I've always been curious about where the +C comes from. I fully understand its use in calculus, but recently I've been studying physics and realized that integrating time we tend to consider the lower bound as 0 and define the respective lower bound for the dependent variable as $y_0$. I am curious if my math work below is accurate for representing +C $$dy=dx$$ $$\int_{y_0}^y{dy}=\int_0^xdx$$ $$y-y_0=x$$ $$y=x+y_0$$

I will give this work done in a real example.

$$\frac{dy}{y}=6x^2dx$$ find solution with initial conditions y=0 x=4 $$\int_{y_0}^y\frac{dy}{y}=\int_0^x 6x^2dx$$ $$\ln|y||_{y_0}^y=2x^3|_0^s$$ $$\ln|y|-\ln|y_0|=2x^3-2(0)^3$$ $$\ln|y|=2x^3+\ln|y_0|$$ $$e^{\ln|y|}=e^{2x^3+\ln|y_0|}$$ $$y=y_0e^{2x^3}$$ $$y=4e^{2x^3}$$

My main question for this arose as integrating the dy side didn't give a +C but when we integrate the dx side that side gets a +C. I wonder if my math work where we consider the lower bound of the dependent variable as 0 to be the reason that +C only shows on the dx side. And because we define y as dependent on the x value, we simply consider the respective lower bound as $y_0$ which then gets eventually moved to the x side as +C.

My one exception I’ve found is the case of functions where having the lower bound as 0 for x returns two values. For example $e^x$ which integrated from 0 to x becomes $e^x-1$

  • 1
    $\begingroup$ The constant of integration is added for indefinite integrals. For definite integrals, like the ones in your example, it is not needed. Indeed, the indefinite integral of $6x^2$ is $2x^3+C$, but if you use that to compute the definite integral $(2x^3+C)|_0^x$ you get $(2x^3+C) - (2\cdot0^3+C) = 2x^3$ and the $C$'s cancel out. $\endgroup$
    – user856
    Apr 11, 2019 at 6:46
  • $\begingroup$ By the way you shouldn't use the same variable name for both the variable being integrated and the limits of the integration. $\endgroup$
    – user856
    Apr 11, 2019 at 6:47
  • $\begingroup$ The "constant of integration" on the LHS is $-\ln |y_0|.$ On the RHS it's $-2(0)^3=0 .$ $\endgroup$ Apr 11, 2019 at 7:02
  • 1
    $\begingroup$ This +C in an indefinite integral is added because the antiderivative is not uniquely defined. The definite integral is uniquely defined by its limit of integration. True, you can find an antiderivative using a definite integral, setting $F(x) = \int_{x_0}^x f(x')dx'$, with arbitrary $x_0$, and the freedom of choice of $x_0$ is related to the freedom of choice of constant $C$. However, there's no guarantee that for an arbitrary antiderivative $F(x)$ you can find $x_0$ such that $F(x)$ is given by the formula above. $\endgroup$ Apr 11, 2019 at 10:44


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