Solving for $x$ in $\int_0^2x^tdt=3$

Question

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

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1. We are solving for the unknown variable $x$.
2. $\int_0^2x^tdt=3$ 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

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$$\int_0^2x^tdt=\frac{x^3}{3}-x=3\text{ Since }\int_a^bx^tdt=\left(\frac{x^{a+1}}{(a+1)}\right)-\left(\frac{x^{b+1}}{(b+1)}\right)$$$$\text{Which we can rewrite our equation that we have now as }\frac{x^3}{3}-x=3$$$$=x^3-3x=9$$$$\text{Or }x^3-3x-9=0$$$$\text{And plugging this into the cubic formula gets us:}$$

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$$\text{Real solution (exact form)}$$

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$$x=\frac{1}{3}\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}+\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}$$

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$$\text{Real solution (approximate form)}$$

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$$x\approx2.5541$$

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$$\text{Complex solutions}$$

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$$\text{Complex solution #1 (exact form)}$$

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$$x=-\frac{1}{6}\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}-\frac{1}{2}\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}+i\left(\frac{\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}}{2\sqrt3}-\frac{1}{2}\sqrt3\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}\right)$$

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$$\text{Complex solution #2 (exact form)}$$

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$$x=-\frac{1}{6}\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}-\frac{1}{2}\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}+i\left(\frac{1}{2}\sqrt3\sqrt[3]{\frac{1}{2}(9+\sqrt{77})}-\frac{\sqrt[3]{\frac{243}{2}-\frac{27\sqrt{77}}{2}}}{2\sqrt3}\right)$$

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$$\text{Complex solutions (approximate form)}$$

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$$x\approx-1.2771\pm1.3758i$$Meaning that the solutions for $x$ are:$$x\approx-1.2771\pm1.3758i\text{, }2.5541$$

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$$\text{My question}$$

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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?

Answer 1

Actually both of your attempts are wrong because you have not done the integration properly.

$\int_0^2 x^t dt =3$

Now, $\int_0^2 x^t dt = \int_0^2 (e^{\ln(x)})^t dt = \int_0^2 e^{\ln(x)t} dt =\left[\dfrac{e^{\ln(x)t}}{\ln(x)}\right]_{t=0}^{t=2}=\dfrac{e^{2\ln(x)}-e^{0\ln(x)}}{\ln(x)}=\dfrac{e^{\ln(x^2)}-e^0}{\ln(x)}=\dfrac{x^2-1}{\ln(x)}$

So, we have to solve $\dfrac{x^2-1}{\ln(x)}=3$ for $x$.

Thus, now we want a solution to the equation $x^2-1=3\ln(x)$ for $x>0$ and $x\ne 1$.

According to WolframAlpha, $x\approx 1.46425$

Answer 2

Hint: Remember, the variable of integration is $t$, not $x$. So the integrand is a function of the form $f(t) = a^t$, not a function of the form $g(t) = t^n$.

In other words it is the base, not the exponent, that is constant.

It is helpful then to write $a^t$ as $e^{(\ln a)t}$ and then consider how to integrate functions of the form $h(t) = e^{ct}$.