# Is integrating both sides of an equation useful? [closed]

This asks whether or not differentiating both sides of an equation is allowed. Which it isn't, however, can you integrate both sides of an equation?

If we have,

$$x^2=x+1$$

Can we apply the integral operator and get,

$${1 \over 3} \cdot x^3 \ |_a^b={1 \over 2} \cdot x+x \ |_a^b$$

Haha, just kidding. That's too easy, I actually mean can we take the anti-derivative of both sides?

$${1 \over 3} \cdot x^3+C_1={1 \over 2} \cdot x+x+C_2$$ $$\Rightarrow {1 \over 3} \cdot x^3-{1 \over 2} \cdot x-x=C$$

Where $C$ is an arbitrary constant. Can doing this ever result in an equation simpler to solve? Perhaps this be used to derive new results? I know that this can be used when the equation is functional, but I'm interested in the cases when it isn't.

Here's one use I came up with,

We have,

$$(1) \quad x^2=x+1$$

$$(2) \quad {1 \over 3} \cdot x^3+C_1={1 \over 2} \cdot x+x+C_2$$ $$\Rightarrow {1 \over 3} \cdot x^3-{1 \over 2} \cdot x-x=C$$

With $C=-\cfrac{5\cdot \sqrt{5}+7}{12}$. Therefore, a root of $(2)$ is ${{\sqrt{5}+1} \over 2}$. This could be done for any order equation where a solution is known. For instance, you could have a quartic equation with known solutions, and then derive a solution to a quantic equation. Assuming you integrate and set $C$ to the right value. This would allow you to find specific solutions of quantic equations, which are not generally solvable.

For a concrete example, consider,

$$(3) \quad (x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4)=0$$

Integrating both sides results in,

$$(4) \quad {{x^5} \over 6}-{{5 \cdot x^4} \over 2}+{{35 \cdot x^3} \over 3}-25 \cdot x^2+24 \cdot x=C$$

If we wish to retain the solution $x=1$ we set $C=251/30$ and then we have,

$$(5) \quad {{x^5} \over 6}-{{5 \cdot x^4} \over 2}+{{35 \cdot x^3} \over 3}-25 \cdot x^2+24 \cdot x-{{251} \over {30}}=0$$

Where we actually know one of the solutions!

## closed as unclear what you're asking by copper.hat, Calle, Simon S, Zach466920, BruceETOct 16 '15 at 19:40

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• This makes no sense. The equation $x^2=x+1$ can only hold for a finite number of values, so how do you integrate? – copper.hat Oct 16 '15 at 16:37
• @copper.hat Set $C=-\cfrac{5 \cdot \sqrt{5}+7}{12}$... – Zach466920 Oct 16 '15 at 16:42
• @copper.hat Retain the solution to the equation...that's the whole point of the arbitrary constant...which is not ad hoc by the way. – Zach466920 Oct 16 '15 at 16:46
• This is meaningless. The equality must (obviously) hold over the range of integration. Which is clearly does not in the example above. – copper.hat Oct 16 '15 at 16:47
• @Zach466920 Apparently copper.hat is a knowledgable fellow when he says meaningless he means mathematically, if I were you I would appreciate his comment and would try to understand why is that the case instead of being defensive and offended. – clark Oct 16 '15 at 16:52

Differentiating on both sides of an equation (and integrating) is allowed when the equation is an equality of functions, as opposed to "what $x$ makes the right-hand side equal the left-hand side?"
For instance, if you have some function $f$ that you know is the product of two other functions, $g$ and $h$, then after setting $f(x) = g(x)h(x)$, you're allowed to differentiate and get $f'(x) = g'(x)h(x) + g(x)h'(x)$, or integrate and get $\int f(x)dx = \int g(x)h(x) dx$.