Using the first three nonzero terms of the Maclaurin series, the approximated value of $\int _ { 1 } ^ { 2 } \frac { e ^ { x ^ { 2 } } } { x } d x$ is

A) $\frac { 27 - 8 \ln 2 } { 8 }$
B) $\frac { 23 + 8 \ln 2 } { 8 }$
C) $\frac { 27 + 8 \ln 2 } { 8 }$
D) $\frac { 23 - 8 \ln 2 } { 8 }$
E) $\frac { 27 + \ln 2 } { 8 }$

Question

Using the first three nonzero terms of the Maclaurin series, the approximated value of $\int _ { 1 } ^ { 2 } \frac { e ^ { x ^ { 2 } } } { x } d x$ is

A) $\frac { 27 - 8 \ln 2 } { 8 }$
B) $\frac { 23 + 8 \ln 2 } { 8 }$
C) $\frac { 27 + 8 \ln 2 } { 8 }$
D) $\frac { 23 - 8 \ln 2 } { 8 }$
E) $\frac { 27 + \ln 2 } { 8 }$

Quizplus · Accepted Answer

The Maclaurin series for $e^{x^2}$ is $1 + x^2 + \frac{x^4}{2} + \cdots$. Integrating $\frac{e^{x^2}}{x}$ using the first three terms gives $\int \left(\frac{1}{x} + x + \frac{x^3}{2}\right) dx$, which evaluates to $\ln x + \frac{x^2}{2} + \frac{x^4}{8}$. Evaluating from 1 to 2 gives $\left(\ln 2 + 2 + 2\right) - \left(0 + \frac{1}{2} + \frac{1}{8}\right) = \ln 2 + \frac{27}{8}$.

Using the First Three Nonzero Terms of the Maclaurin Series ∫12ex2xdx\int _ { 1 } ^ { 2 } \frac { e ^ { x ^ { 2 } } } { x } d x∫12​xex2​dx

Using the First Three Nonzero Terms of the Maclaurin Series $\int _ { 1 } ^ { 2 } \frac { e ^ { x ^ { 2 } } } { x } d x$