I am a junior high school third year. The following is a method I thought of. The content is original. I don’t know if it has become the wisdom of my predecessors.
As an OIer, some of the proofing process will not be very rigorous, please forgive me.
describe
For a given function\(f(x)=a_0x^0+a_1x^1+a_2x^2+\dots +a_nx^n\),beg\(l\le x\le r\)When the function\(f(x)\)and\(x\)The area of the shaft clamp.
Note that the area is positive and negative, that is,\(x\)Above the axis, function\(f(x)\)The area below is positive.\(x\)Below the axis, function\(f(x)\)The area above is negative.
break down
First,\(f(x)\)Disassembled into\(g_0(x)+g_1(x)+g_2(x)+\dots+g_n(x)\),in\(g_i(x)=a_ix^i\)。
So\(f(x)\)The area below can also be disassembled into\(g_0(x),g_1(x),g_2(x),\dots,g_n(x)\)the sum of areas below.
So we just need to ask for each\(g_i(x)\)The area below is enough.
You may wish to set up\(l\ge 0\)（\(l<0\)The same is true for the time being), in\(l\le x\le r\)The area of this section can be used\(0\le r\)Subtract the area of the part\(0\le x\le l\)Part of the area is obtained, so we only need to solve it\(0\le x\le r\)Just a question.
Constructing the model
Now we only need to solve this problem:
- Have functions\(f(x)=x^k\),beg\(0\le x\le r\)hour\(f(x)\)area under.
Consider when\(x=a\)hour,\(f(x)=a^k\), this is equivalent to a side length of\(a\)of\(k\)The volume of the dimension cube,\(k\)The coordinates of the dimension are expressed as\((x_0,x_1,x_2,\dots,x_{k-1})\), then this cube can be represented as\(0\le x_0\le a,0\le x_1\le a,0\le x_2\le a,\dots,0\le x_{k-1}\le a\)。
So for\(0\le x\le r\)The sum of these cubes in this interval can be expressed as one\(k+1\)Dimensional cone\(0\le x_0\le r,0\le x_1\le x_0,0\le x_2\le x_0,\dots 0\le x_k\le x_0\)。
Then the volume of this cone is mentioned above\(f(x)\)area under.
Ask for an answer
Can be constructed\(k+1\)Equivalent cone body (for the convenience of writing, the following default\(x_i\ge 0\)）：
- \(x_0\le r,x_1\le x_0,x_2\le x_0,\dots,x_k\le x_0\)
- \(x_1\le r,x_0< x_1,x_2\le x_1,\dots,x_k\le x_1\)
- \(x_2\le r,x_0<x_2,x_1<x_2,\dots,x_k\le x_2\)
- \(\dots\)
- \(x_k\le r,x_0<x_k,x_1<x_k,\dots,x_{k-1}<x_k\)
Although the situations of the above-mentioned conical bodies with less than the sign and so on are not the same, they are actually the same, because when two numbers are equal, this figure is dimensionality-reduced, it is\(k+1\)The volume under the dimension is\(0\)。
So the above\(k+1\)Graphics can be assembled into high-dimensional cubes\(x_0\le r,x_1\le r,x_2\le r,\dots,x_k\le r\), then this\(k+1\)The sum of the volumes of the cone is\(r^{k+1}\), then the volume of a cone is\(\frac{r^{k+1}}{k+1}\)。
Right now,\(0\le x\le r\)hour,\(f(x)=x^k\)The area below is\(\frac{r^{k+1}}{k+1}\)。
prove
Next prove the above\(k+1\)The cone is the same as the last cube:
Proof: All points in the cone are in the cube
For any cone, it is represented as\(x_{max}\le r,x_0<x_{max},x_1<x_{max},\dots,x_k\le x_{max}\)For each point\((x_0,x_1,x_2,\dots,x_k)\), the maximum value of these coordinates is less than or equal to\(r\), then each dimension is less than or equal to\(r\), then this point must be in the cube.
Proof: All points in the cube are in the cone
For any point in the cube\((x_0,x_1,x_2,\dots,x_k)\), take out the maximum value\(x_{max}\)(If there are multiple selection subscripts with the smallest subscript), then this point must be in the cone body\(x_{max}\le r,x_0<x_{max},x_1<x_{max},\dots,x_k\le x_{max}\)middle.
At this point, the certification is completed.
in conclusion
For functions\(f(x)=x^k\),exist\(0\le x\le r\)Inside,\(f(x)\)The area below is\(\frac{r^{k+1}}{k+1}\)。
For functions\(f(x)=a_0x^0+a_1x^1+a_2x^2+\dots+a_nx^n\),exist\(l\le x\le r\)Inside,\(f(x)\)The area below is\(\sum_{i=0}^n a_i(\frac{r^{i+1}}{i+1}-\frac{l^{i+1}}{i+1})\)。