What are the antiderivatives of y=\frac{\ x^2-47x+33}{10} and y=-.03(x-6)^3+x?

[Ry122]

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

 

[trajan22]

Oh I guess you edited it since I last saw it. The first integral is correct now. For the second just try a substitution.

 

[Gib Z]

Come on guys there are easy ones!

1. $$\int \frac{\ x^2-47x+33}{10} dx = \frac{1}{10} \int x^2 - 47x + 33 dx = \frac{1}{10} ( \int x^2 dx - \int 47x dx + \int 33 dx)$$

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 doesn't seem correct.

$$\int \frac{-3}{100}(x-6)^3 + x dx = \frac{-3}{100} \int (x-6)^3 dx + \int x dx$$.

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