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Mathematics 19 Online
OpenStudy (preetha):

How do you integrate 4x^2 ?

OpenStudy (anonymous):

how do you get a prism as your 99?

OpenStudy (preetha):

I am a subscriber. You should subscribe too! It helps OpenStudy!

OpenStudy (anonymous):

Is that you in the picture?

OpenStudy (anonymous):

I seen you on here in 2012

OpenStudy (anonymous):

2013 something like that

OpenStudy (anonymous):

do you make money being on here?

OpenStudy (usukidoll):

uhhh we need an integral from a to b .... first unless we just take antiderivative of 4x^2 ?

OpenStudy (perl):

$$ \Large \rm { \int 4x^2 ~dx = 4 \cdot \int x^2~ dx }$$

OpenStudy (perl):

Now we can use power rule

OpenStudy (usukidoll):

then take antiderivative of x^2 which is add one to the exponent and divide by the new exponent

OpenStudy (usukidoll):

and don't forget the +C at the end.

OpenStudy (anonymous):

cos^2(4x) = (1 + cos8x)/2 Therefore, integration of cos^2(4x) = Integration of (1+cos8x)/2 = 1/2 [Integration of (1) + Integration of (cos 8x)] = 1/2 [x + sin8x/8 ] = x/2 + (sin8x)/16

OpenStudy (usukidoll):

HUH!

iYuko (iyuko):

\[\int\limits_{?}^{?}4x ^{2} dx=\] \[\frac{ 4x ^{3} }{ 3 }\]

OpenStudy (perl):

$$ \Large \rm { \int x^n ~dx= \frac{x^{n+1}}{n+1} + C \\~\\\therefore\\ 4\int x^2 ~dx= 4\cdot \frac{x^{2+1}}{2+1}+C= 4\cdot \frac{x^3}{3} + C }$$

OpenStudy (preetha):

That is truly epic! Thanks perl!

OpenStudy (perl):

Your welcome ☺ ☺

jabez177 (jabez177):

"You're"

OpenStudy (perl):

woops. You're*

iYuko (iyuko):

Can I subscribe for only 1 month? And do they take Pre paid visa cards?

OpenStudy (usukidoll):

but you're not supposed to give direct answers perl :O

iYuko (iyuko):

>.> <.<

OpenStudy (perl):

That's true.

OpenStudy (perl):

But the question was asking how to do something, it doesn't look like homework.

OpenStudy (usukidoll):

busted....and getting medals too... XD

OpenStudy (usukidoll):

anyone out there know matlab ughhh I am stuck on a code that should be straight forward but gives errors.

OpenStudy (anonymous):

Wow. A moderator giving direct answers. Tsk, tsk. Guess it must be alright for the rest of us too.

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