A little problem with the derivative of variable upper bound integrals

Let f(t) be a primitive function of tf(t).

Then f'(t) = tf(t).

The lower limit is 0, and the upper limit is x)t f(t)dt = f(t)|(Lower limit 0, upper limit x) = f(x)-f(0).

After deriving it, it is f'(x) = xf(x) (Since f(0) is equal to a constant, its derivative is 0).

Let g(t) be a primitive function of (x-t)f(t).

g‘(t)=(x-t)f(t)

and (lower limit 0, upper limit x)(x-t) f(t)dt=g(t)|(Lower limit 0, upper limit x) = g(x)-g(0).

After finding its derivative, it is g'(x)-[g(0)].'=(x-x)f(x)-[g(0)]' =-[g(0)]'

Because g(t) contains x, g(0) is a function related to x, and the derivative of it is not 0).

Since we don't know what the relationship between g(0) and x is, we don't ask for it.

To derive the integral with a variable upper limit, it is necessary to ensure that the integrand is not related to x, but only to the integral variable t, so that the formula 0,x) f(t) dt is used to derive x, and the derivative is f(x).

Therefore, it is necessary to first use methods such as element exchange to adjust x to the upper and lower limits of the integral.

The derivative of the variable upper bound integral is not a Newton-Leibniz formula.

First of all, you need to know the derivative formula: f(x) = upper limit x, lower bound a) f(t)dt, then f'(x) = f(x), this is the basic formula.

If f(x)=x (upper limit x, lower bound a) f(t)dt, then f(x) can be regarded as the multiplication of two functions, one is x, and the other is (upper limit x, lower bound a)f(t)dt, so f(x) derivative is calculated according to the law of product derivative, note (upper limit x, lower limit a)f(t)dt=u(x).

f'(x)=(xu(x))'x)'u(x)+xu'(x)=u(x)+xu'(x) = upper limit x, lower limit a) f(t)dt + xf(x).

There are two results: the former is the x derivative, u(x) is constant, and the latter is x-invariant, u(x) is derivative.

f(x)= (a,x)xf(t)dt, this theorem is the most important property of the variable limit integral, and two points need to be paid attention to to master this theorem: first, the lower limit is a constant, and the upper limit is the parameter variable x (not other expressions containing x);

Second, the integrand f(x) contains only the integral variable t and not the parametric variable x.

Integral variable limit functions are an important class of functions, and their most famous application is in the proof of Newton's Leibniz formula

In fact, the integral variable limit function is an important tool for generating new functions, especially since it can represent non-elementary functions and transform integral problems into differential calculus problems.

1).Derivative is based on the law of chain to

x * 0,x] f(t) dx ) ' = (x) ' * 0,x] f(t) dx + x * 0,x] f(t) dx ) ' = ∫[0,x] f(t) dx + x * f(x)；

2). 0,x] t * f(t) dt ) '= x * f(x), you can set f(t) = t * f(t), so that you can easily understand;

3).That's right, just look at x as a normal number;

4).u=x - t 0, x] f(u) du, here there is t, there is u, what are you counting?

Just ask me

Let x-t=u, then dt=

du, so (0 to x)f(x-t)dt

(x to 0)f(u)du

(0 to x)f(u)du, thus.

x(0 to x)f(x-t)dt

x(0 to x)f(u)du, so.

d[x (0 to x)f(x-t)dt] dxd

x (0 to x) f(u)du

dxxd(0 to x)f(u)du

dxdx/dx

0 to x)f(u)du

Note that here, if the variable upper bound integral function (0 to x)f(u)du is derived from x, the derivative is f(x) and not f(u), so d[x(0 to x)f(x-t)dt] dx=xd

0 to x)f(u)du

dxdx/dx

0 to x)f(u)du=x

f(x)(0 to x)f(u)du

It is not a question of whether xf(u) can be directly replaced with xf(x), but that the derivative of xf(x) is obtained by xf(x).

Let u=x-t

then t=x-u

The upper and lower limits of t are 0---x, so the upper and lower limits of u are x---0, and dt=d(x-u)=

dux (0 to x)f(x-t)dt

x (x to 0) f(u) (-du) =

x (0 to x) f(u) (

du) = x (0 to x) f(u)du

The following derivative is obtained: xf(x)+)0 to x)f(u)du=xf(x)+)0 to x)f(t)dt

[∫[0,x] f(t)dt]'=f(x)，That is, the derivative of the upper limit of change integral to the upper limit of change is equal to bringing the upper limit of change into the integrand. Example:

f(x)= 0,x] sint t dt Although the original function f(x) of sint t cannot be represented by an elementary function, the derivative of f(x) can be calculated according to the Derivative of the Integral of the Upper Limit of Variation: [f(x)].'0,x] sint/t dt ]'sinx/x。

The general form of the [Variation Upper Limit Integral Derivative Rule] is: [ x) ,x)] f(t)dt].' f(φ(x))φx)-f(ψ(x))ψx)

Let the function y=f(x) be integrable over the interval [a,b], and for any x [a,b], y=f(x) be integrable on [a,x], and its value forms a correspondence with x (as shown in ** in the overview), (x) is called a definite integral function with variable upper bounds.

Definite integral of the integral upper limit function:

Let f(x) be continuous over the interval [a,b], then f(x) is integrable over [a,b]. Let f(x) be bounded by the interval [a,b] and have only a finite number of discontinuities, then f(x) is integrable on [a,b]. Let f(x) be monotonic over the bridge sock interval [a,b], then f(x) is integrable on [a,b].

Divide the image [a,b] of the function in a certain interval into n parts, divide it into an infinite number of rectangles with a straight line parallel to the y-axis, and then find the sum of the areas of all these rectangles when n +.

In the case of a proportional function, the quotient between x and y is (x≠0). In the inverse ratio of the Zen search example function, the product of x and y is fixed. In y=kx+b (k,b is constant, k≠0), when x increases m, the function value y increases km, and conversely, when x decreases m, the function value y decreases km.

The variable limit integral of the upper limit infinity, regardless of the upper and lower limits, first write out the original function, and then when the variable takes infinity, it is equivalent to taking the limit as a fixed value.

The lower bound of the integral is a, and the lower bound is g(x) Then to find the derivative of the product of this variable upper limit, g(x) instead of t in f(t), and then multiply g(x) to find the derivative of x.

i.e. g'(x) So the derivative is f[g(x)]*g'(x) The meaning of this Qi hunger is that the lower limit of the integral is a, and the lower limit is g(x), so to find the derivative of the integral function of this variable upper limit, use g(x) instead of t in f(t), and then multiply g(x) to find the derivative of x, that is, g. g'(x) So the derivative is f[g(x)]*g'(x)。

In fact, the integral variable limit function is an important tool for generating new functions, especially since it can represent non-elementary functions and transform integral problems into differential calculus problems. In addition to expanding our understanding of the concept of functions, integral variable limit functions have important applications in many occasions.

If you find the derivative directly in this case, you will make an error. The reason is simple: the basic form of the upper derivative of the variable integral in the book is: <>

In this form, f(t) does not contain x. If f(t) in the problem contains x (e.g., f(tx)), you can only transform the subsequent integral variable t so that it has the same form as the variable of the integrand, and finally use the commutation method to complete the substitution.

For example: <>