Performing derivatives and integrals of vectors must be done by components, just as vectors must be added by components. The reality is that an integral is a sum of terms. That's why the symbol is reminiscent of the letter "S". It was actually related to the Latin word "summa", but luckily that is also where we get the English word sum.[br][br]In any case, I mention that because if vectors must be added (or summed) by components, then they must also be integrated by components. Since derivatives are related to differences, they too must be done by components just as the difference of any two vectors must be found by taking the difference of their components.[br][color=#1e84cc][br]EXAMPLE: Let us suppose, for example, that we can define the position of an object using the following time-dependent vector: [math]\vec{r}=2.0t\hat{x}+1.0\hat{y}-3.0t^2\hat{z}.[/math] Find the derivative of this vector at time t=3.0. Then find the integral of that result for t between t=0 and t=3.0. NOTE: I am being deliberately careless not using units right now because we will soon enough take the time to be very careful to get them exactly right. For now I don't want them to distract from the present focus. If this bothers you, then suppose we are using standard SI units for all terms such that the units work out.[br]SOLUTION: The derivative with respect to time is given by [math]\frac{d\vec{r}}{dt}=2.0\hat{x}+0\hat{y}-6.0t\hat{z},[/math] which is found by just taking the derivatives of the individual terms and then making sure you retain the x,y and z hats. Evaluating at t=3.0 gives [math]2.0\hat{x}+0\hat{y}-18.0\hat{z}.[/math] As mentioned above, we must also integrate functions by components. Taking the function we just got, and integrating gives[center][math][br]\int_0^{3.0}\frac{d\vec{r}}{dt}\;dt = 2.0\hat{x}+0\hat{y}-6.0t \\[br]=2.0t\hat{x}+0\hat{y}-3.0t^2\hat{z}|_0^3. \\[br]=6.0\hat{x}-27\hat{z}.[br][/math][/center][/color][br]Notice that the integral of the derivative (before using the limits of zero to three) is not the original function. We lose any information about constants in the original function once we take the derivative, and we don't recover them when we integrate. So given the definition of an inverse operation 'g' for operation 'f' being that g(f(x))=x, differentiation and integration are not inverse functions.[br][br]What the integral does to f(x) is returns the family of functions whose derivative is f(x), and the members of the family may differ by a constant.[br][br]A similar issue arises with common inverse functions like the inverse trigonometric functions, which exhibit similar behavior due to the periodic nature of the trigonometric functions themselves. The inverse sine of the sine of x doesn't generally return x if x is beyond pi/2, but that behavior is accounted for by defining the range of the inverse sine being between -pi/2 and pi/2. We don't have a handy range argument like that for integration.