integration

**kittycat** · August 17th 2007, 11:38 AM

2 times the integral (from -3 to 3) of (4-x) (sqrt(9-x^2) ) dx.

Thank you very much.

**CaptainBlack** · August 17th 2007, 11:40 AM

Originally Posted by kittycat

2 times the integral (from -3 to 3) of (4-x) (sqrt(9-x^2) ) dx.

Thank you very much.

You question is:

Evaluate:

$ \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx $

RonL

**CaptainBlack** · August 17th 2007, 11:43 AM

Originally Posted by CaptainBlack

You question is:

Evaluate:

$ \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx $

RonL

$ \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx = \int_{-3}^3 4\sqrt{9-x^2}~dx - \int_{-3}^3 x \sqrt{9-x^2}~dx $

and these two integrals can be done by the usual book methods, the first by a trig substitution and the second
by looking at the derivative of $(9-x^2)^{3/2}$ .

RonL

**topsquark** · August 17th 2007, 11:47 AM

$ \int_{-3}^3 (4-x) \sqrt{9-x^2}~dx $

Hint:
$= 4 \int_{-3}^3 \sqrt{9-x^2}~dx - \int_{-3}^3 x\sqrt{9-x^2}~dx$

You can do the second integral with a simple substitution. So let's look at the first term:
$\int_{-3}^3 \sqrt{9-x^2}~dx$

Let $x = 3y \implies dx = 3 dy$

Thus
$\int_{-3}^3 \sqrt{9-x^2}~dx = \int_{-1}^1 \sqrt{9-(3y)^2}~3dy$

$= 3\int_{-1}^1 \sqrt{9-9y^2}~dy$

$= 3\int_{-1}^1 3\sqrt{1-y^2}~dy$

$= 9\int_{-1}^1 \sqrt{1-y^2}~dy$

Does this integral look familiar?

-Dan

**Krizalid** · August 17th 2007, 12:19 PM

Or from the beggining you can use a simple trig. sub. defined by $x=3\sin u$

**Krizalid** · August 17th 2007, 12:24 PM

In general we have that

$\int\sqrt{a-x^2}\,dx=\frac12\left(x\sqrt{a-x^2}+a\arctan\frac x{\sqrt{a-x^2}}\right)+k$

**ThePerfectHacker** · August 17th 2007, 12:43 PM

Originally Posted by Krizalid

In general we have that

$\int\sqrt{a-x^2}\,dx=\frac12\left(x\sqrt{a-x^2}+a\arctan\frac x{\sqrt{a-x^2}}\right)+k$

I think you want to write $a^2$ (also mention that $a\not = 0$

, just to be safe, write $a>0$ ).

**topsquark** · August 17th 2007, 12:45 PM

Originally Posted by ThePerfectHacker

I think you want to write $a^2$ (also mention that $a\not = 0$

, just to be safe, write $a>0$ ).

And mention that a is a real number, and...

-Dan

**kittycat** · August 17th 2007, 01:10 PM

Hi Dan,

how do I end this one?

9 times the integral from -1 to 1 of sqrt(1-y^2) dy - your last step!

Really don't know how

Also,
Can anyone remind me how to do the trig substitution? I forgot it

! Please show me the steps . Thanks.

**Krizalid** · August 17th 2007, 01:13 PM

Set $y=\sin u\implies dy=\cos u\,du$

Now when you have all integrated you use the first substitution to turn the trig. expressions into $y$ expressions.

**kittycat** · August 17th 2007, 01:25 PM

hi Krizalid,

Please show me more ...! really forgot !

If possible , please show me the steps how to show the first integral by trig substitution. thank you very much.

**Krizalid** · August 17th 2007, 01:30 PM

Change of variables according to $y=\sin u$ implies

$\begin{aligned} \int {\sqrt {1 - y^2 } \,dy} &= \int {\sqrt {1 - \sin ^2 u} \cos u\,du}\\ &= \int {\cos u\cos u\,du}\\ &= \int {\cos ^2 u\,du} \end{aligned}$

Which is easy to take.

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