integrate

[Ryan Rigdon]

i have been staring at this example for about an hour now and cant seem to grasp on how they got the answer. Help?

answer is -2.944

the problem is attached.

 

[BigGlenntheHeavy]

$\displaystyle If \ you \ can't \ prove \ trig. \ identities, \ then \ you \ have \ no \ business \ doing \ integrals.$

$\displaystyle Try \ staring \ at \ it \ for \ another \ hour.$

 

[mmm4444bot]

Hi Ryan,

I don't know enough to explain it, but I can provide a software result (from MapleV Release 5) that contains something called The Error Function erf(x).

(Double-click image, if necessary, to expand)

[attachment=0:1drople4]erf.JPG[/attachment:1drople4]

The first two commands confirm the answer.

The next command uses symbolic base "b" instead of constant base 11.

As you can see, erf(x) is defined as the product of 2/sqrt(x) times an integration with respect to t.

Perhaps, you can find an explanation of The Error Function erf(x) by googling. Or, maybe Subhotosh or Glenn will post something.

As always, don't trust software 100%. There may be other approaches to calculate the given integral.

Cheers ~ Mark

 

[galactus]

$\displaystyle \int_{1}^{0}11^{x^{2}}dx$

We can rewrite it as:

$\displaystyle \int_{1}^{0}e^{x^{2}ln(11)}dx$

Try using the series for e. The more terms, the closer to the solution.

$\displaystyle e^{x^{2}ln(11)}=1+ln(11)\left(x^{2}+\frac{x^{4}}{2!}+\frac{x^{6}}{3!}+\frac{x^{8}}{4!}+..............\right)$

Now, integrate term by term. The more terms the closer and closer to -2.944.

Integrating the above series, we get:

$\displaystyle x+ln(11)\frac{x^{3}}{3}+(ln(11))^{2}\frac{x^{5}}{10}+(ln(11))^{3}\frac{x^{7}}{42}+(ln(11))^{4}\frac{x^{9}}{216}+................$

Now, use the limits of integration.

This is actually kind of tough without some numerical method or series. As mmm mentioned, the error function comes into play and that is beyond the scope of an elementary calculus course. That is, it has no nice closed form doable by elementary means.

 

[BigGlenntheHeavy]

$\displaystyle Using \ Simpson's \ Rule \ with \ n \ = \ 10, \ gives \ -2.9451350212268.$

 

[mmm4444bot]

Are you sure, Glenn ?

With n = 10, I would expect more precision than one decimal place, but I'm not sure.

 

[galactus]

That's pretty close for n=10, I think.

Here is my TI-92's answer:

 

[BigGlenntheHeavy]

$\displaystyle \int_{1}^{0} 11^{x^2} dx \ = \ -\int_{0}^{1}11^{x^2}dx$

$\displaystyle Hence, \ using \ Simpson \ with \ n \ = \ 10, \ we \ get:$

\(\displaystyle -\int_{0}^{1}11^{x^2}dx \ = \-\frac{-1}{30}[11^0+4(11^{1/100})+2(11^{1/25})+4(11^{9/100})+2(11^{4/25})\)

$\displaystyle +4(11^{1/4})+2(11^{9/25})+4(11^{49/100})+2(11^{16/25})+4(11^{81/100})+11] \ \dot= \ -2.9451350212268$

$\displaystyle Don't \ forget, \ we \ have \ x^2 \ to \ contend \ with, \ anyways \ good \ enough \ for \ government \ work.$

 

[BigGlenntheHeavy]

$\displaystyle Note: \ If \ we \ let \ n \ = \ 14 \ (using \ Simpson), \ we \ get \ \dot= \ -2.94438016508.$