Suppose $g$ is a strictly growing function on the interval $[a,b]$ and that $g$ and $g^{-1}$ are differentiable. Show that $$\int_{a}^{b} g(x) dx + \int_{g(a)}^{g(b)} g^{-1}(x) dx = bg(b)-ag(a) .$$ Here's how I tried: $$ \int_{a}^{b} g(x) dx = xg(x) \bigg|^{b}_{a} - \int_{a}^{b} x g'(x) dx =bg(b)-ag(a) - \int_{g(a)}^{g(b)} g^{-1}(u) du,$$ $$ u = g(x) $$ And so my idea is that $$\int_{g(a)}^{g(b)} g^{-1}(x) dx - \int_{g(a)}^{g(b)} g^{-1}(u) du = 0$$ and thus we're done but I'm guessing I'm wrong so if anyone could correct me that'd be great.
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Marco Cantarini
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Rousseau
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Your proof is correct. What's the matter in your opinion? – Crostul Feb 18 '15 at 12:45
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Oh cool!. Normally something goes wrong... – Rousseau Feb 18 '15 at 12:46
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Notice that the relation holds even if $g$ is not differentiable. See the Wikipedia article "integral of inverse function" (that I wrote). – MikeTeX Feb 18 '15 at 13:58
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You can also look at this question. – Julián Aguirre Feb 18 '15 at 17:33
