Questions to an integration

$\frac{2x^3}{3}+$ $2x^2 =\frac{4}{3}$

in the end I got the value of x by trials. It is x = 0,732 (ca.). My question in this case is, what I have to do to solve my calculation to x? What should I do to complete my calculation?

In the second case x is 2 and the integration is a function of zero. The value of a is ask for. After an integration rule, namely $\int_{a}^{a}f(x) \, dx = 0$ I suppose that the value of a must be 2 too.

$\frac{40}{3}-$$\frac{2a^3}{3}-$$2a^2 = 0$

And when I put a = 2 in the equation, I get the function of zero for the integration, but my question is the same: what I have to do to solve the equation to a?

Why are the variables x and t, as well the integration operators dt and dx randomly scattered all over the place?

Peace.

Re-arrange your equations into the standard form of a polynomial. Multiply through by 3/2 to simplify it a bit and get integer coefficients.
(...)

Yes, that is one possible value of "a". And the remainder of your working will show that there is no other possible value.

Similar method to the first one. Except in this case you know one of the roots - it's 2. So, by the factor theorem, (a-2) is a factor. If you take that out you're left with a quadratic, which in this case has no roots.

Why I should multiply by 3/2? or did you mean 2/3? as the term is $\frac{2x^2}{3}$ I would multiply by three and then I get $2x^3 + 6x^2 = 4$ and now I subtract the equation by 4 and I get $2x^3 + 6x^2 - 4 = 0$

and then it is possible to use a polynomial division which you told, I would get an quadratic function and I am able to solve the variable by pq-formula. But then I apply these steps, I would calculate a value of a zero of a function. That's why it makes no sense to use a polynomial division in my opinion.

Could you be more precise, please? I don't still know what should I do exactly in both of them.

(...) Which part don't you understand, and need more details for?

I have found out that there is no difference whether I multiply by 3/2 or divide by 2/3

I used a polynomial division by x+2...

When I divide $\frac {-2a^3} {3} - 2a^2 + \frac {40} {3} = 0$ by -2/3, I get $a^3 + 3a^2 - 20 = 0$. And when I use a polynomial division by x-2 I get: $a^2 + 5x + 10$. I see at the first glance that a = 2 is not a zero of a function in the term. What is wrong again?

(...)
The quadratic has no real roots, and hence the only zero is a=2.

In other words: it is useless to use the polynomial division. That is to say I have to solve the cubic function $a^3 + 3a^2 - 20 = 0$ to get the variable. Am I right? then I have to use the solution process of a cubic function. That is the only possibility which makes sense in my opinion.

No.

You know x=2 is a solution to Fa(x)=0

So, (x-2) is a factor.

After polynomial division, the remaining quadratic has no roots, no zeros.

So x=2 is the only solution.

Done.

Maybe this image will help you understand?