Are You Ready to Challenge Your Integral Solving Skills?

[dextercioby]

A correct solution to an integration involving square roots should always include an analysis of the sign of the quadatic under the square root sign.

 

[Schrodinger's Dog]

\int\sqrt{-x^2+2x+17}dx=\int\sqrt{(3\sqrt{2})^2-(1-x)^2}dx

let y=1-x\implies-\int\sqrt{(3\sqrt{2})^2-y^2}dy

which equals

-\frac{y}{2}\sqrt{18-y^2}-9\arcsin\frac{y}{3\sqrt{2}}

how good am i?



ps: sd's solution above is wrong.

no really, NS Sherlock.

I copied that off mathcad what I mean to say was it's (du) -18 sin^{-1} \frac {(x-1)}{(3\sqrt 2)} - (du)

that's what I meant sorry :/.

I'll copy the answer off mathcalc, but it's probably equivalent anyway although it is much different.

\int\sqrt{-x^2+2x+17}dx= \frac{1}{2} (\sqrt -x^2+2x+17x)+--18 sin^{-1} \frac {(x-1)}{(3\sqrt 2)} - \sqrt -x^2+17x+17)+C

Actually looking at your solution? If it's equivalent I'm not seeing it, if you could explain

I need a good lesson, my integration is not great atm.

 

[J77]

A correct solution to an integration involving square roots should always include an analysis of the sign of the quadatic under the square root sign.

pedant

OK - x\in[1-\frac12\sqrt{72},1+\frac12\sqrt{72}]

 

[J77]

no really, NS Sherlock.

NS Watson

 

[Schrodinger's Dog]

NS Watson

 

[murshid_islam]

Nor did I realize Multiplying integrals resulted in a double integral.

as far as i know, multiplying integrals does not always result in a double integral. it does when the two single integrals are of variables that are independent of each other. did i put it correctly? someone?

 

[Gib Z]

I'm kinda scratchy on my Calculus, I am reviewing my Vector Calculus right now, which I am Glad to say I remember most of. I've never encountered something like that double integral method of solving integrals, like the one murshid_islam gave...Maybe Someone could gimme pointers on those?

 

[quasar987]

I think the general idea is that if we can say that* I=\sqrt{I^2}, then Fubini's theorem permits us to interchange the order of integration at will, and then we can make a change of variables, such as r(x,y)=\sqrt{x^2+y^2}, to put the integral in a form that can be evaluated using the FTC.

*This implies knowing that I, if it exists, is real and positive.

 

[tehno]

Is the arcsin one ln (1-\sqrt{1-u}) - \frac{arcsin u}{u} where u = e^x? I was swear I was right, but when i check it it says I am slightly off >.<

You're right that you are wrong about your solution for that integral.
Here's the most important hint for solving it:
Write it as \int e^{-x}\cdot arcsin \ e^x \ dx and do it by a method of partial integration.

 

[Gib Z]

RAWR I silly mistaked! I did it the same as my first attempt, but on that time Near the end, I had ln ( (1+root1-e^2x)/ e^x), and I simplyfied that to
ln 1+root 1-e^2x - ln e^x, and I simplified the last part to 1 instead of x...shoot me...

The Right solution is

\int \frac{\arcsin (e^x)}{e^x} dx = x - ( \frac{arcsin (e^x)}{e^x} + log_e ( 1+\sqrt{1-e^{2x}} ) )

Yay

 

[Warr]

I'm kinda scratchy on my Calculus, I am reviewing my Vector Calculus right now, which I am Glad to say I remember most of. I've never encountered something like that double integral method of solving integrals, like the one murshid_islam gave...Maybe Someone could gimme pointers on those?

Double integrals just mean that more than one variable can change. You wouldn't have double integrals for functions of one variable. The method of solving double integrals (and higher order integrals for that matter) depends mainly on choosing the order of integration (which variable to integrate first), and the implied limits of integration based on this choice. Any calc textbook that goes into multivariable calc. should have them.

 

[Gib Z]

Ahh yes right, The book I am doing my vector calculus in goes through those lol.

Funnily enough the titles - Multivariable Calculus by McCALLUM, HUGHES-HALLETT, GLEASON, ET AL.

What ever that means...yea well the vector bits are on page 87, while the double integrals are further..Ill try to get there lol

 

[Gib Z]

Umm, okie the double integrals I am working on. Someone give me a single variable one again :)

 

[murshid_islam]

try this: \int_{0}^{1}\exp\left(x^2\right)dx

 

[Gib Z]

I don't think that can be done in terms of elementary functions >.< All I could tell from finding its derivative is that it grows really quickly, so the area is probably large for larger limits of integration.

1.4626 is what I get from some simpsons rule. That doesn't look familiar to me >.<

I tried repeated integration by parts, not working.

 

[murshid_islam]

try expanding \exp(x^2) into a series and then integrate term by term. after that put the limits and you should get a series that looks like \sum_{k=0}^{\infty}\frac{1}{(2k+1)k!}

 

[Gib Z]

O thats, I did that and it becomes the series you stated, but how can i evaluate that? I know it will converge, just not what to >.<

 

[murshid_islam]

can anybody help me on this?

\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta

 

[ssd]

\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta<br /> =\int_{0}^{\pi}\frac{\pi\sin\theta}{1 + \cos^4\theta}d\theta<br /> - \int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta<br />
Move the 2nd integral to left to get 2I, I=the original integral.
For the first integral put z=cos\theta

 

[murshid_islam]

how did you get this:

\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta=\int_{0}^{\pi}\frac{\pi\sin\theta}{1 + \cos^4\theta}d\theta- \int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta

 

[ssd]

how did you get this:

This is a standard result
\int_{0}^{a}f(x)dx= \int_{0}^{a}f(a-x)dx<br />

 

[Gib Z]

If you say so?...I've never seen that before...

 

[murshid_islam]

ssd, can you show me how to prove this?:

\int_{0}^{a}f(x)dx= \int_{0}^{a}f(a-x)dx

edit: ok i think i got it. by substituting a - x = u, i get,
\int_{0}^{a}f(a-x)dx = -\int_{a}^{0}f(u)du = \int_{0}^{a}f(x)dx

 

[Gib Z]

And that proof works how? I can't see how the 2nd part equals the 3rd.

 

[cristo]

And that proof works how? I can't see how the 2nd part equals the 3rd.

The variables u and x are just (dummy) integration variables, so from going from the second part to the third, let u=x. Then noting that \int_a^bf(z)=F(b)-F(a)=-[F(a)-F(b)]=-\int_b^a f(z) where F(z) is an antiderivative of f(z), yields the result.

 

[murshid_islam]

\int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta<br /> =\int_{0}^{\pi}\frac{\pi\sin\theta}{1 + \cos^4\theta}d\theta<br /> - \int_{0}^{\pi}\frac{\theta\sin\theta}{1 + \cos^4\theta}d\theta<br />
Move the 2nd integral to left to get 2I, I=the original integral.
For the first integral put z=cos\theta

even then it becomes messy. i get,
I = \pi \int_{0}^{1}\frac{dz}{1 + z^4}
i have read that even leibniz found this cumbersome. i have to complete the squares in the denominator, then split into partial fractions and so on...

 

[siddharth]

I don't think a trig sub will work. Try writing the denominator as (u^2-\sqrt 2u+1)(u^2+\sqrt 2u+1), then use partial fractions. It should work, but it won't be pretty!

There's a pretty way to do that integral. Involves some trickery.

https://www.physicsforums.com/showthread.php?t=127563

 

[ssd]

even then it becomes messy. i get,
I = \pi \int_{0}^{1}\frac{dz}{1 + z^4}
i have read that even leibniz found this cumbersome. i have to complete the squares in the denominator, then split into partial fractions and so on...

The process is cumbersome no doubt but the problem is such...By the way I found it interesting.

 

[ssd]

And that proof works how? I can't see how the 2nd part equals the 3rd.

\int_{a}^{b}f(u)du = \int_{a}^{b}f(x)dx = F(b)-F(a)<br />
where, f(y) is the first derivative of F(y) w.r.t. y.

 

[Gib Z]

K thanks guys, got it. As for that integral you have left, I feel sorry for you >.<

 

[murshid_islam]

The process is cumbersome no doubt but the problem is such...By the way I found it interesting.

i found the integral interesting too, even if it is cumbersome.

 

[Gib Z]

O well, let's revive this thread again. Anybody have another, single variable, indefinite integral? It doesn't even have to be from a textbook, it can be your homework, or one your having trouble with :D

 

[murshid_islam]

here is an extremely easy one:

\int_{-1}^{1}x^3e^{-x^{4}}\cos 2xdx

although it is very easy, i found it interesting

 

[Gib Z]

Ahh I am thinking integration by parts, u=x^3e^{-x^4}, and dv = cos 2x, and integration by parts again on the u, but could you just tell me if I am right before I do it, i don't want to spend so much time on it if it won't get me anywhere lol.

 

[gammamcc]

0

There are hard ways and easy ways. (Try graphing it. Notice any symmetry?)Try

\int_-90000 ^90000 sin^57(x) e^{-\pi^{77} 234535x^88}cos(2453245x^4) (x^{99999994}+1)dx

 

[d_leet]

Ahh I am thinking integration by parts, u=x^3e^{-x^4}, and dv = cos 2x, and integration by parts again on the u, but could you just tell me if I am right before I do it, i don't want to spend so much time on it if it won't get me anywhere lol.

It might be doable that way, but I doubt it. Note that this is a definite integral, what is special about the bounds? What is special about the function?

 

[Gib Z]

:D:D:D

Thanks a heap guys, I can't believe I didn't see that. I already noticed it was an odd function, but for some reason didn't make the connection.

The Answers 0 :).

As for gammamcc's one, I am just going to put the tex brackets that he forgot to over here, because I don't understand it written like that, then ill try it.

\int_{-90000}^{90000} \sin^{57} (x) e^{-\pi^{77} 234535x^{88}}\cos (2453245x^4) (x^{99999994}+1) dx

 

[Gib Z]

Thats an odd function as well...so the answers zero again...

 

[murshid_islam]

Ahh I am thinking integration by parts, u=x^3e^{-x^4}, and dv = cos 2x, and integration by parts again on the u, but could you just tell me if I am right before I do it, i don't want to spend so much time on it if it won't get me anywhere lol.

you don't need any calculations at all. that's why i said it is extremely easy.
just look at the function x^3e^{-x^{4}}\cos 2x. do you see anything? is it odd? even? then you can immediately see what the answer is.

edit: oh, sorry, you have already got help. didn't notice that.

 

[Gib Z]

And as I just realized, that integral is not expressible in terms of elementary functions...

EDIT: INDEFINITE integral i meant to say.

 

[murshid_islam]

And as I just realized, that integral is not expressible in terms of elementary functions...

EDIT: INDEFINITE integral i meant to say.

how did you realize that?

i have another integral. i don't know how to do it. can anybody help?

\int_{0}^{\infty}x^ne^{x^2}dx

 

[Gib Z]

I worked out the taylor series and integrated term by term, this is my best:

\int_0^{\infty} x^{a} e^{x^2} dx =\lim_{x\rightarrow {\infty}} \sum_{n=0}^{\infty} \frac{x^{a+(2n+1)}}{n!\cdot (a+(2n+1))}

That doesn't help much

 

[arildno]

how did you realize that?

i have another integral. i don't know how to do it. can anybody help?

\int_{0}^{\infty}x^ne^{x^2}dx

It diverges.

 

[Gib Z]

Looking at the summation again, It looks like it does lol.
EDIT: Or one could see that both are increasing functions, which i just noticed and am starting to feel stupid about for not noticing sooner.

 

[murshid_islam]

what about \int_{0}^{\infty}x^ne^{-x^2}dx

 

[ssd]

what about \int_{0}^{\infty}x^ne^{-x^2}dx

Putting z=x^2,

\int_{0}^{\infty}x^ne^{-x^2}dx<br /> =\frac{1}{2}\int_{0}^{\infty}z^{\frac{n+1}{2}-1}e^{-z}dz
Apart from the factor 1/2, the integral is a Gamma Integral with parameters (n+1)/2,1.
If (n+1)/2 is a positive integer, then the integral = 0.5[(n-1)/2]! ... (Gamma distribution is known as Erlang distribution if the shape parameter is an integer.)

 

[murshid_islam]

what if (n+1)/2 is not an integer?

 

[ZioX]

It becomes \frac{1}{2}\Gamma(\frac{n+1}{2}).

Which can be approximated...no closed form (in general).

 

[Gib Z]

Closed forms can be generated for any integer n with the following identities:

\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{1-2z} \; \sqrt{\pi} \; \Gamma(2z). \,\!

And

\Gamma\left(\frac{n}{2}+1\right)= \sqrt{\pi}\, \frac{n!}{2^{(n+1)/2}} (n odd)

 

[Gib Z]

Lol its been a while since I've posted here. Anybody got an interesting integral?