How to weigh the abnormal ball in this question There are very few points and there is no reward, so

An age-old question, I'm coming.

Answer: 1 divide the 12 balls into three piles, (the first time with the balance) put the two piles on both sides of the balance, if the balance is balanced, the defective products are in the 4 of the third pile, and the rest are **; (with the balance for the second time) put the two ** and the two of the third pile on both sides of the balance, if the balance is balanced, the defective product is in the third pile of 2, if the balance is unbalanced, if the balance is balanced, the defective product is in the other 2 of the third pile; (the third time with the balance) and the same (the second time with the balance) can determine 1.

2 1 (the first time with the balance) If the balance is unbalanced, then the 4 of the third pile are **; Among the 8 defective products on the scale, 2 (the second time with the balance) among the 8 (the 4 that sink to the side when the balance is used for the first time are called "down"; The 4 on one side are called "up"), and then 2 "up" and 1 "down" are placed on one side of the balance, if the balance is balanced, the defective product is in the other 2 out of 8; (with the balance for the third time) the same as the first.

If you use the balance a second time, if the balance is unbalanced. In any case, there is only one possibility of one side or up or down, we analyze one, and the other possibility can be "up" and "down" reversed. If 2 "down" and 1 "up" are placed on one side of the scale, then the defective product cannot be 2 "down" on this side and "down" on the other side (because there is only one possibility whether the defective product is heavy or light), so the defective product is locked in the range of 1 "up" on this side or 2 "down" on the other side; (The third time with the balance) use the defective product to lock the 1 "up" on this side or the other side of the 2 "down" range for comparison, put the 2 "down" on both sides of the balance, if the balance is balanced, the defective product is balanced 1 "up", if the balance is unbalanced, it is 1 on the sinking side.

It seems that something is missing.

Six on one side. Go down the heavy one.

The light one does not need (six left).

The heavy one is further divided into two parts.

Go to the light one (3 left).

Take just one and don't weigh it.

The rest is then balanced on one side.

It's not measurable.

If the balance is uneven.

It's sinking.

I'm talking about what goes if it's bad.

If it's light, it's the other way around.

Divide it into 3 points equally, and if you don't have 4 points, the first time you find out the lightest of the 3 points, there must be a bad one, and then you can divide it into 2 points, 2 points each, and find the lightest point with the bad one, and finally find the lightest of the last 2 points.

There is no direct formula. Use the partial integral method.

xsinxdx = - xdcosx

xcosx + cosxdx = -xcosx + sinx + c

Hello classmates, because the red line is to give the area d integral, d is a circle with respect to the x-axis and y-axis symmetry, so -x 2 is an odd function with respect to the y-axis, -y 2 is an odd function with respect to the x-axis, and the integral of the odd function is 0, so the term -(x+y) 2 is gone.

I don't understand, I don't feel like it's actually necessary, 2 is not a breakpoint, the estimated answer is pumped, and the answer is actually the same as directly calculating 1 to positive infinity.

Only if the integrand is analytic can have such a representation, because the integral of the analytic function is path-independent. In this case, the Newton-Leibniz formula in the higher definite integral still holds, so I can only say that it really works the same way as the higher definite integral! In order for this kind of integral to become the formula you are talking about, you have to give an arbitrary path.

t=-u,,Of course, the upper and lower limits change.