Definite Integrals: Rate of Change

Instructor: Matthew Bergstresser
Derivatives and integrals are opposites of each other. A definite integral is the area under the equation of a derivative, evaluated between two boundaries.

Slopes and Areas

Let's pretend we are standing on a hill. We set a ball on the ground and let it roll down the slope. We could plot this slope on a graph as how far down the ball moves relative to how far to the side the ball moves. This is known as the derivative. Now let's determine the cross-sectional area between the slope and the level surface at the bottom of the hill. This represents the integral. Diagram 1 shows the relationship between derivatives and integrals.


Diagram 1. Slopes are derivatives and area under slopes are integrals
d1


Let's take a closer look at how derivatives and integrals are related and focus on definite integrals.

Definite Integrals

Let's look back at modified Diagram 1 (we'll call it Diagram 2) and focus on the area under the sloped line, which represents the definite integral. A definite integral is the area between a derivative equation's graph and the x-axis bounded by two x-coordinates.


Diagram 2. The area under the derivative function is bounded by two x-coordinates.
x


We can see from Diagram 2 the definite integral is the area between the sloped line (derivative) and the x-axis between coordinates x-coordinate 1 and x-coordinate 2. The small black rectangle is one of an enormous number of rectangles that are aligned side by side filling in this area. Their widths are dx. We can represent how to determine this area using the expression

eq1

where x2 and x1 are the boundaries along the x-axis previously called x-coordinates 1 and 2. Basically this notation tells us to add up all of the areas of the rectangles with widths dx from x2 to x1. Let's work this example using actual values.

The equation for our sloped line is y = -2x + 4. Let's determine the area between this line and the x-axis between x = 0 and x = 2.


The integral of y = -2x between x = 0 and x = 2 is the integral of -2x(dx) between x = 0 and x = 2
z


The formal expression for this operation is

i1

Since we are reversing a derivative we raise the power on the exponent by 1 and then divide by that new value. This gives us

s2

which simplifies to

s3

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