Expert reviewed • 08 January 2025 • 6 minute read
As discussed in earlier chapters, indefinite integrals are integrals, which are not governed by any limits. In simpler terms, when you take the derivative of an indefinite integral, you get the original function back.
So far in this module, we have explored various integration formulas. However, we must manipulate these formulas to account for the constant , which is introduced as part of an indefinite integral. The formulas are as follows:
There is only one condition where these formulas will not work, and that is when . Solving integrals under this condition, will be explored in a later module.
As seen in the formulas above, we have introduced a new variable to the integration formulas, . This is a constant which occurs when you integrate a indefinite integral. This is due to the fact, that when an integral is not defined by limits, the original equation, or the anti-derivative, can be an infinite amount of different possibilities.
For example, the following expression can have infinite solutions, as there are infinite primitive equations which equal the same derivative.
As such, we simply integrate, including , which can be determined later using various other points, that fit the equation.
Integrate the following function:
Because this equation is in the form , we can use the formula listed above.
If the function found in the previous question, passes through the point , find .
Substituting in points:
When integrating functions with negative and fractional indices, they can seem tricky to approach as they often appear in the forms and . However, there is a simple way of solving these problems.
To solve by using the formulas listed above, simply change the form in which they are displayed, as shown below.
By doing this, the equations given in these forms become simple to integrate. See a hard example completed in the video above.