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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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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?
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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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