Question

Let $f(x)$ be an increasing function with inverse $g(x)$. Explain geometrically: $$ \int_0^a f(x) d x+\int_{f(0)}^{f(a)} g(x) d x=a f(a) $$

   
Let $f(x)$ be an increasing function with inverse $g(x)$. Explain geometrically:

$$
\int_0^a f(x) d x+\int_{f(0)}^{f(a)} g(x) d x=a f(a)
$$
Calculus
Calculus
Jon Rogawski,… 2nd Edition
Chapter 6, Problem 62 ↓
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Let $f(x)$ be an increasing function with inverse $g(x)$. Explain geometrically: $$ \int_0^a f(x) d x+\int_{f(0)}^{f(a)} g(x) d x=a f(a) $$
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Transcript

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00:01 Okay, we're given that f of x is an increasing function, and we have an inverse of f of x, which is g of x inverse.
00:13 And we're asked to show that from 0 to a of f of x t x, that's integral from 0 to f of a, from g of x, t x, that's this a times f of a.
00:26 Okay, so let me draw something else.
00:29 So if we know that f of x is an increasing function, so as we plug in positive numbers of x, we should get positive numbers of y, and that should increase.
00:40 So we have something like this.
00:42 It doesn't need to be a straight line, so it can be something like this.
00:48 Okay, and this is from zero to a, this is what we're considering, correct? so if we take the integral from 0 to a of our f of x function, that should just give us something down here.
01:01 But the inverse of f of x, which is g of x, should look something like this.
01:09 So it should be reflected on the y is equal to x axis.
01:14 So if this is symmetric, i should get something like the same thing here.
01:22 Right...
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