Use both trapezoid and Simpson's rule to find ${\int}_{0}^{\mathrm{\infty}}{e}^{-x}dx$ starting with $h=2$ where $h$ is the length of subintervals $[{x}_{i},{x}_{i+1}]$.

Wisniewool
2022-07-05
Answered

Use both trapezoid and Simpson's rule to find ${\int}_{0}^{\mathrm{\infty}}{e}^{-x}dx$ starting with $h=2$ where $h$ is the length of subintervals $[{x}_{i},{x}_{i+1}]$.

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sniokd

Answered 2022-07-06
Author has **22** answers

A well-known trick is to make the change of variables

$x=\frac{t}{1-t}$

So that the integral becomes

${\int}_{0}^{+\mathrm{\infty}}{e}^{-x}\mathrm{d}x={\int}_{0}^{1}\frac{{e}^{-t/(1-t)}}{(1-t{)}^{2}}\mathrm{d}t$

the trick here is that this argument goes to zero when $t\to 1$, so you can manually set the last point in your quadrature to $0$

$x=\frac{t}{1-t}$

So that the integral becomes

${\int}_{0}^{+\mathrm{\infty}}{e}^{-x}\mathrm{d}x={\int}_{0}^{1}\frac{{e}^{-t/(1-t)}}{(1-t{)}^{2}}\mathrm{d}t$

the trick here is that this argument goes to zero when $t\to 1$, so you can manually set the last point in your quadrature to $0$

asked 2022-06-03

Approximate ${\int}_{0}^{1}\phantom{\rule{mediummathspace}{0ex}}\sqrt{2-{x}^{2}}dx$ using the trapezoidal and simpson's rule for $4$ intervals.

Now we can determine the simpson rule is

$\frac{h}{3}{\textstyle (}f({x}_{0})+4f({x}_{1})+2f({x}_{2})+4f({x}_{3})+f({x}_{4}){\textstyle )}$

and the trapezoidal rule is

$\frac{h}{2}{\textstyle (}f({x}_{0})+2f({x}_{1})+2f({x}_{2})+2f({x}_{3})+f({x}_{4}){\textstyle )}$

and $h=\frac{b-a}{n}$ which I assume is $\frac{1-0}{4}$

but how we add it all together?

Now we can determine the simpson rule is

$\frac{h}{3}{\textstyle (}f({x}_{0})+4f({x}_{1})+2f({x}_{2})+4f({x}_{3})+f({x}_{4}){\textstyle )}$

and the trapezoidal rule is

$\frac{h}{2}{\textstyle (}f({x}_{0})+2f({x}_{1})+2f({x}_{2})+2f({x}_{3})+f({x}_{4}){\textstyle )}$

and $h=\frac{b-a}{n}$ which I assume is $\frac{1-0}{4}$

but how we add it all together?

asked 2022-05-08

Assume that $S(h)$ is equivalent to the (composite) Simpson's rule where h is the size of the step. Correct use of Richardson's extrapolation gives the formula: $R(h)=\frac{16S(h)-S(a)}{b}$. What's a and b?

asked 2022-06-27

Does there exist a monic ${x}^{n}$, $n>4$, for which Simpson’s rule is exact? If not, why?

$S(f)=\frac{b-a}{6}f(a)+\frac{2(b-a)}{3}f(\frac{a+b}{2})+\frac{b-a}{6}f(b)$

$S(f)=\frac{b-a}{6}f(a)+\frac{2(b-a)}{3}f(\frac{a+b}{2})+\frac{b-a}{6}f(b)$

asked 2022-06-13

From a proof of Simpson's rule using Taylor polynomial where $f\in [{x}_{0},{x}_{2}]$ and, for

${x}_{1}={x}_{0}+h$

where

$h=\frac{{x}_{2}-{x}_{0}}{2}$

it got:

${\int}_{{x}_{0}}^{{x}_{2}}f(x)dx\cong 2hf({x}_{1})+{h}^{3}\frac{{f}^{\u2033}({x}_{1})}{3}+{h}^{5}\frac{{f}^{(4)}(\xi )}{60}$

and then, it changed ${f}^{\u2033}({x}_{1})$ by

$\frac{f({x}_{0})-2f({x}_{1})+f({x}_{2})}{{h}^{2}}$

Where it came?

${x}_{1}={x}_{0}+h$

where

$h=\frac{{x}_{2}-{x}_{0}}{2}$

it got:

${\int}_{{x}_{0}}^{{x}_{2}}f(x)dx\cong 2hf({x}_{1})+{h}^{3}\frac{{f}^{\u2033}({x}_{1})}{3}+{h}^{5}\frac{{f}^{(4)}(\xi )}{60}$

and then, it changed ${f}^{\u2033}({x}_{1})$ by

$\frac{f({x}_{0})-2f({x}_{1})+f({x}_{2})}{{h}^{2}}$

Where it came?

asked 2022-06-15

Calculate:

${\int}_{-14}^{-8}ydx$

${\int}_{-14}^{-8}ydx$

asked 2022-05-23

Evaluate ${\int}_{-1}^{0}({x}^{4}-{x}^{2}+2)dx$ using Simpson's Rule

asked 2022-05-10

Show that one extrapolation of the trapezoid rule leads to Simpson's rule.