수치적분

Trapezoidal Rule: $0.9006$

Simpson's Rule: $0.90452$

Trapezoidal Rule: $3.0241$

Simpson's Rule: $2.9315$

Trapezoidal Rule: $13.9604$

Simpson's Rule: $13.9066$

Trapezoidal Rule: $3.0695$

Simpson's Rule: $3.14295$

Trapezoidal Rule: $1.1703$

Simpson's Rule: $1.1873$

Trapezoidal Rule: $2.52971$

Simpson's Rule: $2.5447$

Trapezoidal Rule: $1.0803$

Simpson's Rule: $1.077$

Trapezoidal Rule: $3.5472$

Simpson's Rule: $3.6133$

(a) $n=161$ (using $\max\big(f\,''(x)\big)=1$) (b) $n=12$ (using $\max\big(f\,^{(4)}(x)\big)=1$)

(a) $n=150$ (using $\max\big(f\,''(x)\big)=1$) (b) $n=18$ (using $\max\big(f\,^{(4)}(x)\big)=7$)

(a) $n=1004$ (using $\max\big(f\,''(x)\big)=39$) (b) $n=62$ (using $\max\big(f\,^{(4)}(x)\big)=800$)

(a) $n=5591$ (using $\max\big(f\,''(x)\big)=300$) (b) $n=46$ (using $\max\big(f\,^{(4)}(x)\big)=24$)

(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$^2$

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: $$\approx 8.147\%$$

Midpoint rule:

Trapezoidal rule:

(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.

True. Because $f(x)$ is positive and concave up, the graph of $f(x)$ is always below the top edges of the trapezoids used in the trapezoidal rule.

(a) When $0 \leq x \leq1$, then $x^2 \leq 1$ and $x+1 \geq 1$, so $|f''(x)| = \dfrac{x^2}{|x+1|}\leq \dfrac{1}{1}=1$. (b) To find the maximum value of a function over a closed interval, we test the function's values at the endpoints of the interval and at its critical points inside the interval. The critical points are where the function's derivative is zero or does not exist.

The function we're trying to maximize is $|f''(x)| = \frac{x^2}{|x+1|} = \frac{x^2}{x+1}=f''(x)$ (since our interval only contains nonnegative numbers). So, the critical points occur when $f'''(x) = 0$ or does not exist. We find $f'''(x)$ Using the quotient rule.

$$\begin{aligned} f'''(x)&=\frac{(x+1)(2x)-x^2}{(x+1)^2}=\frac{x^2+2x}{(x+1)^2}\\ 0&=\frac{x(x+2)}{x+1}\\ 0&=x \quad\text{or}\quad x=-1\quad\text{or}\quad x=-2 \end{aligned}$$

The only critical point in $[0,1]$ is $x=0$. So, the extrema of $f''(x)$ over $[0,1]$ will occur at its endpoints. Indeed, since $f'''(x) \geq 0$ for all $x$ in $[0,1]$, $f''(x)$ is increasing over this interval, so its maximum occurs at $x=1$. That is, $$|f''(x)|\leq f''(1)=\frac{1}{2}$$ (c) The absolute error using the midpoint rule is at most $\dfrac{M(b-a)^3}{24n^2}$. Using $M=1$, if we want this to be no more than $10^{-5}$, we find an acceptable value of $n$ with the following calculation:

$$\begin{aligned} \dfrac{M(b-a)^3}{24n^2}&\leq 10^{-5}\\ \dfrac{1}{24n^2}&\leq 10^{-5}&(b-a=1,\,M=1)\\ \frac{10^5}{24} & \leq n^2\\ n & \geq 65 \end{aligned}$$ (d) The absolute error using the midpoint rule is at most $\dfrac{M(b-a)^3}{24n^2}$. Using $M=\frac{1}{2}$, if we want this to be no more than $10^{-5}$, we find an acceptable value of $n$ with the following calculation:

$$\begin{aligned} \dfrac{M(b-a)^3}{24n^2}&\leq 10^{-5}\\ \dfrac{1}{48n^2}&\leq 10^{-5}&(b-a=1,\,M=\frac{1}{2})\\ \frac{10^5}{48} & \leq n^2\\ n & \geq 46 \end{aligned}$$

Remark: how accurate you want to be in these calculations depends a lot on your circumstances. Imagine, for instance, that you were finding $M$ by hand, using this to find $n$ by hand, then programming a computer to evaluate the approximation. For a simple integral like this, the difference between computing time for 65 intervals versus 46 is likely to be miniscule. So, there's not much to be gained by the extra work in (b). However, if your original sloppy $M$ gave you something like $n=1000000$, you might want to put some time into improving it, to shorten computation time. Moreover, if you were finding the approximation by hand, the difference between adding 46 terms and adding 65 terms would be considerable, and you would probably want to put in the effort up front to find the most accurate $M$ possible.