Mathematics Some of the most commonly used integral equations

Comrade, didn't Mr. Jiang talk about it!!I've heard it all!!I also sent a curly and fainted.

It's simple, but it's slow to type out on a computer.

Let's talk about the steps in a nutshell.

Sectional. 2.Even function, 2 times 0 to 2

3. Sectional. 4. Sectional.

5.Use Newton to not be in the Ni Ci Gong Dilemma Style.

6, the number of code pretending to be strange letter Wang mold files, is 0

7 Newton's formula.

8. Let the root number inside the t, and change the yuan to seek.

Find the indefinite integral of (x 3) (x 2 4x 13) and try to use rationalization.

This assumption is "redundant" and unnecessary.

Since it is a definite integral, the result must be constant, and f(x) must be a primary function.

If the upper and lower limits of the integral are not constants, it depends on the specific situation of the upper and lower limits of the integral, if it is only the product from the constant to x, you can find the integral factor (if=integral factor);

If the upper and lower limits are special functions, generally speaking, they cannot be solved.

It can be seen that f is a primary function set to y=x+a

then f(t)dt=(1 2x 2+ax)(0 to 1)=1 2+a=a2

a=-1y=x-1

You can set y=ax2+bx+c because it passes through two points.

You can get c=0, a+b=2

Its axis of symmetry is parallel to the y-axis and the opening is downward, so a<0 and its axis of symmetry y=-b 2a

Because it crosses the origin.

So the root is simple and slag is based on symmetry.

Another point of the shed (-a b,0) can be known

So it is possible to integrate in the range of 0.

can be obtained. a=-b3/3a2+b3/2a2b3/6a2

a+b=2 substitution.

According to the extreme value of the barrier quietly. Come and beg.

f(x) is discontinuous at x=, so the original function of f(x) at [1,3] does not exist.

It cannot be divided into [1, 3], and therefore the definite integral in [1,3] cannot be found.

Using the additivity of integrals: f(x) is converted into three intervals of the piecewise function on the integral of [-1,3] and the definite integration is performed separately.

Finding the Indefinite Integral [(cos t) ( sint)] dx solution: Define domain: sint>0, so 2k So the original formula = [cost(1-sin t) sint]dt=2 (1-U) du=2 (1-2U +U)du=2[u-(2 5)U +(1 9)U +c

2[√sint-(2/5)(sint)^(5/2)+(1/9)(sint)^(9/2)]+c.

1+x）/x=1+1/x

Integrals respectively to obtain (x+1) x dx =x+lnx+c (c is any constant) 1 key hu (2x 2) = 1 2 * x (-2) derivation 1 2 * (-2) * x (-3) = -1 x 3 If the manuscript helps you, it is a virtue to adopt the key sedan chair.