The problem of the basic theorem of definite integral and calculus: there is a thin rod with non-uniform mass distribution. It is known that its linear density is p (x) = X3 (take one end of the thin rod as the origin and the straight line as the x-axis). If the rod length is 1, the mass m of the rod is ()

The problem of the basic theorem of definite integral and calculus: there is a thin rod with non-uniform mass distribution. It is known that its linear density is p (x) = X3 (take one end of the thin rod as the origin and the straight line as the x-axis). If the rod length is 1, the mass m of the rod is ()

Mass element DM = x ^ 3DX
The mass is obtained by integrating from 0 to 1, and the integral result is 1 / 4
M=1/4

Basic theorems of definite integral and calculus Find a function so that its derivative is y = 2 ^ X. please explain in detail,

Z = 2 ^ X / LN2 + C is any constant;
Just calculate Z 'and remember the answer

Prove that ∫ [0, a] DX ∫ [0, x] f (y) dy = ∫ [0, a] (A-X) f (x) DX

It can be solved by using a simple partial integral method, and the order of integral can also be exchanged

Let y = f [(3x-2) / (3x + 2)], f ` (x) = arctan (x ^ 2,) then dy / DX is at x = 0= I calculated 3 / 2 π. The teacher gave 3 / 4 π

y=f[(1-4/(3x+2)],y`={arctan[(1-4/(3x+2)]^2}*[12/(3x+2)^2
When x = 0, y = (arctan1) * (12 / 4) = 3 / 4 π

Dy / DX = 2 * how to find the general solution of Y / x + Y / x under the root sign?

This is what I did. I don't know if it's right:
Let Y / x = t
Then, y = TX
Then dy / DX = t + XDT / DX
According to the conditions, dy / DX = 2 √ T + T
Therefore: XDT / DX = 2 √ t
Sorting: DT / √ t = 2DX / X
Integral on both sides, get: √ t = LNX + C (C is a constant)
t=(lnx+C)^2
And because y = TX
Y = x (LNX + C) ^ 2

Dy / DX = (1-y ^ 2) / (1-x ^ 2) open root sign for general solution

dy/√(1-y ²)= dx/√(1-x ²)
So arcsiny = arcsinx + C
Y = sin (arcsinx + C), where - 2