How to find the original function of integrand in indefinite integral F '(x) = f (x), given f (x), how to find f (x)

How to find the original function of integrand in indefinite integral F '(x) = f (x), given f (x), how to find f (x)

If it's an elementary function, we can directly look up the derivation formula of the elementary function. F (x) is the original function. (that is to say, the derivation of F (x) is f (x). If we have a small f (x) table, we can know the form of the corresponding f (x), but we need to add a constant or some other formats after f (x), Or calculus in college

When x ∈ [1,2], why do we need to calculate the interval of X ∈ [0,1) again? That's why we reduce 1 / 6

Because it is an integral on [0, x], the integral interval should be divided according to the value of X
Why did 1 / 6 appear when you asked? In fact, it was just the sum of two definite integrals
PS: definite integral and indefinite integral must be distinguished, definite integral must see integral upper limit and integral lower limit

The solution of indefinite integral of piecewise function
Seek knowledge

For each section of indefinite integral, and then piecewise write the original function, on the line

Finding definite integrals of several functions,
∫ (x ^ 2-2x) DX, interval [0,1]
∫ 1 / (1 + x) DX interval [0,1]
∫ [1 / radical (1-x ^ 2) DX, the interval is [- 1 / 2,1 / 2]

Find the original function first
1.-2/3
I'm too lazy to type. You can do it yourself
The algorithm of definite integral is to find the original function first, and then substitute the upper limit of integral into the original function minus the lower limit of integral into the original function

Definite integral of one variable function
∫ (lower limit-1, upper limit + 1) x (1 + x ^ 2001) (e ^ x-e ^ - x) DX let's see how to find it, write it in detail, thank you

E ^ x-e ^ - x is an odd function, the product of odd function and even function is an odd function, so the product of x ^ 2002 and e ^ x-e ^ - x is integral 0 on - 1, + 1, and ∫ x (e ^ x-e ^ - x) DX = ∫ - 1, + 1x [D (e ^ x + e ^ - x)]
={x(e^x +e^-x)-1,+1}-∫-1,+1(e^x +e^-x)dx=4/e

High number, is there any relationship between a function and the parity of its original function? The original function here refers to the definite integral of a function

It should be indefinite integral! It has something to do with it

It is proved that the function f (x) = x + 1 is an increasing function in real number

It is proved that: let x2 > x1, (X2, X1 belong to R), f (x2) - f (x1) = x2 + 1 - (x1 + 1) = x2-x1 > 0, so f (x) = x + 1 is an increasing function in real number

It is proved that X-1 of a / X-1 of a is an increasing function on the set of real numbers

Please don't say a ＞ 0, otherwise the title itself is meaningless; and it must be a ＞ 1, otherwise the title is wrong
Let t = a ^ x, t increases with the increase of X
On the derivative of the function f (T) = T-1 / T + 1
f'(t)=1+1/t²＞0
The f (T) increases with the increase of T, and t increases with the increase of X
The value of function increases with the increase of X
That is, the function is an increasing function in R

1. The increasing interval of the function f (x) = x + 2sinx in the interval (0,2 π) 2. F (x) = X
1. The increasing interval of the function f (x) = x + 2sinx in the interval (0,2 π)
2. The increasing interval of F (x) = x & # / 1 + X
3. Find the extreme value of F (x) = [x & # 178; - 2x + 1] / [X-1]

（1）f'(x)=1+2cosx>0
∴ cosx>-1/2
∵ x∈（0,2π）
∴ x∈(0,2π/3)U(4π/3,2π)
The increasing intervals are (0,2 π / 3) and (4 π / 3,2 π)
（2）f'(x)=[2x(1+x)-x²]/(1+x)²>0
∴ x²+2x>0
‖ x > 0 or X

Given the function f (x) = x + 2 / x, it is proved that f (x) is an increasing function in the interval (1, + ∞)

In the interval (1, + ∞), DF (x) = 1-2 / X2, when 1-2 / X2 is greater than 0, f (x) increases monotonically, which is an increasing function, that is, X2 is greater than 2, X is greater than radical 2 or less than negative radical 2 (not the meaning of the problem). Therefore, the problem can not be proved and is a false proposition