Find all antiderivatives for the following derivative dy/dx=80x^3
Ignore the constant factor for now since you know that it just passes through differentiation unharmed. Do you know the antiderivative of \(x^3\)? If not, what about \(4x^3\)?
well I think that for those I would use the rule where it would be 3x^2 and 12x^2?
Those are the *derivatives* not antiderivatives :-) What do you have to take the derivative of to get \(4x^3\)?
make it negative? I'm extremely lost if you cant tell haha
Hm, what about \(x^4\)? $$\frac{d}{dx}x^4=4x^3$$So then it makes sense that:$$\int4x^3\,dx=x^4+C$$... right? Well, now, consider that we can play with our integral a little bit:$$\int80x^3\,dx=20\int4x^3\, dx=20(x^4+C)=20x^4+C$$
This is just those two rules you learned earlier:$$\int kf(x)\,dx=k\int f(x)\,dx$$ i.e. you can pull out a constant factor from your integral, and$$\int x^n\,dx=\frac1{n+1}x^{n+1}+C$$
So we start with \(x^3\to\frac14x^4+C\) and then we multiply by our \(80\) we pulled out earlier to give us \(80(\frac14x^4+C)=20x^4+C\)
Okay so when it tells us to find all of them its really just to trick me up?
The "all" part is that integration constant \(C\) -- it can be literally any number. By putting it there you take care of every possible antiderivative, e.g. \(20x^4\), \(20x^4-\pi\), \(20x^4+333333\), etc.
Oh....why is calculus so confusing!?
Join our real-time social learning platform and learn together with your friends!