- #1
Ry122
- 565
- 2
I need help with finding the antiderivative of
1. [tex]y=\frac{\ x^2-47x+33}{10}[/tex]
and
2. [tex]y=-.03(x-6)^3+x[/tex]
My attempts:
1. [tex]y=.1[(x^3/3)-(47x^2/2)+33x)][/tex]
2. [tex]y=-.03x-.18^3+x[/tex]
[tex]y=(\frac{\-.03x^2}{2}-.18x)^3+(x^2/2)[/tex]
1. [tex]y=\frac{\ x^2-47x+33}{10}[/tex]
and
2. [tex]y=-.03(x-6)^3+x[/tex]
My attempts:
1. [tex]y=.1[(x^3/3)-(47x^2/2)+33x)][/tex]
2. [tex]y=-.03x-.18^3+x[/tex]
[tex]y=(\frac{\-.03x^2}{2}-.18x)^3+(x^2/2)[/tex]
Last edited: