| Thread Closed |
Antiderivatives |
Share Thread | Thread Tools |
| Oct30-05, 11:49 AM | #1 |
|
|
Antiderivatives
I am taking an Architectural Geometry class, and have only had Precal. We just started antiderivatives (I understand regular derivatives), and had a question:
I have to find the antiderivative of (-5/12 x^4) + (10/3 x^3) - (103/12 x^2) + (23/3 x) I think I use the power rule for antiderivates...so far i have the first function to be (-5/12 x^5)/(5) is that right? if so, does it simplify to -.083x^5? This is where i get confused. Later on I will need to plug in a variable for x. I just wanted to make sure that I am doing this right so far. Thank you. |
| Oct30-05, 12:36 PM | #2 |
|
|
yes that is right but if you simpifly it it is just -x^5/12
|
| Oct30-05, 01:06 PM | #3 |
|
|
Ok, thanks for the reply. If that's the case, does that make the rest of the equation:
(-x^5) / 12 + (5x^4)/6 - [(103x^3)/12) / 3] + [(23x^2)/3) /2 ] Thanks again. |
| Oct30-05, 01:19 PM | #4 |
|
|
Antiderivatives
Yes, but you can simplify. I'll take one of your terms.
What's [tex]\frac{103}{12}*\frac{x^3}{3}[/tex]? |
| Oct30-05, 04:12 PM | #5 |
|
|
it would be 103x^3 / 36 . Or do you want it simplified more?
|
| Thread Closed |
| Thread Tools | |
Similar Threads for: Antiderivatives
|
||||
| Thread | Forum | Replies | ||
| What is the antiderivative of x^2 + (5/x^5) +cos9x - e^-7x? | Calculus | 3 | ||
| Antiderivatives | Calculus & Beyond Homework | 6 | ||
| antiderivatives | Calculus & Beyond Homework | 3 | ||
| How to get (secx)(tanx) from (1/cosx)(sinx/cosx)? | Calculus | 8 | ||
| antiderivatives | Calculus & Beyond Homework | 2 | ||