Finding the antiderivative

  • Thread starter Ry122
  • Start date
  • #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]
 
Last edited:

Answers and Replies

  • #2
132
1
Oh I guess you edited it since I last saw it. The first integral is correct now. For the second just try a substitution.
 
Last edited:
  • #3
Gib Z
Homework Helper
3,346
6
Come on guys there are easy ones!!

1. [tex]\int \frac{\ x^2-47x+33}{10} dx = \frac{1}{10} \int x^2 - 47x + 33 dx = \frac{1}{10} ( \int x^2 dx - \int 47x dx + \int 33 dx)[/tex]

Please tell me you know the reverse power rule?

Edit: Sorry I just saw your attempt. It is correct.

2. Your attempt at the 2nd one doesnt seem correct.

[tex]\int \frac{-3}{100}(x-6)^3 + x dx = \frac{-3}{100} \int (x-6)^3 dx + \int x dx[/tex].

If you really need it, for that first integral do substitution u= x-6
 

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