Let f (x) be a continuous even function on (- ∞, + ∞), and prove that f (x) = ∫ (0 → x) f (T) DT is an odd function

Let f (x) be a continuous even function on (- ∞, + ∞), and prove that f (x) = ∫ (0 → x) f (T) DT is an odd function

It is proved that f (x) is a continuous even function on R: F (- x) = f (x) f (x) = ∫ (0 → x) f (T) DTF (- x) = ∫ (0 → x) f (T) DT (let m = - t, t = - M) = ∫ (0 → x) f (- M) d (- M) = - ∫ (0 → x) f (- M) DM = - ∫ (0 → x) f (m) DM = - ∫ (0 → x) f (T) DT = - f (x)

F (x) is continuous on [- L, l] and Φ (x) = ∫ (0, x) f (T) DT, (- L ≤ x ≤ L). If f (x) is an even function, it is proved that Φ (x) is an odd function on [- L, l]
Please write down the part of replacing T with other letters in detail. I'm a little confused about the formula of exchanging upper and lower limits

First, we show that Φ (x) is meaningful in [- L, l], because the integral range is (0, x) contained in [- 1,1], so f (x) is meaningful in every x ∈ [- 1,1], and F is continuous in this interval, so Φ (x) is defined in [- L, l]. Secondly, we need to prove that Φ (- x) = - Φ (x) and f (- x) = f (x) Φ (- x) = ∫ (0, - x) f (T) DT let s = - T

How to prove that any function can be uniquely written as the sum of an odd function and an even function
How to use the counter evidence?

F (x) can be expressed as [f (x) + F (- x)] / 2 + [f (x) - f (- x)] / 2, the former is even function, the latter is odd function
This uniqueness Maybe it can be proved to the contrary
(I don't know how to prove it.) )

It is proved that any function on (- L, l) can be written as the sum of an odd function and an even function

Let f (x) = H (x) + G (x)
f(-x)=h(-x)+g(-x)=-h(x)+g(x)
so h(x)=[f(x)-f(-x)]/2
g(x)=[f(x)+f(-x)]/2
so f(x)=)=[f(x)-f(-x)]/2+[f(x)+f(-x)]/2

How to calculate 9.9 * 99 + 9.9 simple algorithm?

9.9*99+9.9
=9.9*（99+1）
=9.9*100
=990

It is known that a line L passing through the focus of the parabola y * 2 = 2px intersects with P.Q, and a line passing through P and the vertex of the parabola intersects with the collimator M,
Find MQ ‖ on X-axis

The perpendicularity of the vertical line (i.e. the parallel line of X axis) passing through P and Q is g and D. PQ is known to pass through the focus f (P / 2,0). The equation of the line is x = -- P / 2 and DP Tan angle GPD = dg / GP = (yi-y2) / (x1 + P / 2)
Tan angle GPM = Y1 / x1
Tan angle GPD -- Tan angle GPM = = (yi-y2) / (x1 + P / 2) -- Y1 / x1
=[y1x1--y2x1-y1x1--y1p/2]/(x1+p/2)x1
=[--y2x1--y1p/2]/(x1+p/2)x1=--[y2(myi+p/2)+y1p/2]/(x1+p/2)x1
=--[my1y2+p/2*(y1+y2)]/(x1+p/2)x1
=--[m(--p^2)+p/2(+2pm)]/(x1+p/2)x1=--0/(x1+p/2)x1=0
Tan angle GPD -- Tan angle GPM = 0 Tan angle GPD = Tan angle GPM because both angles are acute angles, so angle GPD = angle GPM, that is, MP and PD are collinear, m point coincides with D point MQ ‖ on X axis
Linear equation y / (X-P / 2) = 1 / m my = X-P / 2 parabolic equation y ^ 2 = 2px = 2pmy + P ^ 2 y ^ 2 -- 2pmy -- P ^ 2 = 0
x1=my1+p/2 y1+y2=+2pm y1y2=--p^2

Judge the equal volume of cube, cuboid and cylinder with equal base and height

Yes

56-0.56 * 56-0.4 * 56 simple calculation requires a process

56－0.56×56－0.4×56
=56×（1－0.56－0.4）
=56×0.04
=2.24

The domain of function f (2x + 1) is [- 2,3]. Find the domain of function f (2x-1)
Doubts: does the domain [2,3] refer to the value range of X or (2x + 1)?
Sorry, I read it wrong

Note: the domain of definition [- 2,3] refers to the value range of X
-3

The volume of an oxygen bottle is 10 cubic decimeters, the aerobic density is 2.5 kg per cubic meter, and 5 g of deoxygenation is used to save patients. What is the density of residual oxygen?

10 cubic decimeter = 0.01 cubic meter
When density = m / V, M = density * V = 2.5 * 0.01 = 0.025kg
0.025-0.005=0.02KG
Density = m / v = 0.02 / 0.01 = 2kg / m3