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The area of between the function $f(x)=x^2$ and the $x$-axis from $1\to a$ is the same as the area between $f^{-1}(x)$ and the $y$-axis from $1 \to b$ when $f(a)=b$

It says write two equations of $a$ in terms of $b$, and so far I've got this $f(a)=b \therefore a=\sqrt{b}$

Then solve for $a$

Question: What is the working to get the second equation of a in terms of $b$?

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I don't understand what is your question. – Sigur Jan 16 '13 at 22:20
Tried to word me a bit more spesific now, hope you'll understand it now – user54630 Jan 16 '13 at 22:25
A way to get a second equation would be to calculate those two areas. – Gerry Myerson Jan 16 '13 at 22:38
up vote 2 down vote accepted

For $f:[0,\infty)\to [0,\infty)$ given by $x\mapsto f(x)=x^2$, it is a bijection and its inverse function is $g:[0,\infty)\to [0,\infty)$ given by $x\mapsto g(x)=\sqrt{x}$.

The area below the graph of $f$ and above the $x$-axis between the vertical lines $x=1$ and $x=a$ is given by $$A=\int_1^a x^2dx=\left.\frac{x^3}{3}\right|_1^a=\frac{a^3-1}{3}.$$

Now, can you compute the other area using the inverse function?

enter image description here

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Will it be: $\frac{2b^{\frac{3}{2}}-2}{3}$? – user54630 Jan 16 '13 at 22:48
yes good job, can you solve for a now – kaine Jan 16 '13 at 22:54
$^3\sqrt{a^3}=^3\sqrt{{2b^{\frac{3}{2}}-2}}$. Is this correct? – user54630 Jan 16 '13 at 22:58
$$a={2a^3-^3\sqrt{2}}$$ This doesn't make sense does it? – user54630 Jan 16 '13 at 23:01
no unfortunately not; $\frac{a^3-1}{3}=\frac{2a^3-2}{3}$ is a good place to start; that is A1=A2 with the b replaced with $b=a^2$ – kaine Jan 16 '13 at 23:07

There is no need to do any calculations. This statement holds for all increasing functions.

The simple way to think about it, is that the graph of $f^{-1} (x)$ is the reflection of the graph of $f(x)$ about the line $y=x$. And the reflection of the $y-$axis about the line $y=x$ is just the $x-$axis. Finally, note that $f(1) = 1$ and $f(a) = b$.

Hence, the 2 figures are congruent by this reflection, so the areas are the same.

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