# Calculus problem

1. Mar 1, 2016

### chwala

1. The problem statement, all variables and given/known data given$dm/dt=120(t-3)^2$ find m(t)

2. Relevant equations

3. The attempt at a solution
now i have a problem here the first attempt $∫120(t-3)^2dt$
= $120∫(t^2-6t+9)dt$
=$120(t^3/3-3t^2+9t)+k$
but if you use chain rule you have
$∫120(t-3)^2dt$ let $u = t-3$then it follows that $du=dt$ thus we have
$∫120u^2du$ =$(120u^3)/3 + k$=$(120(t-3)^3)/3$ this second solution clearly is the correct answer having four terms + a constant, why is the first solution wrong ? its having three terms + a constant?

2. Mar 1, 2016

### Samy_A

Both seem correct to me. The fourth term in your second solution is just a number, so you can absorb it in the integration constant.

3. Mar 1, 2016

### Ssnow

The solution are the same if in the second method you call $-9\cdot 120 + k = c$ a constant, both solution are of the same form $120(t^3/3-3t^{2}+9t) +k$ and both differ only by a constant term...

4. Mar 1, 2016

### chwala

can one take the first solution as the answer to the differential equation?

5. Mar 1, 2016

Yes

6. Mar 1, 2016

### chwala

Thanks Sam and snow. regards