Integrate...?

Question

How do you integrate 1/(2y) with respect to y?

Answer 1

(1/2)lny+c

Answer 2

Others have given you the correct answer but there are a couple of learning points to be stressed.
1. Don't let the y worry you. You get so used to being asked about x that when you seen another symbol it can throw you.

2. If there are brackets round the 2y in the question you have been set don't let that throw you either. 1 divided by 2y is the same as 1/2 times 1/y. This helps to see the 1/2 is a constant.

3. That leaves the good old 1/y (or 1/x or 1/P etc.) which is of course In y. DON'T make the mistake of converting it to y to the power minus 1 (y^-1).

Answer 3

Remember, 1/(2y) can be written as the product:

1/2 x 1/y

since 1/2 (half) is just a constant, you can take it outside the integral sign.

So you're left with the integral of 1/y, which is ln (y).

*RULE* integral of 1/x = ln (x).
note that ln = log to base e (natural log).

So, your answer is 1/2 (outside the integral) x ln (y)
Ans = 1/2 ln y (half ln y) + c

where c is an arbitary constant. (constant of integration)

Answer 4

1/(2y)
= (1/2)(1/y)

Integration of (1/2)(1/y)
= 1/2 lny + C

Answer 5

1/y integrated becomes ln(y) +c, so 1/2y integrated becomes (1/2)ln(y) + c

Answer 6

int 1/(2y) = 1/2 int (1/y) = 1/2 ln (y) + C

Answer 7

For y > 0, the antiderivative is (1/2)ln y + c.
But for y < 0, the antiderivative is (1/2)ln(-y) + c.

Answer 8

int(1/(2y)dy
=(1/2)int(dy/y)=lny+C

where C is a constant

and ln is natural logs

i hope that this helps

Answer 9

This is just (1/2)Int[(1/y)dy] which is (1/2)ln|y| + C