2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.

Thank you very much.

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- Nov 3rd 2007, 08:10 AMkittycatintegration - urgent help please
2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.

Thank you very much. - Nov 3rd 2007, 08:25 AMtopsquark
$\displaystyle \int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt$

First let's get that radicand simplified a bit:

Let $\displaystyle x = 3t \implies dx = 3dt$

$\displaystyle \int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt = \int_0^3 \left [ 2 \left ( \frac{x}{3} \right ) - 3 \left ( \frac{x}{3} \right ) ^2 \right ] \sqrt{1 - x^2} ~ \frac{dx}{3}$

$\displaystyle = \frac{1}{9} \int_0^3 (2x - x^2) \sqrt{1 - x^2} ~ dx$

$\displaystyle = \frac{1}{9} \int_0^3 2x \sqrt{1 - x^2} ~ dx - \frac{1}{9} \int_0^3 x^2 \sqrt{1 - x^2} ~ dx$

The first integral may be done using a simple substitution. I'll let you deal with that. For the second integral:

$\displaystyle \int_0^3 x^2 \sqrt{1 - x^2} ~ dx$

Take a moment and look at the graph I've posted below. You can easily see that this integrand does not exist beyond x = 1. So

$\displaystyle \int_0^3 x^2 \sqrt{1 - x^2} ~ dx = \int_0^1 x^2 \sqrt{1 - x^2} ~ dx$ <-- We need to do this, else the substitution below is illegal.

Now Let $\displaystyle x = sin(\theta) \implies dx = cos( \theta) d \theta$

$\displaystyle \int_0^1 x^2 \sqrt{1 - x^2} ~ dx = \int_0^{\pi/2}sin^2(\theta) \sqrt{1 - sin^2(\theta) } ~cos(\theta) ~ d \theta $

$\displaystyle = \int_0^{\pi/2}sin^2(\theta) ~ cos^2(\theta) ~ d \theta $

I'll let you do this one also and then collect all the pieces.

-Dan - Nov 3rd 2007, 08:48 AMSoroban
Hello, kittycat!

A variation on Dan's solution . . .

Quote:

$\displaystyle 2 \int^1_0\,(2t-3t^2)\cdot\sqrt{1-9t^2}\,dt$

. . and: .$\displaystyle \sqrt{1-9t^2} \:=\:\cos\theta$

Substitute: .$\displaystyle 2\int\left[2\left(\frac{1}{3}\sin\theta\right) - 3\left(\frac{1}{3}\sin\theta\right)^2\right]\cdot\cos\theta \left(\frac{1}{3}\cos\theta\,d\theta\right) $

. . $\displaystyle = \;\;\frac{2}{9}\int(2\sin\theta - \sin^2\!\theta)\cos^2\!\theta\,d\theta \;\;=\;\;\frac{4}{9}\int\cos^2\!\theta\sin\theta\, d\theta - \frac{2}{9}\int\sin^2\!\theta\cos^2\!\theta\,d\the ta $

For the first integral, use: $\displaystyle u \,=\,\cos\theta$

For the second integral, we have:

. . $\displaystyle \frac{1}{4}\left(4\sin^2\!\theta\cos^2\!\theta\rig ht) \;=\;\frac{1}{4}\left(2\sin\theta\cos\theta\right) ^2 \;=\;\frac{1}{4}\!\cdot\!\sin^2\!2\theta \;=\;\frac{1}{8}(1 - \cos4\theta)$

Can you finish it?

- Nov 3rd 2007, 08:55 AMtopsquark
- Nov 3rd 2007, 09:21 AMkittycat
what about for this integral,

the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much. - Nov 3rd 2007, 09:30 AMtopsquark
- Nov 3rd 2007, 09:46 AMkittycat
Quote:

Originally Posted by**kittycat**http://www.mathhelpforum.com/math-he...s/viewpost.gif

*what about for this integral,*

the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.

Instead of using the sine function, use the tangent function. (http://www.mathhelpforum.com/math-he...632bea41-1.gif.)

-Dan

Hi Dan,

Could you please show me how to let the tangent function in this question? Sorry to ask you because I forgot the trig substitution. Thank you very much. - Nov 3rd 2007, 09:56 AMtopsquark
The same way that Soroban did, though in this case there are no restrictions due to the limits. So

$\displaystyle \int t^2\sqrt{1 + 9t^2}~dt$

Let $\displaystyle 3t = tan(\theta) \implies 3 dt = sec^2(\theta)~d \theta$

So

$\displaystyle \int t^2\sqrt{1 + 9t^2}~dt = \int \frac{1}{9}tan^2(\theta) \sqrt{1 + tan^2(\theta)} ~ sec^2(\theta)~d\theta$

$\displaystyle = \frac{1}{9} \int tan^2(\theta) sec^3(\theta)~d\theta$

-Dan - Nov 3rd 2007, 07:29 PMkittycathelp me again please
http://www.mathhelpforum.com/math-he...256dda3d-1.gif

Hi,

I might be wrong in the process of calculation. :(But don't know why can't get the answer (given by the textbook) from your approach.

The answer given by the textbook is -11/108 *sqrt(10) -1/36*ln(sqrt(10)-3) -4/27

help me again please:D - Nov 4th 2007, 02:00 AMtopsquark
- Nov 4th 2007, 08:15 AMkittycat
he same way that Soroban did, though in this case there are no restrictions due to the limits. So

http://www.mathhelpforum.com/math-he...4654d32f-1.gif

Let http://www.mathhelpforum.com/math-he...1814665f-1.gif

So

http://www.mathhelpforum.com/math-he...ae7cc688-1.gif

http://www.mathhelpforum.com/math-he...ab7e230e-1.gif

========================

*Referring to : the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.*

I got to the same answer as you posted. But how should i evaluate it if it is =>*the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.*

Could you please guide me to the final answer of this integration?

Thank you very much. (Handshake)