So I was scrolling through the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I found this video by Blackpenredpen. The question in the video was asking to solve for an unknown $x$ in the equation$\int_0^2x^tdt=3$ because people apparently kept saying that the integrals in the videos were too easy in the comment sections. However, I thought that I might be able to solve this equation. This is my attempt at solving$$\int_0^2x^tdt=3$$What I know
- We are solving for the unknown variable $x$.
- $\int_0^2x^tdt=3$ can be rewritten as $$\int_0^2x^tdt=x^2-x^0=x^2-1=3$$Therefore, $x^2$ must be equal to $4$, meaning that $x$ is equal to $2$. However, after plugging that into Desmos, it turns out that I somehow went wrong with the integral simplification, since that would equal $\approx4.32608512267$.
My second attempt at solving the hard integral
$$\int_0^2x^tdt=\frac{x^3}{3}-x=3\text{ Since }\int_a^bx^tdt=\left(\frac{x^{a+1}}{(a+1)}\right)-\left(\frac{x^{b+1}}{(b+1)}\right)$$$$\text{Which we can rewrite our equation that we have now as }\frac{x^3}{3}-x=3$$$$=x^3-3x=9$$$$\text{Or }x^3-3x-9=0$$$$\text{And plugging this into the cubic formula gets us:}$$
$$\text{Real solution (exact form)}$$
$$x=\frac{1}{3}\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}+\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}$$
$$\text{Real solution (approximate form)}$$
$$x\approx2.5541$$
$$\text{Complex solutions}$$
$$\text{Complex solution #1 (exact form)}$$
$$x=-\frac{1}{6}\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}-\frac{1}{2}\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}+i\left(\frac{\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}}{2\sqrt3}-\frac{1}{2}\sqrt3\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}\right)$$
$$\text{Complex solution #2 (exact form)}$$
$$x=-\frac{1}{6}\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}-\frac{1}{2}\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}+i\left(\frac{1}{2}\sqrt3\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}-\frac{\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}}{2\sqrt3}\right)$$
$$\text{Complex solutions (approximate form)}$$
$$x\approx-1.2771\pm1.3758i$$Meaning that the solutions for $x$ are:$$x\approx-1.2771\pm1.3758i\text{, }2.5541$$
$$\text{My question}$$
Is my second attempt solving the integral correct, or was it not a good idea for me to attempt plugging what I got from simplifying the integral into the cubic formula?