Indefinite Integrals

Posted by Jeff Schuler on Oct 11, 2009 in Uncategorized |

What are indefinite integrals used for?

In integral can be thought of as an area underneath a curve.  Integrals are often used to manipulate position, velocity, and acceleration equations to estimate different situations.  If you are given an equation that represents the velocity of a golf cart driving, like so: y \prime = 3 x^{2} + 4 x + 3 (the ‘ in y’ represents the differential element \frac{d}{dy} that results when one performs a derivative on the position function f(x).)  then you can find use an integral (Anti-Derivative) to get an expression for the position of the golf cart, y(x).

The integral of a function is represented like so: \int_{}^{x} f(x) and it can be thought of as a sum of areas like so:

integral-summation

Here the integral is performed on the function f(x) from point a to point b, which would make it a definite integral because the bounds are defined.  Our indefinite integral is the same procedure, except missing the bounds, which makes indefinite integral operation require a little twist.

We all know that the derivative of a constant is zero.  For instance, the derivative of 5 is equal to zero.  Once performing this derivative, we should still be able to perform the anti-derivative on this new function (zero) to obtain the original equation (5).  But how will we know what number permeates from performing an anti-derivative of zero.  An indefinite integral is called indefinite because the bounds are not defined on the integration, like so:  \int_{1}^{4} .  When we perform our indefinite integral we represent this long-lost constant by the letter ‘C’.

Before we integrate our golf cart velocity equation, lets go ahead and look at the laws of integration:

\int_{}^{} f (x)^{n} = \frac{f(x)^{n+1}}{n+1} + C

The integral can also be split up into separate individual integrals if there is addition in the function you are integrating.

\int_{}^{} (5x^{2} + 3x + 4) = \int_{}^{} 5x^{2} + \int_{}^{} 3x + \int_{}^{} 4 and whenever you add these integrals together, you only need to account for 1 of the constants (C).

So the integral of our velocity y \prime = 3 x^{2} + 4 x + 3 will go as follows:

\int_{}^{} 3x^{2} + 4x + 3 = \frac{3x^{3}}{3} + \frac{4x^{2}}{2} + 3x + C

or

y(x) = \frac{3x^{3}}{3} +\frac{4x^{2}}{2} + 3x + C

We can solve for our constant, C, if we are given initial conditions, such as the golf cart was moving at y'(0) = 1 m/s when we began collecting our data.  Otherwise, we leave the integral in this form.  If you would like to learn how to perform a definite integral, refer to our article on definite integrals.

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