It is proved that f (x) is integrable and odd on [- A, a], then ∫ f (x) DX = 0

It is proved that f (x) is integrable and odd on [- A, a], then ∫ f (x) DX = 0

∫ ^a_ {-a)f(x)dx = ∫ ^a_ 0f(x)dx+∫ ^0_ {-a}f(x)dx= ∫ ^a_ 0f(x)dx-∫ ^0_ {a}f(-x)d(-x) = ∫ ^a_ 0f(x)dx+∫ ^a_ 0f(-x)dx = ∫ ^a_ 0f(x)dx-∫ ^a_ 0f(x)dx =0

Let f (x) be a monotonically decreasing continuous function on [0,1]. It is proved that for any Q ∈ [0,1], there is an inequality ∫ 0 → Q f (x) DX ≥ Q ∫ 0 → 1 f (x) DX

q> When 0, divide both sides by Q, and then calculate the derivative to verify that the left end decreases

We know the function f (x) = - x2 + | x | (1) express the function in the form of piecewise function and draw the image of the function; (2) find the monotone interval of the function; (3) find the maximum value of the function

(1) ∵ f (x) = - x2 + | x | = − x2 − x, x < 0 − x2 + X, X ≥ 0 ∵ the image of function f (x) is shown in the following figure: (2) from the image of function in (1), the monotone increasing interval of function f (x) is: (- ∞, - 12] and [0, 12], and the monotone decreasing interval of function f (x) is: [- 12, 0] and [12, + ∞). (3) (2) from the image of function in (1), the monotone increasing interval of function f (x) is: (- ∞, - 12] and [0, 12, + ∞) The maximum is 14

A = {1,2}, B = {x | ax + 1 = 0}, if B ∈ a, find the value of real number a

It should be B ∈ a, so x = 1 or 2
When x = 1, a = - 1
When x = 2, a = - 1 / 2

6.3 △ 6 =? Calculated by vertical formula

Limit of function in high number 1, definition related to neighborhood 1.4.2

A neighborhood is an open interval, and an interval is a set of numbers, which needs to be understood first. An interval is a set of numbers between two points on a one-dimensional number axis. A neighborhood is an open interval, which is a set of numbers between two end points whose center is a point on the number axis and whose distance is a

To solve the equations: 2x − 3Y = 85y − 7x = 5

(1) X 7 + (2) × 2 is - 11y = 66, y = - 6. Substituting y = - 6 into (1) is: 2x + 18 = 8, x = - 5. The solution of the original equations is x = − 5Y = − 6

Java programming pattern 1 12 123 1234 12345
Using java to write patterns
one
twelve
one hundred and twenty-three
one thousand two hundred and thirty-four
twelve thousand three hundred and forty-five

public class Test {
public static void main(String[] args) {
for (int i = 0; i < 5; i++) {
for (int j = 0; j

The derivation of function f [x] at point XO proves that limf [XO + α H] - f [XO – β H] / h = (α + β) f '(x) (α, β are constants)

Because when h is very small, (alpha. H) is also very small, so from the definition of derivative, we can get: limf (XO alpha. H) / (alpha. H) = limf '(XO), and alpha is a constant, so it can be put forward from the limit. Then the other part is the same. Finally, the two on the left side of the equation are added to form the part on the right side of the equation!
I am self-taught, for reference only!

The result of a logical expression operation is
A. Numerical type B. character type C. logical type D. date type

C. Logical type
The result of logical expression operation is only true and false (binary is 0 or 1), no matter how complex the expression is