
$$
\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1935
$$
$$
|R_1| \leq \frac{(2-1)^3}{12} \max_{1 \leq x \leq 2} |f''(x)| = 0.6750
$$
Simpson's Rule:
$$
\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0163
$$
$$
|R_2| \leq \frac{(2-1)^5}{2880} \max_{1 \leq x \leq 2} |f^{(4)}(x)| = 0.06550
$$
G. Trapezoidal Rule:
$$
\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1932
$$
$$
| R_{1} | \leq \frac{(2-1)^{3}}{12} \max_{1 \leq x \leq 2} | f''(x) | = 0.6720
$$
Simpson's Rule:
$$
\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0003
$$
$$
| R_{2} | \leq \frac{(2-1)^{5}}{2880} \max_{1 \leq x \leq 2} | f^{(4)}(x) | = 0.06490
$$
H. Trapezoidal Rule:
$$
\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1735
$$
$$
| R_{!} | \leq \frac{(2-1)^3}{12} \max_{1 \leq x \leq 2} | f''(x) | = 0.6800
$$
Simpson's Rule:
$$
\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0363
$$
$$
| R_{2} | \leq \frac{(2-1)^5}{2880} \max_{1 \leq x \leq 2} | f^{(4)}(x) | = 0.07050
$$
I. Trapezoidal Rule: $\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.2035$
$$
|R_1| \leq \frac{(2-1)^3}{12} \max_{1 \leq x \leq 2} |f^{\prime\prime}(x)| = 0.6832
$$
Simpson's Rule: $\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0563$
$$
|R_2| \leq \frac{(2-1)^5}{2880} \max_{1 \leq x \leq 2} |f^{(4)}(x)| = 0.07170
$$
J. Trapezoidal Rule: $$\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.1235$$
$$|R_{!}| \leq \frac{(2-1)^{3}}{12} \max_{1 \leq x \leq 2} |f''(x)| = 0.6706$$
Simpson's Rule: $$\int_{1}^{2} \mathrm{e}^{\frac{1}{x}} \mathrm{d} x^{2} = 2.0963$$
$$|R_{2}| \leqslant \frac{(2-1)^{5}}{2880} \max_{1 \leqslant x \leqslant 2} |f^{(4)}(x)| = 0.07460$$
Answer: B