### Scribe for Friday

Friday, first we answer some questions in the Iflurtz questionaire for Valentines Day. Then for our class we start by answering this question.

To answer this question is that you make it to a simpler equation first before you antideffrientiate it. So it would be

Then Mr. K showed us this

So if were given this:

However, if the 3 is not there we can put the 3 there but in order to make it equals to 1 we should multiply it with 1/3. And we can use this relationship:

*(x+x^2)^2 dx.***S**To answer this question is that you make it to a simpler equation first before you antideffrientiate it. So it would be

**(x^2+2x^2+x^4) dx. Now its easy to antideffrientiate it, so it would become x^3/3+x^4/2+x^5/5 + C.***S*Then Mr. K showed us this

**F(x)= f(g(x))****F'(x)= f'(g(x)) * g'(x)**So if were given this:

*S***3x^2****sin (x^3)**dx we can say that it is equals to this**- cos (x^3)+ C**, because of the relationship above. Its like undoing the chain rule.However, if the 3 is not there we can put the 3 there but in order to make it equals to 1 we should multiply it with 1/3. And we can use this relationship:

**So if this is given***S*k f(x) dx= K S f(x) dx.**, we can put 3 in front of x^2 so that it would look like its derivative. But we should put 1/3 in front of***S*x^2 sin (x^3) dx**too to make it balance.***S***1/3***S***3x^2 sin(x^3)**dx then the answer would become**- 1/3 cos (x^3) +C .**And that's what we learn last friday....the next scribe is xun...

## 1 Comments:

At 10:44 PM CST, Mr. Kuropatwa said…

Great scribe post Prince! Outstanding use of colour. Bravo!!

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