If f and g are inverse functions and f ' is continuous, prove that: [from a to b] ∫ f(x) dx = b f(b) - a (a) - [from f(a) to f(b)] ∫ g(y) dy Hint: Use part (a) and make the substitution y = f(x) I have been trying to rearrange the equations and have looked online at answers, and still don't quit understand how to solve this question. From this link (http://math.berkeley.edu/~bgillesp/m1b-s12/files/worksheets/ws1-solutions.pdf) it says: First use integration by parts, and then in the remaining integral use integration by substitution with u = f(x). Keep in mind the defining relation x = g(f(x)) = g(u), and also recall the derivative of an inverse function, g′(x) = 1/f′(g(x)), which is obtained by implicit differentiation on f(g(x)) = x. I tried to solve the question using this method, but have not really gotten anywhere. Also, from here (http://answers.yahoo.com/question/index?qid=20090908094826AAQg3Ky) the answer states: ∫(a to b) f(x) dx = xf(x) [for x = a to b] - ∫(a to b) xf '(x) dx = (b f(b) - a f(a)) - ∫(a to b) xf '(x) dx. As for the second term, substitute y = f(x). So, x = g(y) and dy = f'(x) dx. Bounds: x = a ==> y = f(a) and x = b ==> y = f(b). Thus, we get ∫(a to b) f(x) dx = (b f(b) - a f(a)) - ∫(f(a) to f(b)) g(y) dy. which I don't entirely understand since it looks like he just went in a circle and ended up with the original equation without showing anything. I'm just a little confused on how original functions and their inverses relate to each other by this relation when integrated, and any help would be greatly appreciated! Thank you!