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let f be a monotone function on the interval [a,b].
f(a)=[tex]\alpha[/tex] f(b)=[tex]\beta[/tex].
let g be the inverse function of f i.e f^-1.
prove that:
[tex]\int_{\alpha}^{\beta}g(y)dy=b\beta-a\alpha-\int_{a}^{b}f(x)dx[/tex] when you consider the definition of the integral as area.
here what i did:
Sn=[tex]\sum_{i=0}^n g(\zeta_i)(y_i-y_{i-1})[/tex]
f(x)=y f^-1(y)=x=g(y)
y_i-y_i-1=f(x_i)-f(x_i-1)
S'n=[tex]\sum_{i=0}^n \zeta_i(x_i-x_{i-1})[/tex]
[tex] S'_n+\sum_{i=0}^n x_{i-1}\zeta_i=\sum_{i=0}^n x_i\zeta_i=\sum_{i=0}^n g(\zeta_i)y_{i-1}[/tex]
as n approaches infinity S'_n approaches the second integral in with limits a and b, and the first sum of yi times g(zeta_i) minus zeta_i times x_i-1 converges to b*beta- a*alpha, is this correct?
f(a)=[tex]\alpha[/tex] f(b)=[tex]\beta[/tex].
let g be the inverse function of f i.e f^-1.
prove that:
[tex]\int_{\alpha}^{\beta}g(y)dy=b\beta-a\alpha-\int_{a}^{b}f(x)dx[/tex] when you consider the definition of the integral as area.
here what i did:
Sn=[tex]\sum_{i=0}^n g(\zeta_i)(y_i-y_{i-1})[/tex]
f(x)=y f^-1(y)=x=g(y)
y_i-y_i-1=f(x_i)-f(x_i-1)
S'n=[tex]\sum_{i=0}^n \zeta_i(x_i-x_{i-1})[/tex]
[tex] S'_n+\sum_{i=0}^n x_{i-1}\zeta_i=\sum_{i=0}^n x_i\zeta_i=\sum_{i=0}^n g(\zeta_i)y_{i-1}[/tex]
as n approaches infinity S'_n approaches the second integral in with limits a and b, and the first sum of yi times g(zeta_i) minus zeta_i times x_i-1 converges to b*beta- a*alpha, is this correct?