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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!

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This makes no sense. The equation $x^2=x+1$ can only hold for a finite number of values, so how do you integrate? $\endgroup$ – copper.hat Oct 16 '15 at 16:37
  • $\begingroup$ @copper.hat Set $C=-\cfrac{5 \cdot \sqrt{5}+7}{12}$... $\endgroup$ – Zach466920 Oct 16 '15 at 16:42
  • $\begingroup$ @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. $\endgroup$ – Zach466920 Oct 16 '15 at 16:46
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    $\begingroup$ This is meaningless. The equality must (obviously) hold over the range of integration. Which is clearly does not in the example above. $\endgroup$ – copper.hat Oct 16 '15 at 16:47
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    $\begingroup$ @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. $\endgroup$ – clark Oct 16 '15 at 16:52
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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 $\int f(x)dx = \int g(x)h(x) dx$.

So the most prominent case where integrating or differentiating on either side is both allowed and helpful is when the unknowns themselves are functions. This is commonly seen, for instance, in differential equations.

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  • $\begingroup$ This doesn't answer the question "Is integrating both sides of an equation useful?" I already knew this, that's why it was in the question. $\endgroup$ – Zach466920 Oct 16 '15 at 17:25

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