Let the function f (x) = [root sign (x2 + 1)] - ax. When a > = 1, it is proved that the function f (x) is a monotonic function on the interval [0, + infinity]

Let the function f (x) = [root sign (x2 + 1)] - ax. When a > = 1, it is proved that the function f (x) is a monotonic function on the interval [0, + infinity]

Take X1 > x2 > 0
f(x1)-f(x2)=√(x1^2+1)-ax1-√(x2^2+1)-ax2
=(x1^2+1-x2^2-1)÷(√(x1^2=1)+√(x2^2+1))-a(x1-x2)
=(x1-x2)((x1+x2)÷(√(x1^2+1)+√(x2^2+1))-a)
Because X1 > x2 only needs to judge the positivity and negativity in the following brackets
That is, the positivity and negativity of (x1 + x2) ÷ (√ (x1 ^ 2 + 1) + √ (x2 ^ 2 + 1)) - A
Because a > = 1, it is only necessary to judge the size relationship between (x1 + x2) ÷ (√ (x1 ^ 2 + 1) + √ (x2 ^ 2 + 1)) and 1
So compare the size of (x1 + x2) - (√ (x1 ^ 2 + 1) + √ (x2 ^ 2 + 1)) and 0
Because √ (x1 ^ 2 + 1) > x1 √ (x2 ^ 2 + 1) > x2
So (x1 + x2)

It is proved that f (x) = (1 + x) / (root x) is a subtractive function on (0,1]

Set 1 ≥ X1 > x2 > 0
f(x1)-f(x2)
=(1+x1)/(√x1)-(1+x2)/(√x2)
=(√x1-√x2)*[(√(x1*x2)-1]/√(x1*x2)
∵1≥x1>x2>0
∴√x1>√x2,√(x1*x2)0,√(x1*x2)-1

It is proved that under f (x) = (1 + x) / root sign, X is a decreasing function on (0,1] and an increasing function on (1, + ∞) (1+x)/√x This is 1 / x + X under the root sign

F (x) = (1 + x) / x = 1 / √ x + √ x under the root sign
Let's just take the derivative

Given an original function Xe ^ X of F (x), then ∫ (1,0) f (x) DX =?

∫f(x)dx=xe^x+C
So the original formula = (1 * e + C) - (0 * 1 + C) = E

Let an original function of F (x) be Xe ^ x ^ 2 and calculate XF '(x) DX

letxe^(x^2) = ∫ f(x) dxe^(x^2) . [ 1+ 2x^2] = f(x) ∫ xf'(x) dx = ∫ x df(x)= xf(x) - ∫ f(x) dx= xf(x) - xe^(x^2) + C= xe^(x^2) . [ 1+ 2x^2] - xe^(x^2) + C=2x^3.e^(x^2) + C

Find the definite integral, integral 0 to 1, Xe to the x power DX I'm so worried

∫xe^xdx
=∫xde^x
=x*e^x-∫e^xdx
=x*e^x-e^x+C
=(x-1)*e^x+C
So definite integral = (π / 2-1) * e ^ (π / 2) - (- 1) * e ^ 0
=(π/2-1)*e^(π/2)+1