Kneeling down and begging a high level math expert to help analyze a question, I answered 20 points

This problem can be solved by knowing that df(u,v)dudv=p is a fixed-value problem.

Because in a fixed range d, the result is the same regardless of the way of integration.

So f(x,y)=xy+p

df(u,v)dudv= d[xy+p)dxdy=p is an equation of p, and the result can be solved.

The result of a definite integral is a fixed value (e.g., the volume of a curved top cylinder), so let df(u,v)dudv=a

Then: f(x,y)=xy+a(1).

Then bring Eq. (1) to df(x,y)dxdy (the integral variable changes the value unchanged, i.e., df(u,v)dudv= df(x,y)dxdy).

Get: d(xy+a)dxdy= dx (xy+a)dy (the upper and lower limits of the first integral number are 1 and 0, and the upper and lower limits of the first integral number are x and 0 respectively).

That is: the original formula = (x 5 2 + ax 2) dx = x 6 12 + ax 3 3 (the upper and lower limits are 1 and 0 respectively) = 1 12 + a 3 = a

A=1 8 choose c

Full and detailed process rt ......Hope it helps you solve your problem.

1.To find the break point, there is a limit to this problem, and you need to find the limit first.

2.When finding the limit, it is necessary to discuss the situation of x.

3.There are two discontinuities.

and x=-1 are discontinuable points.

Note: |x|> 1, when finding the limit, divide the numerator denominator by the 2n power of x, and then find it again.

1. Find lim(1-x 2n 1+x 2n)x, (n->

f(x)= 0 when x=0 or x= 1

x, when 0 x 1 or x -1

x when -1 x 0 or x 1 (3 cases in total).

2. Then let's find the break:

From the above interval, we can see that there are three "key points", -1;

1) Let's look at 0 first: From the above interval, you can see that limf(0)=limf(x) (x->0+)=limf(x) (x->0-).

So f(x) is continuous at (-1,1), and 0 is not a discontinuity;

2) Look at 1 again: f(1)=0, limf(x)(x->1-)=x=1, limf(x)(x->1+)=-x=-1

f(1)≠limf(x)(x->1-)≠limf(x)(x->1+);So x=1 is the first type of discontinuity;

3) Similarly, -1: f(-1)=0 , limf(x)(x->-1-)=x=-1 , limf(x)(x->-1+)=-x=1

f(-1)≠limf(x)(x->-1-)≠limf(x)(x->-1+);So x=1 is the first type of discontinuity;

3. Conclusion: x=1 and x=-1 are the first type of discontinuities; The continuous intervals of f(x) are (- 1), (1,1), (1,+).

(1)。Find the plane equation through the line l:x 3=(y-1) 2=(z-2) 1 perpendicular to the plane :x+y+z+2=0.

Solution: the direction vector n = of the straight line l; normal vector n = of the plane of ; Let the plane sought be , then

The normal vector n=n n

The equation for the plane is x-2(y-1)+(z-2)=x-2y+z=0;

2). z=f(u,v); u=x²-y；v=y²-x；Find z x, z y;

The step circled by the red pen is to divide the integral interval [0, 2] into [0, 2] and [ 2, the sum of these.

The last step directly substitutes the definite integral formula (Varys formula) of (sint) n, if you don't know this formula, you can convert it into dcost, and convert the remaining (sint) 8 into [1-(cost) 2] 4 reintegrals.