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(a) If [tex]f[/tex] is one-to-one and [tex]f^{\prime}[/tex] is continuous, prove that

[tex]\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy[/tex]

(b) In the case where [tex]f[/tex] is a positive function and [tex]b > a > 0[/tex], draw a diagram to give a geometric interpretation of part (a).

My work:

(a) [tex]\int _a ^b f(x) \: dx[/tex]

Integrating by parts gives

[tex]u = f(x) \Rightarrow \frac{du}{dx} = f ^{\prime} (x) \Rightarrow du = f ^{\prime} (x) \: dx[/tex]

[tex]dv = dx \Rightarrow v = x[/tex]

[tex]\int _a ^b f(x) \: dx = \left. xf(x) \right] _a ^b - \int _a ^b x f ^{\prime} (x) \: dx[/tex]

[tex]\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _a ^b x f ^{\prime} (x) \: dx[/tex]

Applying the Substitution Rule gives

[tex] y = f(x) \Leftrightarrow x = f^{-1} (y) \Rightarrow \frac{dy}{dx} = f ^{\prime} (x) \Rightarrow dx = \frac{dy}{f ^{\prime} (x)}[/tex]

[tex]y(b) = f(b)[/tex]

[tex]y(a) = f(a)[/tex]

[tex]\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy[/tex]

(b) I'm not sure how I should handle this one. The left-hand side is quite easy to visualize: it corresponds to a generic integral from a to b. The right-hand side does not seem to be that simple, and I need some help.

Any help is highly appreciated.