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

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

 

[Vagrant]

For the second one write out the cubic expression then integrate

 

[Ry122]

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

 

[d_leet]

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

Make the substitution u=x-6

 

[Office_Shredder]

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)

 

[Gib Z]

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

 

[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 ?

 

[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?

 

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

 

[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. I am 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.

 

[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?