Finding the integral of the second order partial derivative, about the problem of finding the partia

To the second derivative.

First, find an indefinite integral to get the original function.

Possible first derivatives.

The indefinite integral of the first derivative is obtained by finding the indefinite integral again.

For example, if the second derivative is ax+b, first find an indefinite integral of the second derivative and get the first derivative ax 2+bx+c

Finding an indefinite integral for the first derivative yields its original function as ax 3+bx 2+cx+d, where c and d are arbitrary real numbers. Verification of the second derivative of the original function shows that this result is correct.

In calculus, an indefinite integral of a function f, or the original function, or the antiderivative, is a function f where the derivative is equal to f, i.e., f = f. The relationship between indefinite and definite integrals is determined by the fundamental theorem of calculus.

Are you sure. where f is the indefinite integral of f.

Common derivative formulas:

1. y=c (c is a constant) y'=0

2、y=x^n y'=nx^(n-1)

3、y=a^x y'=a^xlna，y=e^x y'=e^x4、y=logax y'=logae/x，y=lnx y'=1/x5、y=sinx y'=cosx

6、y=cosx y'=-sinx

The second floor is basically a problem!

A double integral is the spatial integral of a binary function, similar to a definite integral, and is the limit of a particular form of sum. The essence is to find the volume of the curved top cylinder. Reintegration has a wide range of applications, such as calculating the area of a surface, the center of gravity of a flat sheet, etc.

The double integral of a planar region can be generalized to the integration on a (directional) surface in a high-dimensional space, called a surface integral.

For the integration of the pre-filial piety derivative, it is only necessary to treat the other variables as constants as the integrals, and to integrate the integrative variables according to the integration method of the unary function.

For example, there is a function as follows (taking the integral of y as an example, finding x is the same as finding y, so I won't go into details):

Let's integrate y, just think of x as a constant in the form of:

To the y points, so get.

Finally, the equation after the integration is set to g<>

Let x = tanu, then dx = secu) 2du

i = 1+x 2) dx = secu 3du = secudtanu

secutanu - secu(tanu)^2du = secutanu - secu[(secu)^2-1]du

secutanu - i + ln|secu+tanu|

i = 1/2)[secutanu + ln|secu+tanu|] c

1 2) [x key wither(1+x 2) +ln|.]x+√(1+x^2)|]c

To require a definite integral of the second derivative of a function over an interval, the following steps can be used:

1.Find the second-order derivative of the function.

2.The second-order derivative function is substituted into the definite integral, and the integral interval is determined.

3.Calculate the points.

The specific solution is as follows:

1.Find the second-order derivative of the function.

Let the function be f(x), then its first-order guide is a field number f'(x), the second-order derivative is f''(x)。

2.The second-order derivative function is substituted into the definite integral, and the integral interval is determined.

Assuming that the definite integral of the second derivative of f(x) over the interval [a,b] is required, the integral expression is:

a,b] f''(x)dx

3.Calculate the points.

Since the interval of the integral is known, only the interval for f is required''(x) Carry out the indefinite integration of the cluster posture, and then substitute the upper and lower limits of the integral pure Zheng shout.

If necessary, the integration can be done by methods such as partial integration or commutation method. The end result is the value of the second derivative of f(x) on [a,b].

Dear: I hope I can help you, if you are satisfied with my service, yo, I wish you all the best!

Mathematical knowledge.

In the partial derivative, finding the partial derivative of x is treated as a constant, and if you find the partial derivative of x for a function that is all y, the result is 0.

Therefore, in the title, we find the partial derivative of v and get 0, which shows that the original function is full of functions about u, which is why 0 becomes f(u) after integration.

The first step here is to find the integral of v, which is the inverse process of finding the partial derivative of v, and the whole process has nothing to do with u, so f(u) appears out of thin air, and this integral only restores the function of u that disappears because of the partial derivative of v.

In the same way, in the second step, after the integration of u, there is a formula about v, just because when finding the partial derivative of u, the formula that is only related to v is derived as a constant to get 0, and here it is just a reduction of the v-related formula that disappears because of this.

1. Generally speaking, an integral without upper and lower bounds is an indefinite integral = indefinite integral;

There is an upper and lower bound integral is a definite integral = definite integral;

2. For the derivative of indefinite integrals, the result is integrand = integrand;

For definite integrals where the upper and lower bounds are both definite values, the derivative result is 0;

For the upper or lower bound, at least one is the derivative of the definite integral of the function, and the result is the function;

For the upper or lower bounds, not only the function, but also the integrand or the upper and lower bounds contain parameters, the derivative result is the function with parameters.

3. All double integrals must be definite integrals, but if the integral interval of this kind of definite integrals is fixed, the result after derivation is 0;

If the integral interval is not fixed, the result of the derivative is a function;

The specific derivation must be determined according to the specific integrand function and the change law of the specific integral region.

4. Generally speaking, it is necessary to turn double integral = double integral into cumulative integral = iterated

integral, only the appropriate successive integrals can be derived. Otherwise, there is no way to start.