- #1

- 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]

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