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.
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