The proof of 〔∫ (a, b) f (x) DX 〕 & # 178 in double integral; =∫(a,b)f(x)dx∫(a,b)f(y)dy

The proof of 〔∫ (a, b) f (x) DX 〕 & # 178 in double integral; =∫(a,b)f(x)dx∫(a,b)f(y)dy


As shown in the picture, if not clear, please ask. Please evaluate in time



Let f (x) be continuous on [a, b], and f (b) = a, f (a) = B. It is proved that ∫ (upper B, lower a) f (x) f '(x) DX = 1 / 2 (A & # 178; - B & # 178;)


Integral = ∫ f (x) DF (x) = [f (x)] ^ 2 / 2 = [f (b)] ^ 2 / 2 - [f (a)] ^ 2 / 2 = (a ^ 2-B ^ 2) / 2



It is proved by double integral that: [ʃ (a to b) f (x) DX] ²


On the square [a, b] * [a, b], there is [f (x) -- f (y)] ^ 2 > = 0
The double integral of f ^ 2 (x) + f ^ 2 (y) - 2F (x) * f (y) > = 0
It is easy to calculate that the double integrals of f ^ 2 (x) and f ^ 2 (y) are (B -- a) * integrals (from a to b) f ^ 2 (x) DX
The double integral of F (x) * f (y) is equal to (integral (from a to b) f (x) DX) ^ 2



If ∫ (1 → x) f (T ^ 2) DT = x ^ 3, then ∫ (0 → 1) f (x) DX=
If we know ∫ (1 → x) f (T ^ 2) DT = x ^ 3, then ∫ (0 → 1) f (x) DX = 3 / 2


∫ f (T ^ 2) DT = x ^ 3 = = > F (x ^ 2) = 3x ^ 2
∴f(x)=3x
So ∫ f (x) DX = 3 ∫ xdx
=3*(1/2)
=3/2.



Make a vertical line from a vertex of a triangle to its opposite side. The line between the vertex and the perpendicular foot is called the line of the triangle______ .


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From one () to its () make a line (). The line () between () and () is called the height of the triangle. This line () is called the bottom of the triangle
I've been struggling with this problem for a long time,


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Isosceles triangle



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Given several elements of a triangle, the process of finding other elements is called solving a triangle



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The basic element of a triangle: the segments that make up the triangle are called (the sides of the triangle)