Why do you add dx after an indefinite integral?

dx means: the increment of the independent variable x.

The indefinite integral contains dx because the indefinite integral is the inverse of the differential, not the inverse of the derivative.

2xdx=x +c because d(x +c)=2xdx.

Similar to the summation notation, dx is infinitesimal.

The sum of infinitesimal infinitesimal is the integral, and when it meets d, it is what follows d.

The operation of DX is the operation of differentiation. DX is fully capable of performing four arithmetic.

For example, make up the differentiation: y'dx

y'=dy dx, so y'dx=dy

Another example is to change the differentiation, x=f(t).

dx=dx/dt*dt=f'(t)dt

What kind of dx, can you write it out and take a look?

We know y'＝dy/dx.

In other words, dy dx means to derive y!

Now d dx is followed by the definite integral, which means to find the derivative of the definite integral, the definite integral is a constant, and the derivative of the constant function is 0!

If d dx is followed by an indefinite integral, e.g. d dx f(x)dx, what is the result? We can do this by letting the original function of f(x) be f(x) c, then f(x) c f(x)dx, then d dx f(x)dx d dx f(x) c f'(x) 0 f(x), i.e. d dx f(x) dx f(x).

Note: Don't confuse definite integrals with variable cap integrals, definite integrals are constants, and variable cap integrals are functions!

What you add is the upper limit integral of the Yanshi variable: d dx (0,x)f(t)dt=f(x), and the derivative rule is to replace the t in the integrand with the upper limit x. For example: d dx (0,x)sintdt=sinx

However, if the upper limit is not x, and the bucket is another dispersion function, such as x 2, then you need to multiply x 2 by the derivative of x 2 instead of t, i.e., multiply by 2x, e.g., d dx (0,x 2)sintdt=sinx 2*2x=2xsinx 2

To give you a formula: (x),g(x)) f(x)dx f(g(x))*g'(x)－f（ψ(x)）*x).

Because the definite integral is a constant, the derivative of the definite integral is zero.

Namely: <>

To be continued. Not to be confused with differential differentiation with indefinite accumulation.

For reference, please smile.

If the upper and lower limits of the definite integral are a function of x, then the departing of the variable integral (i.e., the definite integral you are talking about) is the derivative of the blank line.

If the upper and lower bounds of the definite integral are constants, then the derivative of the definite integral is 0.

The differentiation of the integral variable x of an indefinite integral, denoted as dx.

dx means that variable x is divided into infinitely many parts, each of which is infinitely small.

Similar to the summation notation, dx is infinitesimal.

The sum of infinitesimal infinitesimal is integral, and when it meets d, it is what follows d, and the operation of dx is the operation of differentiation. DX is fully capable of performing four arithmetic.

For example, make up the differentiation: y'dx

y'=dy dx, so y'dx=dy

Another example is to change the differentiation, x=f(t).

dx=dx/dt*dt=f'(t)dt

f(x) is a primitive function of the function f(x), and we put all the original functions of the function f(x)+

c (c is an arbitrary constant) is called the indefinite integral of the function f(x) and is denoted as, i.e., f(x)dx=f(x)+c.Antiderivative.

where is called the integral sign, f(x) is called the integrand, x is called the integral variable, f(x)dx is called the integrand, c is called the integral constant, and the process of finding the indefinite integral of a known function is called integrating this function.

Here f(x) is the derivative of f(x), and f(x)dx is the differentiation of f(x), i.e., f(x)dx=df(x), then f(x)dx= df(x), and f(x)dx is to find the original function f(x) of f(x)dx, and df(x) is the differential f(x)dx of the original function f(x), which means that d and are inverse operations, and the two inverse operators can be canceled out together, i.e. f(x)dx= df(x)=f(x), It's the same as a b b = a, isn't it canceled? You just think of d as it, the essence is the same, the opposite of each other.

The purpose of indefinite integrals is to train you to find the original function so that you can make definite integrals.

Therefore, indefinite integrals have no geometric meaning, you have to understand the meaning of definite integrals, and indefinite integrals are to remove the integral limit.

If you want to use the first commutation method, you need to use the mutual transformation between the differentials, dy=f(x)dx, where f(x) is the derivative of the function. This is an estimation of the change in a function, which is not necessarily equal to the change in the independent variable multiplied by the derivative (i.e., the slope of the image), but when the change in x tends to be 0, it can be approximated instead, which is the idea of differentiation. So dx and f(x) are one, and approximate the value of the dependent variable when the independent variable changes certainly.

Divide the blind part of the hall to give points: the finger of God.