Integration of power functions
So far, we can only integrate polynomial functions. Of course, we also want to integrate all other functions that we have used, such as calculating the integral
When integrating such 'more' difficult functions, it is worth knowing that although we can differentiate (with pen and paper) almost any function, the integration does not always succeed. Actually, it only works in special cases, but thankfully, these are exactly the kind of cases that are always chosen for the tasks.
The first term of the above integral can already be calculated, so let's move on to the next one. Root expressions and fractional potency expressions are learned together because they are calculated in the same way, the root expression is first converted to a fractional potency. We can use the previously used integral formula of the power function with small refinements
Previously, r was a positive integer. Now r can basically be any real number except -1 (we will return to r = -1 later), but we are still limited to rational numbers, i.e. r is in a fractional form. The integral is then defined in the same domain as the function to be integrated. We have 3 different cases for the function domain:
r = 1,2,3, ... Defined for all values of x.
r = 0, -2, -3, -4… Defined for all values except x = 0.
r is not an integer. Defined only when x is positive.
So there is nothing miraculous about these integrals, as long as you remember to take the domain into account.
The root function is written as fractional power, the integral is calculated, and the result is converted back to the root function.
Important note about the domain! In the example above, the integral domain is a square root expression domain that includes the point x = 0. Although during integration we use a fractional power where x = 0 is not included, however, x = 0 is again included in the domain of the response function.
If we integrate the expression for the cubic root of x instead of the square root, which is defined for all values of x, the result would also be defined for all values of x. This is despite the fact that fractional potency functions are used during the calculation. It is therefore worth checking the domain for the original function as well as the final result (the domains are usually the same, but in some cases the latter is smaller).
Another inconvenience arises in situations where the function to be integrated consists of two branches. In this case, the integral function must be searched for in each branch separately. The integral function is usually of the same form, but the integration constant can change. Thus, this only happens in situations where the functions have a hole [the function can still be continuous in its domain, as in the following example].
Example 2: Find the integral function of the function f(x) that passes through points (-1,1) and (1,2).
The function to be integrated is not defined when x = 0 and is divided into two branches on both sides of the y-axis
However, there is nothing difficult in calculating the integral
But the integral functions must now be defined separately in the range x < 0 and x > 0. The result is otherwise the same, but different regions have different integration constants. [The same applies to all piecewise defined functions, i.e., they are assigned their own integration constants for each definition interval.]
In the region x < 0 we have the condition F(-1) = 1, from which we can solve C
And from the condition F(1) = 2 we can solve D
F(x) is a piecewise function
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