java.lang.Math is just a port of what the C math library does.
For C, I think it comes down to the fact that CPU have special instructions to do Math.pow for floating point numbers (but not for integers).
Of course, the language could still add an int implementation. BigInteger has one, in fact. It makes sense there, too, because pow tends to result in rather big numbers.
ceil and floor by definition return integers, so how come they don't return ints
Floating point numbers can represent integers outside of the range of int. So if you take a double argument that is too big to fit into an int, there is no good way for floor to deal with it.
Videos
java.lang.Math is just a port of what the C math library does.
For C, I think it comes down to the fact that CPU have special instructions to do Math.pow for floating point numbers (but not for integers).
Of course, the language could still add an int implementation. BigInteger has one, in fact. It makes sense there, too, because pow tends to result in rather big numbers.
ceil and floor by definition return integers, so how come they don't return ints
Floating point numbers can represent integers outside of the range of int. So if you take a double argument that is too big to fit into an int, there is no good way for floor to deal with it.
From a mathematical perspective, you're going to overflow your integer if it's larger than 231-1, and overflow your long if it's larger than 264-1. It doesn't take much to overflow it, either.
Doubles are nice in that they can represent numbers from ~10-308 to ~10308 with 53 bits of precision. There may be some fringe conversion issues (such as the next full integer in a double may not exactly be representable), but by and large you're going to get a much larger range of numbers than you would if you strictly dealt with integers or longs.
On a similar topic, ceil and floor by definition return integers, so how come they don't return ints?
For the same reason outlined above - overflow. If I have an integral value that's larger than what I can represent in a long, I'd have to use something that could represent it. A similar thing occurs when I have an integral value that's smaller than what I can represent in a long.
When it's a power of 2 keep in mind that you can use a simple and fast shift expression: 1 << exponent
For example:
22 = 1 << 2 = (int) Math.pow(2, 2)
210 = 1 << 10 = (int) Math.pow(2, 10)
For larger exponents (over 31) use long instead:
232 = 1L << 32 = (long) Math.pow(2, 32)
BTW, in Kotlin you have shl instead of <<:
(Java) 1L << 32 = 1L shl 32 (Kotlin)
Integers are only 32 bits. This means that its max value is 2^31 -1. As you see, for very small numbers, you quickly have a result which can't be represented by an integer anymore. That's why Math.pow uses double.
If you want arbitrary integer precision, use BigInteger.pow. But it's of course less efficient.