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

A)3
B) $\frac { 8 } { 3 }$
C) $\frac { 7 } { 3 }$
D) $\frac { 11 } { 3 }$
E) $\frac { 13 } { 3 }$

Question

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

A)3
B) $\frac { 8 } { 3 }$
C) $\frac { 7 } { 3 }$
D) $\frac { 11 } { 3 }$
E) $\frac { 13 } { 3 }$

Quizplus · Accepted Answer

The Maclaurin series for $e^x$ is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$. To find the series for $e^{x^2}$, substitute $x^2$ for $x$ in the series: $e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \cdots$. The first three nonzero terms are $1, x^2, \frac{x^4}{2}$. To integrate $\frac{e^{x^2}}{x^2}$ from 1 to 2 using these terms, we get $\int_1^2 \left(\frac{1}{x^2} + 1 + \frac{x^2}{2}\right) dx$. Integrating term by term gives $-\frac{1}{x}\Big|_1^2 + x\Big|_1^2 + \frac{x^3}{6}\Big|_1^2 = (-\frac{1}{2} + 1) + (2 - 1) + \left(\frac{8}{6} - \frac{1}{6}\right) = \frac{1}{2} + 1 + \frac{7}{6} = \frac{3}{2} + \frac{7}{6} = \frac{9}{6} + \frac{7}{6} = \frac{16}{6} = \frac{8}{3}$.

Using the First Three Nonzero Terms of the Maclaurin Series ∫12ex2x2dx\int _ { 1 } ^ { 2 } \frac { e ^ { x ^ { 2 } } } { x ^ { 2 } } d x∫12​x2ex2​dx

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