적분 기법 전체 문제

about 89,250 m²

$-\frac{\sqrt{4-{\sin }^{2}(x)}}{\sin (x)}-\frac{x}{2}+C$

Answers may vary. Ex: $9.405$ km

(a) Area is $30.8667$ cm$^2$. (b) Area is $308,667$ yd$^2$.

(a) Area is 25.0667 cm$^2$ (b) Area is 250,667 yd$^2$

If a quadratic function can be factored as $(ax+b)(cx+d)$ for some constants $a,b,c,d$, then it has roots $-\frac{b}{a}$ and $-\frac{d}{c}$. So, if a quadratic function has no roots, it is irreducible: this is the case for the function in graph (d).

If a quadratic function has two different roots, then $(ax+b) \neq \alpha(cx+d)$ for any constant $\alpha$. That is, the quadratic function is the product of distinct linear factors. This is the case for the functions graphed in (b) and (c), since these each have two distinct places where they cross the $x$-axis.

Finally, if a quadratic function has precisely one root, then $\frac{b}{a}=\frac{d}{c}$, so:

$$\begin{aligned} (ax+b)(cx+d)&=a(x+\tfrac{b}{a})(cx+d) = a(x+\tfrac{d}{c})(cx+d) = \tfrac{a}{c}(cx+d)(cx+d) \end{aligned}$$

That is, the quadratic function is the product of a repeated linear factor, and a constant $\frac{a}{c}$ (which might simply be $\frac{a}{c}=1$).

Midpoint rule:

Trapezoidal rule:

(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.