The problem of the function of the upper limit of the integral and its derivative in higher mathemat

This question can only be truly understood if it is understood in principle.

First of all, from the definition of derivative, the most important thing for you to know a derivative is to pay attention to what it is about.

Derivation. That is, he is looking for the rate of change of something. For example, if you find a derivative of x, you are essentially finding the rate of change of this function on the x-axis, and finding the derivative of y is the rate of change on y.

That is, it doesn't matter if you're right to x, right to y. In fact, they are all of practical significance.

Clause. 1. The title tells you that the function is f(x), not f(t), that is, this function varies with x, that is, it changes with the transformation of x, and has nothing to do with t, so it can be explained from this point that it is certain that its derivative must be f(x), not f(t).

Clause. Second, the variable upper limit function is also a function, but his.

Argument. Appears at the upper or lower limit, just like the x in f(x). So in the integrand, such as t, u, x, y, etc.

None of them are independent variables, just symbols. This symbol takes the place of the independent variable and accumulates in the interval determined by the independent variable. So from the point of view of the independent variable, it must also be f(x), not f(t).

(x) is a function about x.

His derivative, of course, is about x.

You have to know the difference between indefinite integrals and definite integrals.

a, x)f(t)dt are definite integrals.

What he ended up with was about X.

Because you're going to bring it in.

A and X then subtract each other.

Can you understand that?

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 bound 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.

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Significance of integral monetization function:If the function f(x) is integrable over the interval [a,b], then the integral variable upper bound function is continuous on [a,b]. If the function f(x) is continuous over the interval [a,b], then the integral variable upper bound function has a derivative on [a,b].

If the function f(x) is continuous over the interval [a,b], then the integral variable upper bound function is a primitive function of f(x) on [a,b]. The integrand f(x) contains only the integral variable t and not the parameter variable x.

∫（g（x），c）f（x）dx]'=f(g(x)))cherry date*g'(x), g(x) is the upper limit function of the integral. The formula for finding the derivative of the integral upper bound is the function = the integral function is multiplied by the value of the function whose integral upper bound is the independent variable multiplied by the derivative of the upper integral spine.

∫（g（x）,p（x））f（x）dx]'=f（g（x））*g'（x）-f（p（x））*p'(x), g(x) is the upper limit of the integral, and p(x) is the lower limit of the integral. The formula for finding the upper and lower limits of the integral is the derivative of the function = the value of the function with the upper limit of the integral multiplied by the derivative of the upper limit of the integral - the value of the function with the lower bound of the integral multiplied by the derivative of the lower bound of the integral.

The derivative is as follows: in the commutation 2x-t=u, t is the original integral variable, u is the new integral variable after the commutation, u is a function of t, and u is not a function of x. The first integral after the conversion is equivalent to A to 2A [2af(u)] du.

Replace the left x in f with y, replace the left side of the equal sign with the full guide about x, and finally assign y to x....If you don't have some way to write out the steps, this formula can be taken for granted, just one step, and there is no magic process to speak of. A physics student who has not rigorously proved consciousness in the slightest way would think so.

The derivative of the variable limit integral is to first bring the integral limit into the integral function, and then find a derivative of the integral limit, and if the integral function has an independent variable, find a way to get it out of the integral number.

The upper limit of the integral function, let the function be continuous on the interval and set to a point on the point, and the definite integral is examined.

The independent variable of the integral upper bound function (or variable upper limit integral) is the upper limit variable , and in the derivative of x, it is about x, but in the integration of x, x is treated as a constant, and the integral variable t varies over the integral interval . The result of the integral upper bound function after the derivative of x is f(x).

Answer: The derivative of the function determined by the integral of the variable upper limit is reduced to the integrand itself, and the variable upper bound u=xy is the multivariate function, and the partial derivative of the composite function z=(x,y) is obtained according to the derivative of the composite multivariate function as follows:

For integrals, t is the integral variable, x is the upper limit of the integral, and x is considered a constant.

In the commutation 2x-t=u, t is the original integral variable, u is the new integral variable after the commutation, u is a function of t, and u is not a function of x.

The first integral after the exchange is equivalent to a to 2a [2af(u)] du, so a [i.e. x] can be proposed.

Your leaders of the cap function are basically in different places on the spot.

[∫[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.

Continue to separate integral variables from function variables.

z=xy f(t)dt- tf(t)dt+ tf(t)dt-xy f(t)dt, and then the derivative.

zx=y∫f(t)dt+xyf(xy)y-xyf(xy)y+xyf(xy)(-y)-y∫f(t)dt-xyf(xy)(-y)

y (0 to xy) f(t) dt-y (xy to 1) f(t) dtzxx = yf (xy) y-yf (xy) (-y) = 2y f (xy).

[∫[0,x]

f(t)dt]'=f(x)

That is, the change of the maximum number of points.

to change the upper limit.

, which is equal to bringing the upper limit of the change into the integrand.

Example: f(x) = [0,x].

sint/t

dt notwithstanding. sint/t

of the original function. f(x)

It cannot be expressed as an elementary function, but 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] is:

[φx)ψ(x)]

f(t)dt】'

f(φ(x))φ'(x)-f(ψ(x))ψ'(x)

(x) is a function of x, and its derivative, of course, is about x, and you have to know the difference between an indefinite integral and a definite integral. (a, x)f(t)dt is a definite integral, and what he ends up with is about x, because you're going to take in a and x and subtract it, so you can understand it.

f(x)=∫<0,x>f(t)(x-t)dt=x∫<0,x>f(t)dt-∫<0,x>tf(t)dt

f'(x)=∫<0,x>f(t)dt+x[∫<0,x>f(t)dt]'-[∫0,x>tf(t)dt]'

<0,x>f(t)dt+xf(x)-xf(x)=∫<0,x>f(t)dt

Think of <0,x>f(t)dt as a function of x.

d[∫(0,x)

t*f(2x-t)dt]/dx

∫(0,x+δx)

t*f(2x+2δx-t)dt

(0,x)t*f(2x-t)dt]/δx

δx[∫(x,x+δx)

t*f(2x+2δx-t)dt]/δx

And because [f(2x+2δx-t)-f(2x-t)] δx=2f'(2x-t)

x=∫(0,x)

2t*f'(2x-t)dt

Let g(t)=

t*f(2x+2δx-t), the original function of g(t) is g(t) then [ (x,x+δx)

t*f(2x+2δx-t)dt]/δx

g(x+δx)-g(x)]/δx

g'(x)g(x)

xf(x) (δx is infinitesimal).

Original = (0,x).

2t*f'(2x-t)dt

xf(x) cannot be considered a composite function because when using the formula for deriving a composite function, one of the arguments of the composite function must be within a function.

As in f[g(x)], the derivative of x is f'[g(x)]*g'(x) and the independent variables are not in the same function, such as f[g(x),x], then the derivative formula of the composite function cannot be used, that is, the derivative of f[g(x),x] is not equal to f'[g(x)]*g'(x)。

If we think of the original formula as a composite function, let g(x)dx

Upper limit s, lower limit t) =

h[g(x),s,t]

Then t*f(2x-t)dt(upper x and lower 0)=h[t*f(2x-t),x,0], the argument x is not in the same function.