수치적분

The absolute error is the difference between the two values: $$|1.387-1.5| = 0.113$$ The relative error is the absolute error divided by the exact value: $$\frac{0.113}{1.387}\approx 0.08147$$ The percent error is 100 times the relative error: $$\approx 8.147\%$$

(a) Differentiating, we find $f''(x) = -x^2+7x-6$. Since $f''(x)$ is quadratic, we have a pretty good idea of what it looks like.

• It factors as $f(x) = -(x-6)(x-1)$, so its two roots are at $x=6$ and $x=1$.
• The “flat part" of the parabola is at $x=3.5$ (since this is exactly half way between $x=1$ and $x=6$; alternately, we can check that $f'''(3.5)=0$).
• Since the coefficient of $x^2$ is negative, $f(x)$ is increasing from $-\infty$ to $3.5$, then decreasing from $3.5$ to $\infty$.

Therefore, over the interval $[1,6]$, the largest positive value of $f''(x)$ occurs when $x=3.5$, and this is $f''(3.5) = -(3.5-6)(3.5-1)=6.25$.

So, we take $M=6.25$. (b) We differentiate further to find $f^{(4)}(x)=-2$. This is constant everywhere, so we take $L=|-2|=2$.

Under any reasonable assumptions (Anyone caught trying to come up with a scenario in which I am older than my mother will be sent to maximum security grad school.), my mother is older than I am.

Before we can take our Simpson's rule approximation of $\displaystyle\int_1^x \dfrac{1}{t}~\mathrm{d}t$, we need to know how many intervals to use. That means we need to bound our error, which means we need to bound $\frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}$.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left\{\frac{1}{t}\right\}&=-\frac{1}{t^2}& \frac{\mathrm{d}^2}{\mathrm{d}{t}^2}\left\{\frac{1}{t}\right\}&=\frac{2}{t^3}\\ \frac{\mathrm{d}^3}{\mathrm{d}{t}^3}\left\{\frac{1}{t}\right\}&=-\frac{6}{t^4}& \frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}&=\frac{24}{t^5} \end{aligned}$$

So, over the interval $[1,3]$, $\displaystyle\left|\frac{\mathrm{d}^4}{\mathrm{d}{t}^4}\left\{\frac{1}{t}\right\}\right| \leq 24$.

Now, we can find an appropriate $n$ to ensure our error will be be less than 0.1 for any $x$ in $[1,3]$:

$$\begin{aligned} \frac{L(b-a)^5}{180 n^4}&<0 .1 \\\frac{24(x-1)^5}{180 n^4}&< \frac{1}{10} \\n^4 &> \frac{24\cdot (x-1)^5}{18} \\ \text{Because $x-1 \leq 2$ for every $x$ in $[1,3]$, if $n^4 > \dfrac{24\cdot 2^5}{18}$, then $n^4 > \dfrac{24\cdot (x-1)^5}{18}$ for every allowed $x$.} \\ n^4&> \frac{24\cdot 2^5}{18}= \frac{128}{3}\\ n&>\sqrt[4]{\frac{128}{3}}\approx 2.6 \end{aligned}$$

Since $n$ must be even, $n=4$ is enough intervals to guarantee our error is not too high for any $x$ in $[1,3]$. Now we find our Simpson's rule approximation with $n=4$, $a=1$, $b=x$, and $\Delta x = \dfrac{x-1}{4}$. The points where we evaluate $\frac{1}{t}$ are:

$$\begin{aligned} x_0 =& 1 & x_1&=1+ \frac{x-1}{4}& x_2&=1+2\frac{x-1}{4}& x_3&=1+3\frac{x-1}{4}& x_4&=1+4\frac{x-1}{4}\\ & & &=\frac{x+3}{4}& &=\frac{x+1}{2}&&=\frac{3x+1}{4}& &=x \end{aligned}$$

$$\begin{aligned} \log x=\int_1^x \frac{1}{t}~\mathrm{d}t&\approx \frac{\Delta x}{3}\left[\frac{1}{x_0}+\frac{4}{x_1}+\frac{2}{x_2}+\frac{4}{x_3}+\frac{1}{x_4}\right]\\ &=\frac{x-1}{12}\left[1+\frac{16}{x+3}+\frac{4}{x+1}+\frac{16}{3x+1}+\frac{1}{x}\right]\\ &=f(x) \end{aligned}$$

Below is a graph of our approximation $f(x)$ and natural logarithm on the same axes. The natural logarithm function is shown red and dashed, while our approximating function is solid blue. Our approximation appears to be quite accurate for small, positive values of $x$.

(a) We can't use $u$-substitution since there's not a function-derivative pair present (nor even a composite function). To show that integration by parts doesn't work is more complicated, but comes down to the fact that if we use the natural choice of $u = x$ and $dv = \tan(x) \, dx$, to do so requires us to evaluate the integral $\int \ln(\sec(x)) \, dx$, which appears not to have an elementary antiderivative. If we try instead $u = \tan(x)$ and $dv = x \, dx$, then we have to evaluate the integral $\int x^2 \cdot \sec^2(x) \, dx$ which is more complicated than the integral we started with and would seem to send us back where we started if we attempt integration by parts. (b) We use computational technology to generate the following estimates, which are reported to $5$ decimal places of accuracy:

• $L_4 = 0.25892$
• $R_4 = 0.64827$
• $M_4 = 0.41550$
• $T_4 = \frac{L_4 + R_4}{2} = 0.45360$
• $S_8 = \frac{2M_4 + T_4}{3} = 0.42820$ (c) From the values of these estimates or from the graph of $f(x) = x\tan(x)$, we can observe that $f$ is always increasing and concave up on the interval. Thus, $L_4$ and $M_4$ are underestimates while $R_4$ and $M_4$ are overestimates.

(a) Because $L_4 = 7.23 > 6.75 = R_4$ and we know the function is always increasing or always decreasing, for the left endpoint rule to give a value greater than the right endpoint rule, $f$ must be always decreasing. (b) From the given data, we know that $$T_4 = \frac{L_4 + R_4}{2} = \frac{7.23 + 6.75}{2} = 6.99\,.$$ Since $M_4 = 7.05$, it follows that $$R_4 < T_4 < M_4 < L_4\,.$$ In particular, we know that the function is not only always decreasing on the interval, but it also concave down. Because the concavity is constant, it follows that if the trapezoid rule under-estimates the true value, then the midpoint rule must over-estimate the true value (or vice-versa). Since $T_4 < M_4$, this means that $$T_4 < \int_3^6 f(x) \, dx < M_4$$ and that $f$ must be concave down on the interval. (c) We use Simpson's Rule to determine the best possible estimate for the definite integral. That is, $$\begin{aligned}\int_3^6 f(x) \, dx &\approx S_8 \\ &= \frac{2M_4 + T_4}{3} \\ &= \frac{2 \cdot 7.05 + 6.99}{3} \\ &=7.03\end{aligned}\,.$$