Integral

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April 29th 2009, 04:31 PM
treetheta

Integral

If $x\sin (\pi x) = \int\limits_1^{{x^2}} {f(t)dt}$ where f is a continuous function find f(4) and f'(4)

use Fundamental therom of Calculus

so I know that I'm going to let
$u = x^2$

and then differentiate both sides but how do i defferentiate

$x\sin (\pi x)$

and it's derivate how do i do that

---------------------------------------------------------------------------------------------------

Another question with this F.T.C stuff

Use F.T.C find a function and a number c such that

$6 + \int\limits_c^x {\frac{{f(t)}}{{{t^2}}}} dt = 2\sqrt x$ for x>0

So I know i gotta differentiate both sides but how would i find a number C when F.T.C would just eliminate it since derivative of a number is 0
April 29th 2009, 04:59 PM
skeeter

Quote:

Originally Posted by treetheta

If $x\sin (\pi x) = \int\limits_1^{{x^2}} {f(t)dt}$ where f is a continuous function find f(4) and f'(4)

use Fundamental therom of Calculus

so I know that I'm going to let
$u = x^2$

and then differentiate both sides but how do i defferentiate

$x\sin (\pi x)$

and it's derivate how do i do that

$x\sin(\pi x) = \int_1^{x^2} f(t) \, dt$

derivative ...

$x \cdot \pi \cos(\pi x) + \sin(\pi x) = f(x^2) \cdot 2x$

let $x = 2$ ...

$2 \cdot \pi \cos(2\pi) + \sin(2\pi) = f(4) \cdot 4$

$f(4) = \frac{\pi}{2}$

take the derivative again ...

$\frac{d}{dx}[x \cdot \pi \cos(\pi x) + \sin(\pi x) = f(x^2) \cdot 2x]$

$-x \pi^2 \sin(\pi x) + \pi \cos(\pi x) + \pi \cos(\pi x) = 2 \cdot f(x^2) + 4x^2 \cdot f'(x^2)$

substitute $x = 2$ again, and determine $f'(4)$
April 29th 2009, 05:47 PM
treetheta

lol I don't know why I thought the derivative would be complicated I forgot pi is just a number
Thanks (:

can you help me out with the other one?
April 29th 2009, 06:00 PM
redsoxfan325

Quote:

Originally Posted by treetheta

Another question with this F.T.C stuff

Use F.T.C find a function and a number c such that

$6 + \int\limits_c^x {\frac{{f(t)}}{{{t^2}}}} dt = 2\sqrt x$ for x>0

So I know i gotta differentiate both sides but how would i find a number C when F.T.C would just eliminate it since derivative of a number is 0

Take the derivative:

$\frac{f(x)}{x^2} = \frac{1}{\sqrt{x}}$

$f(x) = \frac{x^2}{\sqrt{x}} = x^{3/2}$

Plug $f(x)=x^{3/2}$ back into original equation to find $c$ .

Spoiler:

$6+\int_c^x\frac{t^{3/2}}{t^2}\,dt = 2\sqrt{x}$

$6+2\sqrt{x}-2\sqrt{c} = 2\sqrt{x} \implies 6=2\sqrt{c} \implies \boxed{c = 9}$
April 29th 2009, 06:01 PM
treetheta

Oh snap, wow, who new it would work out so nicely, wow thank you so much
April 29th 2009, 06:02 PM
redsoxfan325

Your welcome.
April 29th 2009, 06:03 PM
skeeter

Quote:

Use F.T.C find a function and a number c such that

$6 + \int\limits_c^x {\frac{{f(t)}}{{{t^2}}}} dt = 2\sqrt x$ for x>0

So I know i gotta differentiate both sides but how would i find a number C when F.T.C would just eliminate it since derivative of a number is 0

$6 + \int_c^x \frac{f(t)}{t^2} \, dt = 2\sqrt{x}$

derivative ...

$\frac{f(x)}{x^2} = \frac{1}{\sqrt{x}}$

$f(x) = x^{\frac{3}{2}}$

go back to the original integral equation and sub in $f(t)$ ...

$6 + \int_c^x t^{-\frac{1}{2}} \, dt = 2\sqrt{x}$

$6 + [2\sqrt{t}]_c^x = 2\sqrt{x}$

$6 + [2\sqrt{x} - 2\sqrt{c}] = 2 \sqrt{x}$

$c = 9$
April 29th 2009, 07:53 PM
treetheta

How would I go about doing

$\mathop {\lim }\limits_{x \to - \infty } \sqrt {x + {x^2}} - x$

I multiplied by the conjugate, to get

$\mathop {\lim }\limits_{x \to - \infty } \frac{x}{{\sqrt {x + {x^2}} + x}}$

just don't know how to continue I'm sure I could factor out x, but I don't know how to do that, could someone show me or would L'hospital work
April 29th 2009, 08:11 PM
redsoxfan325

Quote:

Originally Posted by treetheta

How would I go about doing

$\mathop {\lim }\limits_{x \to - \infty } \sqrt {x + {x^2}} - x$

I multiplied by the conjugate, to get

$\mathop {\lim }\limits_{x \to - \infty } \frac{x}{{\sqrt {x + {x^2}} + x}}$

just don't know how to continue I'm sure I could factor out x, but I don't know how to do that, could someone show me or would L'hospital work

Well, $\sqrt{x+x^2}\to\infty$ as $x\to-\infty$ , and $-x\to\infty$ as $x\to-\infty$ , so it makes sense that $\lim_{x\to-\infty}\sqrt{x+x^2}-x = \infty$

Maple confirms that $\infty$ is the correct answer.

---------------------------------------

However, that raises a question of my own: Where did I go wrong below:

$\lim_{x\to-\infty}\sqrt{x+x^2}-x = \lim_{x\to-\infty}\left((\sqrt{x+x^2}-x)\cdot\frac{\sqrt{x+x^2}+x}{\sqrt{x+x^2}+x}\right )$ $=\lim_{x\to-\infty}\frac{x}{\sqrt{x+x^2}+x} = \lim_{x\to-\infty}\frac{x}{\sqrt{x^2(1/x+1)}+x} = \lim_{x\to-\infty}\frac{x}{x\sqrt{1/x+1}+x}$ $= \lim_{x\to-\infty}\frac{x}{x(\sqrt{1/x+1}+1)} = \lim_{x\to-\infty}\frac{1}{\sqrt{1/x+1}+1}$

Now we can take the limit as $x\to-\infty$ to get $\frac{1}{\sqrt{0+1}+1} = \frac{1}{2}$

I feel like I made a really fundamental error and am just not seeing it. I know that if $x\to+\infty$ , the limit is $\frac{1}{2}$ , so where is my reasoning faulty?
April 29th 2009, 08:17 PM
treetheta

Hmm yes, I got this answer as well but however in the choices i have to pick from it's not there...
however -1/2 is so i convinced myself that maybe -infnity had something to do with it

oh btw, redsoxfan325 you are godlike math genius (:

could you help me out with this one as well:

$\int {\frac{1}{{x\sqrt {{x^6} - 4} }}}$

I really don't know where to go with this I attempted to trigsubstiution but then that got really messy, I don't think parts will work, and I dont think theres anything i can normally substitute ):
April 29th 2009, 08:28 PM
redsoxfan325

Quote:

Originally Posted by treetheta

Hmm yes, I got this answer as well but however in the choices i have to pick from it's not there...
however -1/2 is so i convinced myself that maybe -infnity had something to do with it

oh btw, redsoxfan325 you are godlike math genius (:

could you help me out with this one as well:

$\int {\frac{1}{{x\sqrt {{x^6} - 4} }}}$

I really don't know where to go with this I attempted to trigsubstiution but then that got really messy, I don't think parts will work, and I dont think theres anything i can normally substitute ):

$\int\frac{dx}{x\sqrt{x^6-4}} = \int\frac{dx}{x\sqrt{x^6(1-\frac{4}{x^6})}} = \int\frac{dx}{x^4\sqrt{1-\left(\frac{2}{x^3}\right)^2}}$

Let $u=\frac{2}{x^3}$ . Thus $du = -\frac{6}{x^4}$

So our integral now looks like: $-\frac{1}{6}\int\frac{du}{\sqrt{1-u^2}} = -\frac{1}{6}\sin^{-1}(u) = \boxed{-\frac{1}{6}\sin^{-1}\left(\frac{2}{x^3}\right)}$

Thanks for the compliment!
April 29th 2009, 08:52 PM
treetheta

you make everything look so easy,

is there a trig formula for
$-\frac{1}{6}\int\frac{du}{\sqrt{1-u^2}} = -\frac{1}{6}\sin^{-1}(u)$

to equal eachother?

I think this will be the last one I shall ask you becuase i'm starting to feel bad ><

$\int {\frac{{{x^5} - 1}}{{{x^4} + {x^2}}}}$
April 29th 2009, 09:36 PM
redsoxfan325

Quote:

Originally Posted by treetheta

you make everything look so easy,

is there a trig formula for
$-\frac{1}{6}\int\frac{du}{\sqrt{1-u^2}} = -\frac{1}{6}\sin^{-1}(u)$

to equal eachother?

I think this will be the last one I shall ask you becuase i'm starting to feel bad ><

$\int {\frac{{{x^5} - 1}}{{{x^4} + {x^2}}}}$

First, the inverse sine integral should be found in any table of integrals. Second, break the integral up into two parts:

$\int\frac{x^5-1}{x^4+x^2} = \int\frac{x^5}{x^4+x^2}\,dx - \int\frac{1}{x^4+x^2}\,dx$

First integral:

$\int\frac{x^5}{x^4+x^2}\,dx = \int\frac{x^3}{x^2+1}\,dx$

Let $x=\sqrt{u-1}$ . Thus $dx=\frac{1}{2\sqrt{u-1}}$

The integral is now $\int\frac{(u-1)^{3/2}}{u}\cdot\frac{1}{2\sqrt{u-1}}\,du = \frac{1}{2}\int\frac{u-1}{u}\,du$ $= \frac{1}{2}\int 1-\frac{1}{u}\,du = \frac{u}{2}-\frac{1}{2}\ln|u| = \frac{x^2+1}{2}-\frac{1}{2}\ln|x^2+1|$

Second integral:

$\int\frac{1}{x^4+x^2}\,dx = \int\frac{1}{x^2(x^2+1)}\,dx$

$\frac{1}{x^2(x^2+1)} = \frac{Ax+B}{x^2}+\frac{Cx+D}{x^2+1} \implies (A+C)x^3+(B+D)x^2+Ax+B = 1$

Thus, $A=0, B=1, C=0, D=-1$

So $\int\frac{1}{x^2(x^2+1)} = \int\frac{1}{x^2}\,dx-\int\frac{1}{x^2+1}\,dx = -\frac{1}{x}-\tan^{-1}(x)$

Subtracting the first and second integrals gives $\boxed{\frac{x^2+1}{2}-\frac{1}{2}\ln|x^2+1| + \frac{1}{x} + \tan^{-1}(x) + C}$

Sorry it took so long. That one was tough.
April 29th 2009, 10:06 PM
treetheta

Don't worry about time man, you can get anything I would have never thought to sperate it good call there, can you suggest any practice type questions (: I got an exam tmrw been scrambling all over the place try to do as much as possible
April 29th 2009, 10:18 PM
redsoxfan325

1.) $\int\frac{1}{x^4+1}\,dx$ (Good luck, it's a hard one!)

2.) $\int-\frac{x^4}{(1+x^2)^{3/2}}\,dx$

3.) $\int\sqrt{1+e^x}\,dx$

4.) $\int\sin(\ln(x))\,dx$

5.) $\int\frac{1}{x\ln(x)\ln(\ln(x))}\,dx$ (This is an easy u-substitution...if you know what to use for u!)

6.) $\int\frac{x^3}{x+\sqrt{x}}\,dx$

I think that's enough to keep you busy for a while, as most of these aren't easy.