Among the following physical quantities, the vector is () A. Centripetal acceleration B. work C. power D. kinetic energy

Among the following physical quantities, the vector is () A. Centripetal acceleration B. work C. power D. kinetic energy

Centripetal acceleration has both magnitude and direction. It is a vector, while work, power and kinetic energy have only magnitude but no direction. They are scalars, so a

In the following physical quantities, the vector is a. length B. time C. mass D. displacement
What belongs to vectors is
A. Length
B. Time
C. Quality
D. Displacement

D [in the title, only displacement includes size and direction, so it is a vector, and others only include size, so it is a scalar] [including size and direction, so it is a vector, such as force, velocity, displacement, etc.]

Is the concept of position vector displacement vector velocity acceleration relative or absolute? Why?

Only one acceleration is absolute & nbsp; in relativity, only the speed of light is absolute, the others are relative & nbsp;

∫ tanxcos2xdx is the definite integral from 0 to Pai

Indefinite integral ∫ tanxcos2xdx = ∫ TaNx [2 (cosx) ^ 2-1] DX
=∫2sinxcosxdx-∫tanxdx=-1/2*cos2x-(-ln|cosx|)+C=-1/2*cos2x+ln|cosx|+C
∫(0,π)tanxcos2xdx=[-1/2*cos2π+ln|cosπ|+C]-[-1/2*cos0+ln|cos0|+C]=0

Definite integral of absolute value of X-1 from 0 to 2

=∫ (0,1) (1-x) dx+∫(1,2) (x-1)dx
=(x-x square / 2) | (0,1) + (x square / 2-x) | (1,2)
=1/2+1/2
=1

Why is definite integral with absolute value solved in this way
Why do absolute definite integrals need to be solved like that?
First find the x0 of integrand = 0, and then divide the integral into the sum of such two limits as (B, x0) (x0, a)
For example, ∫| 1-x| DX
=∫(1-x)dx + ∫(x-1)dx
Based on what?

According to 0-1, | 1-x | > 0
At 3, |1-x|

The absolute value integral problem proves that the absolute value of the sum of definite integrals is less than or equal to the sum of absolute values

|∫ (F + G) DX|

To find a definite integral, the upper and lower limits of the integral are - 4 and 3 respectively, and the integrand is the absolute value of X + A. I don't understand the process,

This is the first time that we want to be able to get the "x + a-dx (upper limit-4, lower limit 3, lower limit 3) as we want to be able to (x + a) DX (upper limit-4, lower limit 3, lower limit 3, lower limit 3, and lower limit-4, lower limit 3, lower limit 3) as we want to get the (x + a) DX (upper limit-4, lower limit 3, lower limit 3) as we want to (x + a) DX (upper limit-4, lower limit-4, lower limit 3, lower limit 3, lower limit 3, lower limit 3, lower limit 3, lower limit 3) as we want to be able (x (x + a) as we are (x (x (x + a) as we are (x (x (x) in this) in the (x (x (x (x) 8747when: original

The proof of "the definite integral of odd function in the region [- A, a] is equal to 0"

The integral is divided into two parts
The formula of integral interval [- A, 0]. (1)
The formula of integral interval [0, a]. (2)
Replace (1) with x = = > - X
Applying the properties of odd function, you will find that it becomes negative (2)
So sum is zero

Which is the greater of the definite integral of the absolute value of the difference between two functions and the absolute value of the definite integral of the difference between two functions?

Don't be misled. The definite integral of absolute value of difference is greater than or equal to the absolute value of definite integral of difference. It can be proved by the definition of definite integral in mathematical analysis. For a simple example, f (x) = six + 1, G (x) = 1, integral interval 0 to 2pi, definite integral of absolute value of difference is 4, and absolute value of definite integral of difference is 0