Scribe for Friday
To answer this question is that you make it to a simpler equation first before you antideffrientiate it. So it would be S (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.
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: S k f(x) dx= K S f(x) dx. So if this is given S x^2 sin (x^3) 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 too to make it balance. 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...