Among the following functions, () A. f（x）=sin2xB. f（x）=xexC. f（x）=x3-xD. f（x）=-x+lnx

Among the following functions, () A. f（x）=sin2xB. f（x）=xexC. f（x）=x3-xD. f（x）=-x+lnx

For a, f (x) = sin2x is a periodic function, which has no monotonicity on (0, + ∞), but does not satisfy the problem; for B, ∵ f (x) = xex, ∵ f '(x) = (1 + x) ex, ∵ f' (x) > 0 when x ∈ (0, + ∞), f (x) is an increasing function on (0, + ∞); for C, ∵ f (x) = x3-x, ∵ f '(x) = 3x2-1, ∵ when x ∈ (0, 13), f' (x) < 0, f (x) = 0 (x) For D, ∵ f (x) = - x + LNX, ∵ f '(x) = - 1 + LX = 1 − XX, when x ∈ (0,1), f' (x) > 0, f (x) is an increasing function, when x ∈ (1, + ∞), f '(x) < 0, f (x) is a decreasing function, ∵ does not satisfy the problem ）B

(Xe ^ x) / radical e ^ X-2 indefinite integral

Let f (x) = x ^ 2 - ∫ (0, a) f (x) DX (a ≠ - 1), try to find the function f (x)

Let B = ∫ (0, a) f (x) DX be a constant
Then f (x) = x ^ 2-b
Therefore, there are:
b=∫﹙0,a﹚f(x)dx
=∫﹙0,a﹚(x^2-b)dx
=﹙0,a﹚(x^3/3-bx)
=a^3/3-ab
We get: B = a ^ 3 / [3 (1 + a)]
So f (x) = x ^ 2-A ^ 3 / [3 (1 + a)]

Let f be continuous on a finite interval I and f be an original function of F on I, then ∫ → XF '(x) DX = f (x)
Let f be continuous on a finite interval I and f be an original function of F on I, then ∫ a → XF '(x) DX = f (x)

∫a→xF'(x)dx=F(x) - F(a)
Generally not. It only holds when f (a) = 0

Let f (x) satisfy the equation ∫ XF (x) DX = x + ∫ x ^ 2F (x) DX, and find ∫ f (x) DX

Both sides of the derivation: XF (x) = 1 + X & # 178; f (x), the solution: F (x) = 1 / (x-x & # 178;) ∫ f (x) DX = ∫ 1 / (x-x & # 178;) DX = - ∫ 1 / (X & # 178; - x) DX = - ∫ 1 / (x-1) DX + ∫ 1 / xdx = ln| X / (x-1) | + C, hope to help you, if you solve the problem, please click "select as satisfactory answer"

Let f (x) be a primitive function of F (x), then ∫ XF (1-x ^) DX = (?)
^It's Square

∫xf(1-x^)=[-∫f(1-x^)d(1-x^)]/2
Because f (x) is a primitive function of F (x), so
∫xf(1-x^)=[-∫f(1-x^)d(1-x^)]/2=-F(1-x^)/2+C

Let f (x) be an original function of F (x), and prove that the integral XF (x ^ 2) DX = 1 / 2F (x ^ 2) + C

Integral XF (x ^ 2) DX
=Integral 1 / 2 * f (x ^ 2) d (x ^ 2)
=1/2*[F(x^2)+C]
=1/2F(x^2)+C

If sec ^ 2 x is a primitive function of F (x), then ∫ XF (x) DX =?
How to do it?

SEC & sup2; X is a primitive function of F (x), that is, D (SEC & sup2; x) / DX = f (x), so f (x) DX = D (SEC & sup2; x)
Integral by parts:
∫xf(x)dx
=∫xd(sec²x)
=x·sec²x-∫sec²xdx
=xsec²x-tan x+C.

Finding definite integral ∫ upper limit is 1, lower limit is 0 ∫ Xe ^ (2x) DX

Quarter times (e ^ 2 + 1)

Definite integral [a, b] f '(3x) DX = f (b) - f (a)
Seek answers

∫(a→b) f'(3x) dx
= ∫(a→b) f'(3x) (1/3)d(3x)
= (1/3)f(3x) |(a→b)
= (1/3)[f(3b) - f(3a)]
Or let u = 3x, Du = 3 DX
∫(a→b) f'(3x) dx
= ∫(3a→3b) f'(u) * (1/3) du
= (1/3)f(u) |(3a→3b)
= (1/3)[f(3b) - f(3a)]