Question

Find a continuous function $f$, where $(x^4-4x^2)\le f\left(x\right)\le (2x^2-x^3)$ such that the area bounded by $y=f\left(x\right),y=x^4-4x^2$, y-axis, and the line $x=t$, where $(0\le t\le 2)$ is $k$ times the area bounded by $y=f\left(x\right),y=2x^2-x^3$, y-axis, and line $x=t$ (where $0\le t\le 2)$.

Answer 1

$0\le t\le 2\&x=t$

Because ${\int }_0^2[f\left(x\right)-(x^4-4x^2)]dx=k\left[{\int }_0^2(2x^2-x^3)-f\left(x\right)\right]dx$

$(x^4-4x^3)\le f\left(x\right)\le (2x^2-x^3)$

${\int }_0^{2}f\left(x\right)dx-{\int }_0^2(x^4-4x^2)dx=k{\int }_0^2(2x^2-x^3)dx-k{\int }_0^{2}f\left(x\right)dx$

$\Rightarrow (1+k){\int }_0^{2}f\left(x\right)dx=k{(\frac{2x^3}{3}-\frac{x^4}{4})}_0^2+{(\frac{x^5}{5}-\frac{4x^3}{3})}_0^2$

$\Rightarrow (1+k){\int }_0^{2}f\left(x\right)dx=k(\frac{2^4}{3}-\frac{2^4}{4})+(\frac{2^5}{5}-\frac{4\times 2^3}{3})$

$=k\left(\frac{16}{12}\right)-\frac{64}{15}$

$(1+k){\int }_0^{2}f\left(x\right)dx=\frac{4k}{3}-\frac{64}{15}$

${\int }_0^{2}f\left(x\right)dx=\left(\frac{k}{k+1}\right)\frac{4}{3}-\left(\frac{1}{1+k}\right)\frac{64}{15}$

The f(x) can be predicted by this as

$f\left(x\right)=\left(\frac{k}{k+1}\right)\left(\frac{x^2}{2}\right)-\frac{1}{(1+k)}\frac{2x^4}{3}$