Integrating...am I doing this right?

[Lizzie]

Ok, I am trying to learn how to integrate by myself. So, I found out how to do it.
\(\displaystyle \int{}{x^n}{dx}{=}\frac{x^{n+1}}{_{n+1}}+c\)

so, if I am trying to integrate \(\displaystyle y=f(x)=16-x^2\), I know that the x² ends up being \(\displaystyle \frac{-x^3}{3}\)
But what do I do with the 16?

*Ha, I am getting the hang of this LaTex thing!

 

[Lizzie]

AHA! Would it be \(\displaystyle 128-\frac{x^3}{3}+C\)??

 

[Lizzie]

No, I went onto the integrator site that I've been using and typed in what I was trying to integrate, just to see if I had it right, but it said that the correct integration was \(\displaystyle 16x-\frac{x^3}{3}\), so now I know how to do that.

 

[stapel]

. . .\(\displaystyle \large{\int{\mbox{ }16\mbox{ }}dx}\)

. . . . .\(\displaystyle \large{=\mbox{ }\int{\mbox{ }16\mbox{ }x^0\mbox{ }}dx}\)

. . . . .\(\displaystyle \large{=\mbox{ }16\mbox{ }\frac{x^1}{1}\mbox{ }+\mbox{ }C}\)

. . . . .\(\displaystyle \large{=\mbox{ }\frac{16}{1}\mbox{ }x^1\mbox{ }+\mbox{ }C}\)

. . . . .\(\displaystyle \large{=\mbox{ }16x\mbox{ }+\mbox{ }C}\)

Eliz.

 

[Lizzie]

Yay! I understand that! lol.

 

[mad_mathematician]

16, once integrated, becomes 16x