When determining the area under a curve in a given interval by using rectangles, we found that the more rectangles we used, the better the approximation.

It would seem that if we could use an infinite number of rectangles in the interval, the approximate area calculated using the rectangles would give us the value of the actual area.

We now are able to use this limit as the basis for our definition of the definite integral.

If a function f (x) is continuous on the closed interval [a, b], then we can define a definite integral from a to b.

So far, we have only looked at areas that lie below the curve and above the x-axis. That is, cases where f (x) is positive.

If f (x) is negative, the definite integral becomes the negative area below the x-axis and above the curve.

We also have the case where f (x) takes on both positive and negative values, as shown here.

The area of the given interval is the sum of the positive area above the x-axis and the negative area below the x-axis.

Now, consider this graph. The area of the given interval has regions above and below the x-axis.

The net area is given by the definite integral shown.

We can determine the area of the positive region using the formula for the area of a triangle.

We now can determine the value of the definite integral, which is the net area.

In this case, the definite integral can be thought of as a net area, or difference of areas.

In the notation of the definite integral, the function f (x) is called the integrand, a term we used in a similar manner with regard to the indefinite integral. The lower limit a and the upper limit b are referred to as the limits of integration.

Although the notation of definite integral looks similar to that of the indefinite integral, the two are fundamentally different. The indefinite integral is a function, whereas the definite integral is a number.

Now, consider this graph.

We can find the area under the curve on the closed interval [0,4] by evaluating the given definite integral.

We divide the interval into n subintervals. The value of Dx is given by the size of the subinterval.

We now can find the limit of the sum using the values of xi and Dx.

To evaluate the sum term, we use the formula for the sum of the squares of the first n positive integers.

After rearranging terms, we have the limit expression in a form that we can easily solve.

Evaluating the limit gives us the value of our original definite integral, and thus, the area under the curve.

We can choose the left endpoint, right endpoint, or any point in between for ci. Let's choose the right endpoint. In this case, ci becomes xi.

Which of the following shows this conclusion in the form of a limit? Click the "Submit" button after selecting your answer.

What is the value of Dx in terms of the number of subintervals n? Click the "Submit" button after selecting your answer.

Now, what is the value of xi that we should use in our limit equation? Click the "Submit" button after selecting your answer.

What is the value of the limit? Click the "Submit" button after selecting your answer.

What is the area of the region above the x-axis? Click the "Submit" button after entering your answer.

Now, what is the area of the region below the x-axis? Click the "Submit" button after entering your answer.

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