Integration by substitution (1 Viewer)

Aysce · Dec 17, 2011

Hey guys, question is:

$\int_{0}^{4} x\sqrt{16-x^2}dx$ using the substitution, u= 16-x^2

Answer is 64/3. Thanks in advance

hayabusaboston · Dec 17, 2011

wtf spiral hasnt answered it already...? that's a surprise.... Just wait for him, he puts it in latex, or whatever it's called, which makes it easier to understand.

Alkanes · Dec 17, 2011

I will try to beat Spi to this question.

hayabusaboston · Dec 17, 2011

Alkanes said:
I will try to beat Spi to this question.

Yes but are you fluent in latex man? haha. Where do you find this latex program anyway? How does spiral put it in such a neat and professional way?

Alkanes · Dec 17, 2011

Let u = $16-x^2$
$\frac{du}{dx}=-2x$
$dx = \frac{du}{-2x}$
$\int_{0}^{4}x\sqrt[]{u}.\frac{du}{-2x}$
$\frac{-1}{2}.[\frac{2u^{\tfrac{3}{2}}}{3}]_{16}^{0}$

And just minus u should get the answer from there.

hayabusaboston · Dec 17, 2011

Alkanes said:
Let u = $16-x^2$
$\frac{du}{dx}=-2x$
$dx = \frac{du}{-2x}$
$\int_{0}^{4}x\sqrt[]{u}.\frac{du}{-2x}$
$\frac{-1}{2}.[\frac{2u^{\tfrac{3}{2}}}{3}]_{0}^{4}$

And just minus u should get the answer from there.

(Okay I see you're a fluent latex man too.... jealous much....

hayabusaboston · Dec 17, 2011

Alkanes said:
Let u = $16-x^2$
$\frac{du}{dx}=-2x$
$dx = \frac{du}{-2x}$
$\int_{0}^{4}x\sqrt[]{u}.\frac{du}{-2x}$
$\frac{-1}{2}.[\frac{2u^{\tfrac{3}{2}}}{3}]_{0}^{4}$

And just minus u should get the answer from there.

$\int_{7}^{77}x\sqrt[]{ap^2}.\frac{....}{hello}$

Alkanes · Dec 17, 2011

Umm alternatively you can make 1/2 become positive by switching the limits around.

hayabusaboston · Dec 17, 2011

hayabusaboston said:
$\int_{7}^{77}x\sqrt[]{ap^2}.\frac{....}{hello}$

OMG that's how you do it!!! YEAAA

Aysce · Dec 17, 2011

I did that and got -8/3 :S

Alkanes · Dec 17, 2011

uhh

$\frac{-1}{2}[0-\frac{128}{3}]$

$\frac{64}{3}$

Maybe calculator error for you

pokka · Dec 17, 2011

pokka · Dec 17, 2011

Aysce said:
I did that and got -8/3 :S

this is because you forgot to change the limits

Alkanes · Dec 17, 2011

Oh dammit i forgot about the limits. My bad.

Alkanes · Dec 17, 2011

Terrible mistake by me... Ignore what i said i was lucky to get the answer for this question though

pokka · Dec 17, 2011

It's ok we can all learn from our mistakes

Aysce · Dec 17, 2011

NO WONDER =.= thanks guys

SpiralFlex · Dec 17, 2011

hayabusaboston said:
wtf spiral hasnt answered it already...? that's a surprise.... Just wait for him, he puts it in latex, or whatever it's called, which makes it easier to understand.

I am not on this forum 24/7 you know....

asianese · Dec 18, 2011

SpiralFlex said:
I am not on this forum 24/7 you know....

You used to be =D

Integration by substitution (1 Viewer)

Aysce

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