수치적분

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, $$\begin{aligned}\int_3^6 f(x) \, dx &\approx S_8 \\ &= \frac{2M_4 + T_4}{3} \\ &= \frac{2 \cdot 7.05 + 6.99}{3} \\ &=7.03\end{aligned}\,.$$

(a) Because $r(t)$ is measured in cubic feet per second and $\triangle t$ is measured in seconds, the definite integral $$\int_0^{60} r(t) \, dt$$ measures the total amount of water (in cubic feet) that flow through the dam in 60 seconds. (b) With $7$ function values and thus $6$ subintervals as the maximum number of intervals we can use, we can compute $M_3$ since we need to use every other data point to evaluate the function. Doing so, $$\begin{aligned}\int_0^{60} r(t) \, dt &\approx M_3 \\ &= r(10) \cdot 20 + r(30) \cdot 20 + r(50) \cdot 20 \\ &= (2100 + 3000 + 5100) \cdot 20 \\ &= 204000\end{aligned}\,.$$ Thus, approximately $204 000$ cubic feet of water pass through the dam on the interval $0 \le t \le 60$. We observe that $r$ appears to be increasing at an increasing rate on $0 \le t \le 60$. In particular, the data suggests that $r$ is concave up on the interval. As such, the trapezoid rule will over-estimate the true value of the definite integral and the midpoint rule will under-estimate the value. Thus, $$\int_0^{60} r(t) \, dt > M_3 = 204000\,.$$ (c) In order to estimate the integral using $S_6$, we need to know the value of $T_3$ along with $M_3$. To find $T_3$, we compute $L_3$ and $R_3$. First $$L_3 = r(0) \cdot 20 + r(20) \cdot 20 + r(40) \cdot 20 = (2000 + 2400 + 3900) \cdot 20 = 166000\,.$$ Similarly, $$R_3 = r(20) \cdot 20 + r(40) \cdot 20 + r(60) \cdot 20 = (2400 + 3900 + 6500) \cdot 20 = 256000\,.$$ Thus, $$T_3 = \frac{L_3 + R_3}{2} = \frac{166000 + 256000}{2} = 211000\,.$$ Finally, we now have the information needed to compute $S_6$. We thus find that $$\int_0^{60} r(t) \, dt \approx S_6 = \frac{2M_3 + T_3}{3} = \frac{2 \cdot 204000 + 211000}{3} = \frac{619000}{3} \approx 206333.33\,.$$ (d) First, $$\frac{1}{60} S_6 = \frac{206333.33}{60} = \frac{30950}{9} \approx 3438.89\,.$$ This is one estimate for the average value of $r$ on $[0,60]$ since $$r_{\operatorname{AVG} [0,60]} = \frac{1}{60-0} \int_0^{60} r(t) \, dt\,,$$ and $\int_0^{60} r(t) \, dt \approx S_6$. Thus, on average, water is flowing through the dam at a rate of $3438.89$ cubic feet per second. Next, $$\frac{2000+2100+2400+3000+3900+5100+6500}{7} = \frac{25000}{7} \approx 3571.43$$ is the average of the $7$ known values of $r(t)$ on the interval $[0,60]$, and thus is one estimate of the average value of $r$. This average treats all of them equally, as opposed to $S_6$ which counts midpoints more than endpoints and is developed in such a way to remove some of the error inherent in using simple estimates like $L_6$ and $R_6$. This simple average of $7$ rate values is analogous to using a left or right Riemann sum, and thus we expect that the estimate of the average value of $r$ that comes from $S_6$ is more accurate for computing the average rate at which water flows through the dam.