It is cosh(x) + c where c is a constant of integration.
Assuming integration is with respect to a variable, x, the answer is 34x + c where c is the constant of integration.
The solution to a differential equation requires integration. With any integration, there is a constant of integration. This constant can only be found by using additional conditions: initial or boundary.
The indefinite integral is the anti-derivative - so the question is, "What function has this given function as a derivative". And if you add a constant to a function, the derivative of the function doesn't change. Thus, for example, if the derivative is y' = 2x, the original function might be y = x squared. However, any function of the form y = x squared + c (for any constant c) also has the SAME derivative (2x in this case). Therefore, to completely specify all possible solutions, this constant should be added.
When you find an indefinite integral of a function (ie, the integral of a function without integration limits) you are actually finding the antiderivative of that function. In other words, you are finding the function whose derivative is the function 'inside' the integral sign. Recall that the derivative of a constant is zero. The point here is that you add the 'c' to acknowledge the fact that when the derivative of the result of your integration effort is taken to get the original function it could, or would, have been followed by some unknown constant value that disappeared upon differentiation. That constant is denoted by the 'c'.
It is cosh(x) + c where c is a constant of integration.
Assuming integration is with respect to a variable, x, the answer is 34x + c where c is the constant of integration.
The solution to a differential equation requires integration. With any integration, there is a constant of integration. This constant can only be found by using additional conditions: initial or boundary.
The indefinite integral is the anti-derivative - so the question is, "What function has this given function as a derivative". And if you add a constant to a function, the derivative of the function doesn't change. Thus, for example, if the derivative is y' = 2x, the original function might be y = x squared. However, any function of the form y = x squared + c (for any constant c) also has the SAME derivative (2x in this case). Therefore, to completely specify all possible solutions, this constant should be added.
The answer depends on what the constant is: the y-intercept in a linear graph, constant of proportionality, constant of integration, physical [universal] constant.
When you find an indefinite integral of a function (ie, the integral of a function without integration limits) you are actually finding the antiderivative of that function. In other words, you are finding the function whose derivative is the function 'inside' the integral sign. Recall that the derivative of a constant is zero. The point here is that you add the 'c' to acknowledge the fact that when the derivative of the result of your integration effort is taken to get the original function it could, or would, have been followed by some unknown constant value that disappeared upon differentiation. That constant is denoted by the 'c'.
When you find an indefinite integral of a function (ie, the integral of a function without integration limits) you are actually finding the antiderivative of that function. In other words, you are finding the function whose derivative is the function 'inside' the integral sign. Recall that the derivative of a constant is zero. The point here is that you add the 'c' to acknowledge the fact that when the derivative of the result of your integration effort is taken to get the original function it could, or would, have been followed by some unknown constant value that disappeared upon differentiation. That constant is denoted by the 'c'.
Where you refer to a particular integral I will assume you mean a definite integral. To illustrate why there is no constant of integration in the result of a definite integral let me take a simple example. Consider the definite integral of 1 from 0 to 1. The antiderivative of this function is x + C, where C is the so-called constant of integration. Now to evaluate the definite integral we calculate the difference between the value of the antiderivative at the upper limit of integration and the value of it at the lower limit of integration: (1 + C) - (0 + C) = 1 The C's cancel out. Furthermore, they will cancel out no matter what the either antiderivatives happen to be or what the limits of integration happen to be.
1/2 y2 + any constant
integration is necessary to make collective sense.
I'm assuming you are asking why you cannot work through your simplification and only put a constant on the last line. The simplest answer is that mathematicians are picky people, and when working through a problem EVERY line must make absolute mathematical sense. Leaving the constant off until the last line makes every line between the point where the integration occurs and the last line false. (Unless you are lucky and the constant of integration is 0, however this still needs to be proven)
When you do an integration, you are (implicitly or explicitly) recognizing that what you are integrating is a "rate of change". Your integration over a particular interval provides you with the answer to the question "what is the total change over this interval?". To get the total value of this quantity you must add the initial amount or value. That is represented by the constant of integration. When you integrate between specific limits and you are asking the question "how much is the total change" the initial value is not needed, and in fact does not appear when you insert the initial and final values of the variable over which you are integrating. So you must distinguish between finding the total change, or finding the final value. Re-reading this, I could have been a bit clearer. I'll give an example. Suppose something is accelerating at a constant acceleration designated by "a". Between the times t1 and t2 the velocity changes by a(t2-t1) which you get by integrating "a" and applying the limits t2 and t1. But the change in velocity is not the same as the velocity itself, which is equal to the initial velocity, "vo", plus the change in velocity a(t2-t1). This shows that the integral between limits just gives the accumulated change. but if you want the final VALUE, you have to add on the initial value. You might see a statement like "the integral of a with respect to time, when a is constant is vo + at ". You can check this by differentiating with respect to t, and you find the constant vo disappears. In summary, the integral evaluated by simply applying the limits gives the accumulated change, but to get the final value you have to add on the pre-existing value, and in this context the pre-existing value also carries the name of "constant of integration".