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F (x) is a function with t as the period. Is the definite integral of F (x) from 0 to a equal to the definite integral of F (x) from t to a + T? Why
be equal to.
Because f (x) is a function with t as the period, so
f(x-T)=f(x)
So the definite integral of F (x) from t to a + T is equal to the definite integral of F (x-t) from t to a + t,
Let t = x-t, then the integral limit becomes from 0 to a, DX = DT,
The definite integral of F (x-t) from t to a + T is equal to the definite integral of F (T) from 0 to a
To sum up, the definite integral of F (x) from 0 to a is equal to the definite integral of F (x) from t to a + t
The definite integral of periodic function on (a, a + T) is independent of A. how to prove it? The definite integral of periodic function on (a, a + T) is independent of A. how is it proved? F (x) = f (x + T). Find the definite integral of F (x) on [a, a + T]. The conclusion is that the definite integral is a definite value: that is, the definite integral of F on [0, t].
The range of commutation x = y + a into (0, t) is independent of A
But I still don't understand what the landlord means
Let f (x) be a continuous function with t as the period, then the value of the definite integral ∫ (a, a + TF (x)) DX A: independent of T, B: independent of a and T, C: independent of A
First C, for example, |sin x| is a periodic continuous function. If you take pie as the period and pie 2 as the period, the result is different, but it is independent of the value of A
Let f (x) be a continuous function with t as the period and a as the variable. Is it still true that the definite integral of F (x) on a to a + T = the definite integral of F (x) on 0 to t? Are there any requirements for a? Is it a constant or a variable? Or is it all right?
The essence of the definite integral that is not required for a is the area of the graph surrounded by the curve f (x), the x-axis and the line at the two ends of x = and f (x) is the period of T. A to a + T is equivalent to shifting a unit length from 0 to t to the right, and the corresponding graph is equivalent to shifting a unit length to the right, so the shape of the graph is not
Please prove that if the function f (x) is a periodic function with t greater than 0 as the cycle, then f (AX) (a greater than 0) is a periodic function with T / A as the cycle,
prove:
Because f (x) = f (x + T)
So f (AX) = f (AX + T) = f [a (x + T / a)]
How to prove that if the period of periodic function f (x) is t and a is not equal to 0, then the period of F (AX + b) is t / A Super difficult problem of urgently seeking periodic function
f(x)=f(x+T)
f(ax+b)=f(ax+b+T)=f(a(x+T/a)+b)
So the period of F (AX + b) is t / A
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