Calculus topic: if ∫ f '(2x) DX = sin2x + C, find the function f (x)

Calculus topic: if ∫ f '(2x) DX = sin2x + C, find the function f (x)

Analysis: ∫ f '(2x) DX = sin2x + C
∴1/2∫f'(2x)d2x＝sin2x＋C
∴1/2f（2x）＝sin2x＋C
Let t = 2x, then
1/2f（t）＝sint＋C
∴1/2f（x）＝sinx＋C
‡ f (x) = 2sinx + C '(where C' = 2C.)

Why did calculus end up with DX?

DX is the differential of the independent variable, that is Δ x. D / DX is the derivative of the following formula to X. you can also write the following formula after D of the molecule, which means the same

Calculate ∫ Xe ^ x DX using the partial integral method requirement: Write step by step, don't skip, attach the explanation and the formula used

∫xe^x dx
=∫xde^x
=xe^x-∫e^xdx
=xe^x-e^x+C

Calculation of definite integral by partial integral method: ∫ (1,0) Xe ^-x DX

Original formula = - ∫ xde ^ (- x)
=-xe^(-x)+∫e^(-x)dx
=-xe^(-x)-e^(-x) (1,0)
=(-1/e-1/e)-(0-1)
=1-2/e

Integral: ∫∞, a Xe ^ [- (x-a)] DX Range ∞, a

Indefinite integral
∫ xe^[-(x-a)] dx
= ∫ xe^(a-x)dx
= -∫ xe^(a-x)d(a-x)
= -∫ xd(e^(a-x))
= -xe^(a-x) + ∫ e^(a-x)dx
= -xe^(a-x) - ∫ e^(a-x)∫d(a-x)
= -xe^(a-x) - e^(a-x) + C
For definite integral, there is
Original integral = LIM (x → + ∞) [(- X - 1) / e ^ (x-a)] - (0-1) / e ^ 0 = 1

Let f (x) = 2 √ x, find f '(4) according to the definition of derivative

solution
f'(x)=（2√x）'=（2x＾1/2）'=2*1/2*x＾-1/2=x＾-1/2=1/√x
f'(4)=1/√4=1/2