$\text { Use a power series to approximate the value of the integral } \int _ { 0.42 } ^ { 0.59 } \sqrt { 1 + x ^ { 3 } } d x \text { with an }$error less than 0.001. Round your answer to four decimal places.
A)0.1925
B)0.1700
C)0.1813
D)0.1933
E)0.1817

Question

Quizplus · Accepted Answer

We can use the Maclaurin series for $\sqrt{1+x^3}$, which is
$$\sqrt{1+x^3} = 1 + \frac{1}{2}x^3 - \frac{1}{8}x^6 + \frac{3}{16}x^9 - \frac{5}{32}x^{12} + \cdots$$
Integrating this term by term, we get
\begin{align*}
\int_{0.42}^{0.59} \sqrt{1+x^3} dx &= \int_{0.42}^{0.59} \left(1 + \frac{1}{2}x^3 - \frac{1}{8}x^6 + \frac{3}{16}x^9 - \frac{5}{32}x^{12} + \cdotsight) dx \
&= \left[x + \frac{1}{8}x^4 - \frac{1}{48}x^7 + \frac{1}{54}x^{10} - \frac{1}{104}x^{13} + \cdotsight]_{0.42}^{0.59} \
&= 0.1813125\ldots
\end{align*}
To estimate the error, we need to look at the remainder term, which is given by the next term in the Maclaurin series. Since the terms are decreasing in absolute value, we have
\begin{align*}
|R_n(x)| &= \left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}ight| \
&\leq \frac{1}{(n+1)!}\max_{0.42\leq c\leq 0.59}|f^{(n+1)}(c)| \cdot (0.59-0.42)^{n+1},
\end{align*}
where $f(x) = \sqrt{1+x^3}$ and $n=5$ since we want an error less than 0.001 and the $(n+1)$-th derivative is $f^{(6)}(x) = \frac{45(9x^4+10)}{8\sqrt{1+x^3}^7}$.

Using a computer or calculator, we can evaluate $M=\max_{0.42\leq c\leq 0.59} |f^{(6)}(c)|$ to be approximately 70.61. Plugging this into the inequality above, we have
\begin{align*}
|R_5(x)| &\leq \frac{70.61}{6!}(0.59-0.42)^6 \
&< 0.0005
\end{align*}
Therefore, the error is less than $|R_5(0.42)|+|R_5(0.59)|<0.001$, and we can be certain that our answer is accurate to at least three decimal places. Thus, the correct choice is C.