# Finding antiderivative of (x^2 - 47x + 33)/10

1. May 19, 2007

### 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. $$y=(-.03x-.18)^3+x$$
$$y=(\frac{\-.03x^2}{2}-.18x)^3+(x^2/2)$$

Last edited: May 19, 2007
2. May 19, 2007

### Vagrant

For the second one write out the cubic expression then integrate

3. May 19, 2007

### Ry122

do you mean simplify it? Is there a quicker way?

4. May 19, 2007

### d_leet

Make the substitution u=x-6

5. May 19, 2007

### Office_Shredder

Staff Emeritus
Something important (though not terribly relevant to integration per se) is your first step in number 2 is wrong, because you forgot to take the cube root of -.03 before bringing it inside the parentheses. Beyond that, your method is wrong anyway, and you need to do what was suggested above (either suggestion works)

6. May 19, 2007

### Gib Z

~sigh~ Slight double post there Ry122, I already posted on your other thread name integration in this section...

7. May 19, 2007

### Ry122

y=-.03(x-6)^3
u=x-6
y=-.03u^3
integral of u = (x^2/2)-6x+C
integral of y=(-.03u^4/4)
Since the derivative is dy/dx=du/dx * dy/du
is the integral du/dx divide dy/du ?

Last edited: May 19, 2007
8. May 19, 2007

### Ry122

After using this calculator http://integrals.wolfram.com/index.jsp I found that
in y=.03(x-6)^3 the exponent becomes 4 and .03 is divided by 4.
This is the first part of the chain rule but nothing is done within the brackets.
Can someone please explain why this is?

9. May 19, 2007

### Gib Z

Your integral should be worked out like this :

$$\int \frac{-3}{100} (x-6)^3 dx = \frac{-3}{100} \int (x-6)^3 dx$$
Let u = x-6, then du = dx
$$\frac{-3}{100} \int u^3 du = \frac{-3}{100} \frac{u^4}{4} + C = \frac{ -3}{100} \frac{(x-6)^4}{4} + C$$

10. May 22, 2007

### Ry122

That's not correct Gib Z. I know because I just attempted to calculate a definite integral using it. My one gives the correct answer. Im just wondering, is there a term used to describe the chain rule when it is used in reverse with integration? Would it be correct to just call it the chain rule?
Edit: Sorry my mistake, they both give the same answers.

Last edited: May 22, 2007
11. May 22, 2007

### Gib Z

Well I know my solution to the integral i posted in my previous post is correct. When you tried to calculate the definite integral, did you take into account my integral only accounts for the first term in question 2?