Application of definite integrals, find solution 20

Refer to the next volume of University Functions.

Solve for irregular graph area, work done by objects, etc.

Many problems in real life can be solved by definite integrals, such as solving the area of the figure and the work done by the object. In this paper, several simple applications of definite integrals in economics and geometry are given. Fixed points in an application of the economy factories regularly order raw materials, deposit them in warehouses for production use, etc.

Defined by a definite integral, its essence is the summation of continuous functions. In solving physical problems, the method of "segmentation, approximation, summing, and limiting" of definite integrals is appropriately infiltrated, and the physical problems are transformed into Wenshi spring potato problems for calculating definite integrals, which is helpful to improve the accuracy of the calculation of physical problems.

Analysis of definite integrals:

1. If a definite integral exists, it is a specific value (the area of a curved trapezoid), and an indefinite integral is a functional expression.

2. Functions, there can be indefinite integrals, but there are no definite integrals; It is also possible to have definite integrals without indefinite integrals. For a continuous function, there must be definite and indefinite integrals; If there are only a finite number of discontinuities, then a definite integral exists; If there is a hop break, then the original function must not exist, i.e., the indefinite integral must not exist.

3. Find the area surrounded by the image in the interval [a, b] of the forest shed function f(x). That is, the area of the graph enclosed by y=0,x=a,x=b,y=f(x).

There are too many to mention.

Here are 20 examples:

zhi1, the circumference formula dao

within the certificate; Appearance.

2. The circle area is common.

proof of the formula; 3. Proof of the formula of the volume of the sphere;

4. Proof of the formula for the surface area of the sphere;

5. Calculation of the centroid position of an arbitrarily shaped object;

6. Calculation of arbitrary curve length;

7. Calculation of ellipse area;

8. Calculation of ellipse circumference;

9. Ellipsoid volume three calculations;

10. Calculation of ellipsoid surface area;

11. Changing force to do work;

12. Formula for calculating spring potential energy;

13. Calculation of moment of inertia;

14. Calculation of capacitance of capacitors of various shapes;

15. Calculation of the potential energy division of the forehead around the charged body;

16. Calculation of magnetic field distribution around current-carrying wires;

17. Calculation of average current, voltage and average power of alternating current;

18. Mass calculation of objects with uneven mass density;

19. Calculation of electricity of objects with uneven charge density;

20. Calculation of enthalpy change in chemical reactions;

Thousands and thousands of them, and they can never be finished.

This question is done using the thin shell method.

2 pie x is the length of the enclosed cube, dx is the width, and f(x) is the height.

That is, take a long strip in any (x,x+dx) interval and cut the hollow cylinder obtained by rotating it around the y-axis to get something similar to a cuboid The volume of this cuboid is 2 xf(x)dx and then take the definite integral. Didn't the teacher talk about it?

The curves y = x, y = 1 x, x = 2 and y = 0 are flat shapes that are:

The area below the line y = x, the curve y = 1 x, above the x axis, and to the left of the line x = 2 can be drawn by yourself. The line y = x and the curve y = 1 x intersect at the point (1, 1).

v = π 0, 1>x^2dx + 1, 2>(1/x^2)dx ]

[x^3/3]<0, 1> +1/x]<1, 2>

1、s=∫[lna,lnb] e^ydy=e^y[[lna,lnb]=e^(lnb)-e^(lna)=b-a.

2. y'=-2x+4, when x=0, y'=4, tangent equation.

is: (y+3) x=4, y=4x-3, x=3, y'=-2, the tangent equation is: y (x-3)=-2, y=-2x+6, and the intersection of the two tangents is:

3/2，3），s1=∫[0,3/2](4x-3-(-x^2+4x-3))dx=∫[0,3/2]x^2dx=x^3/3[0,3/2]=9/8,s2=∫[3/2,3][(2x+6)-(x^2+4x-3)]dx=∫[3/2,3](x^2-6x+9)dx

x^3/3-3x^2+9x)[3/2,3]

9-27+27-(9/8-27/4+27/2)=9/8,s=s1+s2=9/8+9/8=9/4.

3. There are two tangent lines, take the first quadrant.

y'=e x, set the tangent point.

is (x0,y0), the tangent equation y x=e x0, e x0 x0=e x0, x0=1, y0=e, tangent equation: y=ex, s= [0,1](e x-ex)dx

e^x-ex^2/2)[0,1]=e-e/2-(1-0)=e/2-1.

4、v=∫πy^2dx=π∫[0,x0](4ax)dx

4πax^2/2[0,x0]= 2πax0^2.

5、(1)s=(1/2)∫[0,1]y^2dy=y^3/6[0,1]=1/6.

2)v=π∫[0,1/2]y^2dx=π∫[0,1/2]2xdx=πx^2[0,1/2]= π/4.

6、x=-lny,v1= π*1^2*1/e =π/e,v2=π∫[1/e,1]x^2dy=∫[1/e,1](-lny)^2dy

y(lny)^2-2ylny+2y][1/e,1]

2π-4π/e,)

For (lny) 2dy, two partial integrals are used, v= e+2 -4 e=2 -3 e

If y=0, then it should be 2 -3 e, because there is also a part of the cylindrical volume, its radius is 1, the height is 1 e, if y=1, then v=2 -4 e, please correct the question.

This is the first question, if you can see it, I will send the answer later.

The drawing is y=e x, but it does not affect the final conclusion.

Please ask the landlord about me. I won't add more than one**. You have to ask once to send one more **.

When rotated about the x-axis, the shape of the rotating body resembles a ring (solid at x = 0); The outer diameter is r = 2 - x and the inner diameter is r = x; The cross-sectional area is s = (r -r) = [2 - x) x) so it is minus, but the integral interval is the same.

v = ∫₋2 - x²)²x²)²dx

When rotating around the y-axis, the rotating body is a solid, and it is easier to use y as an independent variable. The cross-sectional area at y is different within [0, 1], 1, 2].

At [0, 1], the cross-section radius is determined by y = x, r = x = y

At y at [1, 2], the cross-section radius is determined by y = 2 - x and r = x = (2 - y).

So to divide the integrals.

I've been studying for a long time, and I've almost forgotten it, but I still understand it after looking at it. Remember that the formula is about x and the integral of x = (a,b)f(x) dx.

1. When rotating around the x axis, the integration of x, y=2-x will not form an ellipsoid at (-1,1), of course, it will cut off both sides, and the middle y=x will rotate out an hourglass-like blank, of course, to subtract.

2. When rotating around Y, it is divided into two parts, which are actually equal, and it is not the same to multiply 2 directly, which is obviously symmetrical. The textbook writes that a line is drawn in the middle, and the upper half of y=2-x and y=1 besieged the figure of y integral, y of course in (1,2) and the second half (0,1).

The center of the ellipse is the center of the circle, the major axis is the x-axis, and the short base axis is the y-axis, and the spatial coordinate system is established.

v = f(x,y,z) dv = dx ||f(x,y,z) dydz

Consider x as known, and find the area s(x) of the "regular triangle".

s = 9 root number 3 (1 - x 2 16).

However, after the judgment is changed, the points are awarded for s(x) digging, and x is from -4 to 4