If the function f (x) is continuous in the closed interval [a, b], the definite integral is from a to BF (x) DX = (a-b) the definite integral is from 0 to 1F (a + (B-A) x) DX

If the function f (x) is continuous in the closed interval [a, b], the definite integral is from a to BF (x) DX = (a-b) the definite integral is from 0 to 1F (a + (B-A) x) DX

The latter is replaced by variables: a + (B-A) x = t, x from 0 to 1 corresponds to t from a to B, DX = DT / (B-A), and the right integral is from a to B, f (T) DT is the same as the left integral

Let the function f (x) be continuous on the interval [a, b]. It is proved that ∫ f (x) DX = f (a + b-X) DX Functions are online as B and offline as a

Prove that if a + b-X = t is replaced by variables, then DX = - DT, when x = B, t = a, when x = a, t = B, then ∫ (a, b) f (a + b-X) DX = - ∫ (B, a) f (T) DT = ∫ (a, b) f (T) DT = ∫ (a, b) f (x) DX, that is, ∫ (a, b) f (x) DX = ∫ (a, b) f (a + b-X) DX proposition is proved

F (x) is continuous on [0,1], and the definite integral f (x) DX = 0 proves that there is at least one point ξ, Make f (1)- ξ)=- f( ξ) Definite integral [0,1]

Let x = 1-T, so 0

Please explain the high number definite integral proof 1. If f (x) is continuous and even function on [- A, a], then ∫ (upper a, lower-a) f (x) DX = 2 ∫ (upper a, lower 0) f (x) DX Proof 1. If f (x) is continuous and even function on [- A, a], then ∫ (upper a, lower-a) f (x) DX = 2 ∫ (upper a, lower 0) f (x) DX 2. If f (x) is continuous and odd on [- A, a], then ∫ (upper a, lower-a) f (x) DX = 0 Proof: because ∫ (upper a lower-a) f (x) DX = ∫ (upper 0 lower-a) f (x) DX + ∫ (upper a lower 0) f (x) DX Replace the integral ∫ (upper 0 lower - a) f (x) DX with x = - T ∫ (upper 0 lower - a) f (x) DX = - ∫ (upper 0 lower a) f (- t) DT = ∫ (upper a lower 0) f (- t) DT = ∫ (upper a lower 0) f (- x) DX So ∫ (upper a lower-a) f (x) DX = ∫ (upper a lower 0) f (- x) DX + ∫ (upper a lower 0) f (x) DX =∫ (upper a lower 0) [f (x) + F (- x)] DX (1) If f (x) is an even function, that is, f (- x) = f (x), then f (x) + F (- x) = 2F (x) So ∫ (upper a lower - a) f (x) DX = 2 ∫ (upper a lower 0) f (x) DX (2) (1) if f (x) is an odd function, that is, f (- x) = - f (x), then f (x) + F (- x) = 0 Thus ∫ (upper a lower-a) f (x) DX = 0 What is the key step Replace the integral ∫ (upper 0 lower - a) f (x) DX with x = - T ∫ (upper 0 lower - a) f (x) DX = - ∫ (upper 0 lower a) f (- t) DT = ∫ (upper a lower 0) f (- t) DT = ∫ (upper a lower 0) f (- x) DX I can't understand it. I feel it's not equal at all, especially ∫ (up 0 down - a) f (x) DX = - ∫ (up 0 down a) f (- t) DT and ∫ (upper a lower 0) f (- t) DT = ∫ (upper a lower 0) f (- x) DX how do you wait?

I don't know whether the picture is clear, because I won't directly type out the formula, so I use the mathematical editor to compile the graph cut with QQ first. Ha ha~

∫√ (1-x ^ 2) DX integral upper limit 1 lower limit 0 calculate definite integral

Let x = Sina
Then √ (1-x) ²)= cosa
dx=cosada
x=1,a=π/2
x=0,a=0
Original formula = ∫ (0 → π / 2) cos ² ada
=∫(0→π/2)(1+cos2a)/2da
=1/4∫(0→π/2)(1+cos2a)d2a
=1/4*(2a+sin2a)(0→π/2)
=1/4*(2*π/2+sinπ)-1/4*(2*0+sin0)
=π/4

Find the definite integral ∫ x [f (x) + F (- x)] DX. The upper limit of the integral is a and the lower limit of the integral is - A The answer seems to be 0. Let f (x) = x, SiNx and cosx, the result seems to be 0. But I don't know how to prove it. There must be a process

Integral is an argument
Split first
∫x[f(x)+f(-x)]dx
=∫[-a,0]xf(x)dx+∫[0,a]xf(x)dx+∫[-a,0]xf(-x)dx+∫[0,a]xf(-x)dx
For the third and fourth argument y = - x, note that the upper and lower bounds also change
=∫[-a,0]xf(x)dx+∫[0,a]xf(x)dx+∫[a,0]-yf(y)(-dy)+∫[0,-a]-yf(y)(-dy)
=∫[-a,0]xf(x)dx+∫[0,a]xf(x)dx+∫[a,0]yf(y)dy+∫[0,-a]yf(y)dy
Another x = y
=∫[-a,0]xf(x)dx+∫[0,-a]xf(x)dx+∫[a,0]xf(x)dx+∫[0,a]xf(x)dx
Upper and lower limit exchange, more negative signs
=∫[-a,0]xf(x)dx-∫[-a,0]xf(x)dx+∫[a,0]xf(x)dx-∫[a,0]xf(x)dx
=0