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**Soroban** Hello, CHARLIEDANCE!

Assuming that the first graph is a down-opening parabola,

. . symmetric to the $\displaystyle y$-axis, we have: .$\displaystyle y \:=\:c - ax^2$

The $\displaystyle x$-intercepts are $\displaystyle (\pm1,0)$

. . So we have: .$\displaystyle c - a(\pm1)^2 \:=\:0 \quad\Rightarrow\quad c \:=\:a$

The function (so far) is: .$\displaystyle f(x) \:=\:a(1-x^2)$

The area on $\displaystyle [\text{-}1,1]$ is 1:

. . $\displaystyle \displaystyle A \;=\;2a\int^1_0(1-x^2)\,dx \;=\;1$

. . . . . . . $\displaystyle 2a\left(x - \frac{1}{3}x^3\right)\,\bigg]^1_0 \;=\;1 $

. . . . . . . . . . . . . . $\displaystyle 2a\cdot\frac{2}{3} \;=\;1$

. . . . . . . . . . . . . . . . . $\displaystyle a \;=\;\frac{3}{4}$

Therefore: .$\displaystyle f(x) \;=\;\frac{3}{4}\left(1-x^2\right)$

Assuming that the second graph is a down-opening *quartic,*

. . symmetric to the $\displaystyle y$-axis, we have: .$\displaystyle y \:=\:c - ax^4$

The $\displaystyle x$-intercepts are $\displaystyle (\pm1,0)$

. . So we have: .$\displaystyle c - a(\pm1)^4 \:=\:0 \quad\Rightarrow\quad c \:=\:a$

The function (so far) is: .$\displaystyle g(x) \:=\:a(1-x^4)$

The area on $\displaystyle [\text{-}1,1]$ is 1:

. . $\displaystyle \displaystyle A \;=\;2a\int^1_0(1-x^4)\,dx \;=\;1$

. . . . . . . $\displaystyle 2a\left(x - \frac{1}{5}x^5\right)\,\bigg]^1_0 \;=\;1 $

. . . . . . . . . . . . . . $\displaystyle 2a\cdot\frac{4}{5} \;=\;1$

. . . . . . . . . . . . . . . . . $\displaystyle a \;=\;\frac{5}{8}$

Therefore: .$\displaystyle g(x) \;=\;\frac{5}{8}\left(1-x^4\right)$

I'll let *you* show that the arc lengths are different . . .