Change the integral order, and calculate the integral value ∫ (upper limit is 1, lower limit is 0) DX ∫ (upper x, lower x ^ 2) (x ^ 2 + y ^ 2) ^ - 1 / 2dy

Change the integral order, and calculate the integral value ∫ (upper limit is 1, lower limit is 0) DX ∫ (upper x, lower x ^ 2) (x ^ 2 + y ^ 2) ^ - 1 / 2dy

The original formula = ∫ dy ∫ DX / √ (X & # 178; + Y & # 178;) (changing the order of integration)
=∫ D θ∫ RDR / R
=∫dθ∫dr
=∫(sinθ/cos²θ)dθ
=-∫d(cosθ)/cos²θ
=-(-1)[1/cos(π/4)-1/cos0]
=√2-1.

Calculate ∫ (upper limit Π lower limit 0) xsinx / (1 + cos ^ 2 (x)) DX

See figure

Calculation: ∫ 1 / [x * / (x ^ 2-1) ^ (1 / 2)] DX, the integral limit is (- 1) to (- 2), that is, the lower limit is - 2, and the upper limit is - 1, which requires a detailed process
Thank you`

∫ 1 / [x √ (X & # 178; - 1)] DX let x = SecY, DX = secytany dy when x = - 2, y = arcsec (- 2) = 2 π / 3 when x = - 1, y = arcsec (- 1) = π √ (X & # 178; - 1) = √ (SEC & # 178; Y-1) = √ (Tan & # 178; y), ∫ XA = (1 / a) sec (x / a) + ARCC, when x

In the acceleration section, how to understand the deceleration motion? Why does the acceleration V decrease faster? If the initial velocity X of the object is in phase

If the acceleration is large, it means that the speed decreases quickly, such as A1 = - 2m / S ^ 2, A2 = - 1m / S ^ 2, - 2m / S ^ 2, the speed in table 1s decreases by 2m / S. - 1m / S ^ 2, the speed in table 1s decreases by 1m / s

As shown in the figure, there are two baffles A and B on the straight road, which are 7 meters apart. An object starts from the place close to plate a at an initial speed of 6 MGS, and reciprocates and decelerates between plates a and B. each time the object collides with plates a and B, it bounces back at the original speed. Now it is required that the object finally stops at 2 m away from plate B, then the acceleration of the object should be_____________________________ .
18／（4n+9) (n=0,1,2,3,...) 18/(4n+5)(n=1,2)
It should be 18 / (14N + 9) n = 0,1,2,3
18/(14n+5) n=1,2,3..

Because of the "original speed rebound" every time, the object finally stops at 2m away from the B plate, that is to say, the object must move back and forth with constant acceleration between the baffles, which is similar to doing uniform deceleration linear motion. Then the total distance must be an integral multiple of the distance between the baffles plus 2m. If the acceleration is set as a, then
2 * a * (7n + 2) = 36 or 2 * a * (7n + 7-2) = 36
By the way, my answer is different from that given by LZ. I suggest LZ check whether the baffle distance is wrong
Yeah~

When the particle moves along the x-axis, the relationship between acceleration and position is a = 2 + 6x ^ 2, and the velocity of the particle is 10 meters per second when x = 0. Calculate the velocity of the particle at any coordinate

a=2+6x^2
dv/dx*dx/dt=2+6x^2
vdv=(2+6x^2)dx
∫vdv=∫(2+6x^2)dx
Formula V ^ 2 = 4x + 4x ^ 3 + C (1)
X = 0, v = 10, C = 100
(1) The formula of formula
V = 2 radical x + x ^ 3 + 25

For a particle moving in a straight line on the x-axis, it is known that its acceleration direction is opposite to the velocity direction, and its magnitude is directly proportional to the square of the velocity, that is DV / dt = - kV ^ 2,
The acceleration is proportional to the square of the velocity of the particle in the direction of X,
That is DV / dt = - kV ^ 2, and let t = 0, the velocity is v0. After 10 seconds, the velocity of the particle becomes V0 / 2, and the velocity of the particle at any time can be obtained
The relationship between displacement and time

Application of Rolle's theorem, Lagrange's theorem and Cauchy's theorem

This kind of theorem is generally used to prove some things, as a theoretical basis for deducing other things. For example, the proof of lobida's law uses Cauchy's theorem. We generally use the conclusion, and the lobida's law here is equivalent to a conclusion. The theorem generally plays a role in the formation and perfection of the knowledge architecture and the support of the theoretical architecture

What is Cauchy's mean value theorem?

Cauchy mean value theorem
It's a continuous function
Let x = a, x = b be a function
In that section of the function
Spring in an X ∈ {ah, B}
Let f '(x) = f (a) - f (b) / A-B hold

Ask Master to teach me how to use Cauchy mean value theorem
Baidu Encyclopedia on the look is not very understanding, it can not speak in detail, the best solution to extreme value problem examples

Although Cauchy's mean value theorem is the generalization of Lagrange's mean value theorem, you can find that they can be proved independently through Rolle's theorem by observing their most common proving methods, but the auxiliary functions constructed are different, It is obvious that Rolle's theorem can be deduced from Lagrange's mean value theorem. The derivation can be concluded as follows: Rolle's theorem → Cauchy's mean value theorem (Lagrange's mean value theorem) → Lagrange's mean value theorem → Rolle's theorem. Therefore, they are logically equivalent, but they are different in complexity when used to solve problems, Rolle theorem can also be used to solve, but the idea may be more complex
For example, let B ＞ a ＞ 0, f (x) be continuous in [a, b], and (a, b) be derivable, and it is proved that C ∈ (a, b), such that 2C [f (b) - f (a)] = (b ^ 2-A ^ 2) f '(c)
Proof: refer to the standard form of Cauchy mean value theorem, let g (x) = x ^ 2. Note that b > a > 0 ensures that G '(x) = 2x ≠ 0 and B ^ 2-A ^ 2 ≠ 0
The above problem is of course very simple (that is, direct formula). It can also be proved by the constructor f (x) = (b ^ 2-A ^ 2) f (x) - [f (b) - f (a)] x ^ 2. Then f (a) = f (a) B ^ 2-F (b) a ^ 2 = H (b). From Rolle's theorem, we know that there exists C ∈ (a, b) such that H '(c) = 0, which is to be proved. This solution just applies the proof process of Cauchy mean value theorem
The function of formula is to save people's thinking or calculation time, but it requires skillful use of the formula. The most important thing is observation. It is easy to see the clue of the existence problems such as the above simple example. The key is to observe which thing is more similar to G (x) in the formula (see g (x) and G '(x)). Common functions, such as LNX, e ^ (x) and even e ^ (H (x)), It is possible that Cauchy mean value theorem will be used in the intermediate process of some inequality problems. As for the extremum problem you mentioned,
The differential mean value theorem (Lagrange, Cauchy) is of great significance to the development and perfection of the system of unitary differential calculus, and its significance is theoretical, Observation and proficiency. I think it's helpful for you to have a look at the process of [proving lobita's law with Cauchy mean value theorem]. I think this is another wonderful place of Cauchy mean value theorem besides geometric explanation