Find ∫ l DX dy + ydz, where l is the directed closed broken line ABCA, where a, B and C are (1,0,0), (0,1,0), (0,0,1) respectively

Find ∫ l DX dy + ydz, where l is the directed closed broken line ABCA, where a, B and C are (1,0,0), (0,1,0), (0,0,1) respectively

∫＝∫AB＋∫BC+∫CA.
In AB: DZ = 0. X + y = 1. Dy = - DX
∫ AB = ∫ [1,0] 2DX = 2x in [1,0] value difference = - 2
In BC: DX = 0. Y + Z = 1dy = - DZ. Y = 1-z
∫ BC = ∫ [0,1] (1 + 1-z) DZ = (2z-z & sup2 / 2) in [0,1] value difference = 3 / 2
In CA: y = 0. Dy = 0. X + Z = 1. DX = - DZ
∫ CA = ∫ [0,1] (1-0) DX = x in [0,1] value difference = 1
∫l dx-dy+ydz＝-2+3/2+1＝1/2.

Find ∫ SiNx / (1 + SiNx) DX

∫ SiNx / (1 + SiNx) DX = ∫ (SiNx + 1-1) / (1 + SiNx) DX = ∫ 1 DX - ∫ 1 / (1 + SiNx) DX, the numerator denominator of the next integral is divided by cosx = x - ∫ secx / (secx + TaNx) DX = x - ∫ 1 / (secx + TaNx) d (TaNx) = x - ∫ 1 / (1 + 2tanx) d (TaNx) = x - (1 / √ 2) ∫ 1 / (1 + 2tanx) d (√ 2tanx) = x - (1 / √ 2) arctan (√ 2tanx) + C

F (x) = = ∫ SiNx (e ^ SiNx) DX (upper limit x + 2 π, lower limit x)?
A is positive. B is negative. C is always zero. D is not constant

The integrand has a period of 2 π, so f (x) = ∫ (x → x + 2 π) SiNx (e ^ SiNx) DX = ∫ (0 → 2 π) SiNx (e ^ SiNx) DX = ∫ (- π → π) SiNx (e ^ SiNx) DX = ∫ (- π → 0) SiNx (e ^ SiNx) DX + ∫ (0 → π) SiNx (e ^ SiNx) DX the former is replaced by T = - x = - ∫ (0 → π) Sint (e ^ (...)

If the vertex of the parabola is at the origin, the focus is on the y-axis, and the distance from a point P (m, - 3) on the parabola to the focus is 5, then the Quasilinear equation of the parabola is ()
A. y=4B. y=-4C. y=2D. y=-2

According to the vertex of the parabola is at the origin and the focus is on the y-axis, we can see that the opening of the parabola is downward. Let the parabola equation x2 = - 2PY. According to the definition of the parabola, we can see that 3 + P2 = 5, х P = 4; х the parabola equation is x2 = - 8y, х the Quasilinear equation of the parabola is y = 2, so we choose C

The distance from a point a (- 3, m) on the parabola y ^ 2 = - 2px to the focus is 5

The Quasilinear of parabola is x = P / 2
Distance from a to focus = distance from a to guide line = 3 + P / 2 = 5
So p = 4
So the parabolic equation is y ^ 2 = - 8x
Quasilinear equation x = 2
Because a is on a parabola
So m ^ 2 = - 8 × (- 3) = 24
That is, M = ± 2 √ 6

The focal point is f (0, - 8), the Quasilinear is y = 8, and the standard equation of the parabola is f (0, - 8)______ ．

Let the standard equation of parabola be: x2 = - 2PY (P > 0), ∵ the Quasilinear equation of parabola is y = 8, ∵ P2 = 8, ∵ P = 16, ∵ the standard equation of parabola is: x2 = - 32y. So the answer is: x2 = - 32y

If the Quasilinear equation of parabola y = ax ^ 2 is y = - 1 / 8, then a =?
Is the answer 2 or 1 / 2? I chose 1 / 2. Why is it wrong?
Isn't 1 / 8 multiplied by 4 1 / 2?

∵y=ax^2
∴x^2=y/a,
∴2p=1/a,∴p/2=1/4a
The Quasilinear equation is y = - 1 / 8,
∴p/2=1/8,∴1/4a=1/8,∴a=2

Quasilinear equation x = 2 parabolic equation

The Quasilinear equation is x = - P / 2 = 2 → P = - 4
So the standard equation of parabola is Y & # 178; = 2px = - 8x

The Quasilinear equation of parabola x ^ 2 = - 6y is---------

p=-3
Guide line: y = 3 / 2

What is the Quasilinear equation of a parabola x = y squared

x=y^2
2p=1
p=1/2
p/2=1/4
Because the focus equation is (1 / 4,0)
The guide line is on the other side of the focus
So the alignment is
x=-p/2=-1/4