Calculation of anomalous integral ∫ (∞, 1) e ^ - √ x DX

Calculation of anomalous integral ∫ (∞, 1) e ^ - √ x DX

Let √ x = u, then x = u & #, DX = 2udu, u: 1 → + ∞
The original formula = ∫ (1 → + ∞) 2UE ^ (- U) Du
=-∫ (1→+∞) 2u de^(-u)
=-2ue^(-u)+2∫ (1→+∞) e^(-u)du
=-2ue^(-u)-2e^(-u) |(1→+∞) (1→+∞)
=2e^(-1)+2e^(-1)
=4/e

∫ (upper limit 1, lower limit 0) DX ∫ (upper limit 1, lower limit x) x ^ 2 * siny ^ 2dy
It's not the whole sin y squared, it's the square of Y in sin y

The integral region is a triangle: 0 ≤ x ≤ 1, X ≤ y ≤ 1. Transform the integral region into 0 ≤ y ≤ 1, 0 ≤ x ≤ y, then ∫ (0,1) DX ∫ (x, 1) x & sup2; siny & sup2; dy = ∫ (0,1) dy ∫ (0, y) x & sup2; siny & sup2; DX = ∫ (0,1) dy * (Y & sup3; siny & sup2;) / 3 = - 1 / 6 * ∫ (0,1) y & sup2

Solving problem 13 of exercise 2-2 of Tongji Sixth Edition
Let f (x) and G (x) be defined in a neighborhood of X., f (x) be differentiable at X., f (X.) = 0, and G (x) be continuous at X.. This paper discusses the differentiability of F (x) g (x) at X

Note that f (x0) = 0, so
[f(x)g(x)-f(x0)g(x0)]/(x-x0)
　　= g(x)*[f(x)-f(x0)]/(x-x0)
If f (x) is derivable at x0 and G (x) is continuous at x0, then x -- > x0, the above formula has a limit
g(x0)*f‘(x0),
That is, the derivative of F (x) g (x) at x0 is g (x0) * f '(x0)

What is the meaning of the infinite discontinuity and oscillatory discontinuity where the left and right limits do not exist?

A function whose limit tends to infinity, such as Tan function. An oscillating discontinuity, such as sincos function, whose value varies from - 1 to 1, is called oscillation

Why is DV equal to D * DX / dt * t in college physics

v = velocity,s= displacement
v = ds/dt
dv = d(ds/dt)

What are the corresponding values of dvdt in a = DV / dt?

DV should be the change of speed
DT should be the change of time

College physics has v = DX / DT, so can DT = DX / V or DT = DX / DV?
Slightly M = =, when V is not a constant velocity, that is, when doing linear motion with acceleration change, can DT = DX / v be used?
But in principle, DT should be DX divided by the instantaneous velocity at this time, which should be DV?
And if it is an acceleration motion with acceleration change, there is only one relation between velocity and displacement. If you want to find time, you have to find DT. Is there any other good method?

The formula DT = DX / V can be used at any time. No matter what kind of motion, when the time is recorded as DT, that is, when the time interval tends to zero, V has no time to change. In DT, the motion of an object should be viewed at a constant speed. DT = DX / DV is definitely wrong. DV is the increment of velocity. Dividing the increment of velocity by the increment of displacement is definitely not time

(College Physics) why is small a, that is, acceleration, not equal to DV (lower case, that is, velocity) / DT, but equal to | DV (upper case, that is, velocity) / dt |,
The main answer is why it's not the former. If you can, it's better to give an example
Why is at equal to the former

Because velocity is a vector, it can change both the size and the direction. The derivative of velocity is the tangential acceleration, that is to say, the acceleration that only changes the size. The total acceleration is the derivative of vector velocity, which includes the tangential acceleration that only changes the direction. The former simply calculates the derivative of velocity, but does not include the change of velocity direction

When the engine of an electric boat is turned off, its acceleration direction is opposite to the speed direction, and its magnitude is proportional to the square of the speed, DV / dt = - kV * v
Where k is a constant, it is proved that the speed of driving x distance after the engine is shut down is the - KX power of V1 times e, where V1 is the speed when the engine is shut down

When DV / dt = - KVV is multiplied by DX on both sides, DX (DV / DT) = - K (DX) VV is obtained, DV (DX / DT) = VDV = (1 / 2) d (VV) = - K (DX) VV is deformed, D (VV) / (VV) = D [ln (VV)] = - 2K (DX), ln (VV) = - 2kx + C "is obtained, VV = c'e ^ (- 2kx), i.e. V ^ 2 = C '[e ^ (- KX)] ^ 2, i.e. v = CE ^ (- KX)

Judge and give an example, when the acceleration and speed direction are the same and decrease, the object decelerates!

The acceleration and velocity are in the same direction. Even if the acceleration decreases, the acceleration will not slow down as long as the acceleration increases and decreases before it is not zero. Only when the acceleration and velocity are in the opposite direction can the acceleration slow down. After the velocity becomes zero, if there is any acceleration, the acceleration will be in the opposite direction