What a coincidence, I am also a computer major. I'll hand in a copy as soon as I finish it.

First copy:

One.

1.D. Analysis: the integral value is constant, and the differential value of the constant is 0.

2.b. Analysis: X 3+X is an odd function, so the original function is even. And the integral region is symmetrical, so the value is 0.

3. Answer. Analysis: The image is a circle with the origin as the center and a radius of 2. And 0

4. C. Analysis: If you directly substitute x=0, the integral area is 0 to 0, and there is a 0/0 type, so use the Robida rule.

5. Answer. Analysis: Let F(t) be the original function of sint^2, then f' (x) = Sinx 2.

6.d. Analysis: When q= 1, the original function is lnx, which diverges.

7.d. analysis: divide two sides by 6 to get the standard equation.

8. B. Analysis: It is a circle on the xoz plane.

Two.

1. 1。 Analysis: the fifth sub-topic takes the first big question.

2.2π^3/2。 Analysis: the original function can be found, and the substitution value is enough. Original function: x 3/3-cosx.

3. 1/2。 Analysis: same as above, original function: -e (-2x)/2.

4. 1。 Analysis: original function: xlnx-x

5. 1。 Analysis: Original function: (x- 1) e x

6.-(x-3)+3(z-2)=0. Analysis: concept, no explanation.

7.3 (x-1)+4 (y-2)-1(z-3) = 0. Analysis: the normal vector of a plane is (3,4,-1).

8. (x-4)/2 = (y+1)/-3 = (z+2)/7. Analysis: The normal vector of a plane is (2,-3,7).

Three.

1. Original formula =(-xcosx+sinx)|0 to π/2 = 1.

2. The original formula =-cos2x/4|0 to π/2 = 1/2.

3. The original formula = (lnx) 2/2 | 1 to e = 1/2.

4. Original formula =arctanx|0 to ∞ = π/2 = (e- 1)/2.

6. The original formula = (x 2 * lnx/2-x 2/4) |1to e = (1+e 2)/4.

7. The original formula =∫0 to1x 2dx+∫1+0 to 2e xdx = x 3/3 | 0 to 65438+e x | 1 to 2 = e 2-e+65438.

Four.

1.S=∫0 to 1 √ xdx = 2x (3/2)/3 | 0 to 1 = 2/3.

2.V=π∫0 to 1 (√ x) 2dx = π x 2/2 | 0 to 1 = π/2.

Five.

1.S=∫0 to1(ex-e (-x)) dx = (ex+e (-x)) | 0 to 1 = e+ 1/e-2.

2.V=∫0 to1[(e x) 2-(e (-x)) 2] dx = (e (2x)+e (-2x))/2 | 0 to 1 = (e 2)

The second episode:

One.

1.C 2。 D 3。 B 4.0 5。 A

Two.

1.x^2+y^2<； 1 and x≠0

2.2(x+y)e^(x^2+y^2)(dx+dy)

3.y/(e^y-x)

4.(2x-yz+2z+xy)dx+(2y-xz+2z+xy)dy

5.-2sin(x^2+y)-4x^2cos(x^2+y)

6.0 to 1 dy y to 1 f(x)dx

Three.

1.

δz/δx = 2 sin(x+y)cos(x+y)= sin[2(x+y)]

δz/δy = 2 sin(x+y)cos(x+y)= sin[2(x+y)]

Symmetry shows that:

(δz)^2/δx^2=(δz)^2/δy^2=(δz)^2/(δxδy)=2cos[2(x+y)]

2.

δz/δx=ye^xf 1'+f2'/y^2

δz/δy=e^xf 1'-2xf2'/y^3

3.

-ye^(-xy)-2δz/δx+e^zδz/δx=0

So δ z/δ x = y/[(e z-2) e (xy)]

-xe^(-xy)-2δz/δy+e^zδz/δy

So δ z/δ y = x/[(e z-2) e (xy)]

4.

zx=( 1+2x+4y+2y^2)e^(2x)； zy=2(y+ 1)e^(2x)

zxx=4( 1+x+2y+y^2)e^(2x)； zyy=2e^(2x)； zxy=4(y+ 1)e^(2x)

When zx = 0 and zy = 0, x = 1/2 and y =- 1.

zxx = A = 2e & gt0，zyy=C=2e，zxy=0

So AC-B 2 > 0. So there is a minimum value at (1/2,-1), and z=-e/2 is obtained by substitution.

5.

The original formula =∫0 to 1 dx∫0 to 2e (x+y) dy = e 3-e 2-e+ 1.

6.

The original formula =∫0 to 2 dx∫0 to 2x (x+y)dy = 32/3.

The third copy:

One.

1.D 2。 C 3。 B 4。 A

Two.

1.y=Ce^x+ 1

2.y=-ln(C-e^x)

3.y=C 1e^(4x)+C2e(-x)

4.f(x)=x-x^2/2+x^3/3+...+(- 1)^(n- 1)x^n/n+ ...

Three.

Power series expansion our teacher will talk about it briefly. I am weak, so I won't mislead you here. Skip.

Four.

1.

The characteristic equation is r 2+2r = 0, and the solution is r 1=0, r2=-2.

So the general solution is y = c 1+c2e (-2x).

2.

The characteristic equation is r 2+3r+2 = 0, and r 1=- 1.r2=-2.

So the general solution is y = c 1e (-x)+c2e (-2x).

3.

The characteristic equation is r 2+6r+9 = 0, and the solution is r 1=r2=-3.

So the general solution is y = (C 1+C2X) E (-3x).

4.

Dy/dx =1+x 2, where dy = (1+x 2) dx, and the integral on both sides is:

Y = x+x 3/3+c is a general solution.

When x=0, y= 1, which is replaced by:

C= 1

So y = x+x 3/3+1is a special solution.