Review - Basic Calculus Concepts

This is a very basic review of introductory Calculus concepts. We first look at a quadratic function: y = 9 x² - 50 x + 50, which is graphed in Diagram 2.

This function has a slope at every point. If we take the "derivative" of our quadratic function, we obtain a new function (y' = 18 x - 50), which is graphed in Diagram 1, above Diagram 2. The 'derivative' function gives us the value of the slope of our quadratic function at every point. (Thus at x = 2, the slope of the quadratic function is 18 * 2 - 50 = -14)

If on the other hand we "integrate" our quadratic function, we obtain a new function (y* = 3 x³ - 25 x² + 50 t), which is graphed in Diagram 3 directly below Diagram 1. The 'integrated' function tells us the net area under the quadratic function curve (between the function curve and the x axis). It actually tells us the area between some beginning x-value and ending x-value (when we do what is called a definite integral). For the integrated function above, the initial x value is zero and the ending x-value is what ever value we choose.
That is for x = 2, the area under the quadratic function curve between zero and 2 is: A = 3 (2)³ -25 (2)² + 50 (2) = 24. Thus every y value on the curve in Diagram 3 is equal to the sum of the area under the curve in Diagram 2 up to that point.

 We can also do an "indefinite" integral which results in 'integrated' function involving a constant - which is evaluated by applying a boundary condition.

A second way to think of derivatives and integrals is as inverse functions of each other. That is, the integral asks the question - what function must we take the derivative of to obtain what is inside the integral sign. Before we look at an example of this, let's review some basic derivatives (which are normally covered in the first semester of Calculus).

• d/dx (any constant) = 0
• d/dx (x) = 1
• d/dx (x²) = 2x
• d/dx (x³) = 3x²
• d/dx (xⁿ) = n x^n-1

Now we look at the indefinite integral . This integral asks the question - what function must we take the derivative of to obtain 'x' (what is inside the integral sign). With a little reflection we see that a possible function is: ½ x² . That is, if we take the derivative of ½ x², we do obtain x. However, the function ½ x² + 5 is also a solution, as is the function ½ x² + 1,000,000, or ½ x² + C (where C is any arbitrary constant). Thus, we see that when one does an indefinite integral, the answer is a function plus a constant which must be determined by an appropriate boundary condition.
Some basic integrals, (which we will use in determine bending moment expressions in loaded beams) are:

• 
• 
• 
• 
• , where A is any constant
• that is, the integral of a sum (or difference) is the sum (or difference) of the integrals.

These simple integrals will be enough to solve beam problems involving standard type loads. For an example of the application of integration and use of a boundary condition in solving a problem,

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Example 1
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