How much harder is Calculus II from Calculus I?

Question

I took the AP Calculus AB test passing with 4. IM going to college this year retaking Calculus I. Now I was wandering If I perfect knowledge more or less of Calculus I will Calculus II be some what easier or is totally different from Calculus I?

Answer 1

Don't retake Calc 1. I was in your position last year with a 4 on the AP. I retook it but I wish I had skipped it because it was a waste of time..

As for Calc 2, it draws on some concepts from Calc 1 but it is much harder. It's about the same amount of work though.

Answer 2

It definitely depends where you're taking your courses, but I'm actually doing Calculus II this semester and I'm not find it that much harder than Calculus I. I didn't take the AP tests (my school never offered them), but from what I understand, the AB test covers roughly the same material if not a little more than typical Calculus I courses at college. You did well on that test, so it is assumed that you have at least a decent understanding of the Calc I concepts...and that's all you need for Calc II really. From my experience, there are a few things you should be familiar with for Calc II

1) Trig identities - not the terribly complex one, but you should be able to identify the Pythagorean identities and double-angle formulas.

2) Know differentiation in and out - including the trig derivatives. This does help A LOT - but if you don't know them, that doesn't mean you need to retake Calc I...just brush up, and you'll pick it up through practice.

3) Know the basic integration techniques and be able to identify when to use them. Calc II makes this harder by throwing in new rules and strategies, but as long as you have the foundations, you should be fine.

Granted, I enjoy this kind of stuff, and I don't know what level of interest you have. But if you work at it, you should be fine. Hope that helps some =)

Answer 3

First, take the by-product. f '(x) = (2x²-a million)e^(x³-x) Set this equivalent to 0 to come across the brilliant factors (minutes/maxes) 0 = (2x²-a million)e^(x³-x) First we are able to workout that this exponential function isn't 0 (look on the graph). so we are able to purely look at 0 = 2x²-a million a million = 3x² a million/3 = x² x = a million/sqrt(3) and x = -a million/sqrt(3) because x = -a million/sqrt(3) is between -a million and nil, it extremely is both a optimum or minimum. to study no count number decision if it extremely is a max or min, study factors on the point of to it to the left and precise. f(0) = a million f(-a million/sqrt(3)) = a million.469 f(-a million) = a million once you concentration on that x = -a million/sqrt(3) is larger, it extremely is a optimum. to come across the minimum, because there on the on the spot aren't any more the various severe factors, really study the endpoints (x=0, x=-a million) We already checked that above, and both endpoint yields a fee of one million. So the utmost is f(-a million/sqrt(3)) = a million.469 and the minimum f(-a million) = f(0) = a million Max = a million.469 Min = a million